in a Straight Line chapter key concepts what Effects do Moving objects Have on Society and the Environment?

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1 chapter 1 Motion in a Straight Line key concepts After completing this chapter you will be able to explain how distance, position, and displacement are different explain how speed, velocity, and acceleration are different explain how vectors and scalars are different add and subtract vectors using scale diagrams and algebraic methods obtain motion information from position time, velocity time, and acceleration time graphs solve uniform velocity and uniform acceleration problems using algebraic methods describe how the acceleration due to gravity affects the motion of objects close to the surface of Earth assess the impact on society and the environment of a technology that applies concepts related to kinematics what Effects do Moving objects Have on Society and the Environment? Automobiles have been made in North America for over 1 years. As technology has advanced, automobile designs have changed substantially. For example, a 199 Ford Model T could travel at a maximum speed of approximately 7 km/h. This was considered a frightening speed at the time. Over time, vehicles have become faster. Today, many cars can reach speeds of 2 km/h or more, much higher than the speed limits on any Canadian roads. Scientists and engineers continue to develop a deeper understanding of motion and the factors that affect it. This knowledge, coupled with technological advances, has enabled them to produce extremely fast experimental land vehicles. The ThrustSSC (SuperSonic Car) reached an astounding speed of 1228 km/h. This extraordinary British-built vehicle was driven by a Royal Air Force pilot and powered by two jet engines. The team that built the ThrustSSC and other competing groups are now attempting to build even faster vehicles. Although we benefit greatly from motor vehicles that transport huge amounts of goods daily and make travel much easier, we now realize that burning large amounts of fossil fuels has a negative impact on our environment. Fuel consumption for the ThrustSSC was a mind-boggling 55 L of gasoline per kilometre. In comparison, the passenger vehicles you see on the street consume about.7 L/km. Many researchers have turned their attention to producing practical vehicles with a lower environmental impact. Students at the University of Waterloo are moving the technology of passenger vehicles away from fossil fuels altogether. They have designed and constructed a series of solar vehicles called Midnight Sun, which have reached speeds of 117 km/h. The team captured the world record for the longest journey by a solar-powered car. Faster, more efficient vehicles are an important part of the future of transportation in Canada. Today s students are tomorrow s scientists and engineers. With a sound understanding of the physics of motion, we can improve today s transportation and environmental technologies to help protect our planet. STARTiNg PoInTS Answer the following questions using your current knowledge. You will have a chance to revisit these questions later, applying concepts and skills from the chapter. 1. Give three examples of scientifi c language that can be used to describe the motion of objects. 2. Explain how graphs can provide information about the motion of an object. 3. How have the motion capabilities of automobiles changed over the past century? 4. How have changes in technologies that apply concepts about motion affected society and the environment? 6 Chapter 1 Motion in a Straight Line

2 Mini Investigation The Effect of Gravity on the Motion of Objects Skills: Predicting, Performing, Observing, Analyzing In this activity, you will investigate how mass and shape change the effect of gravity on the motion of objects falling through the air. Equipment and Materials: 2 spherical objects of different mass; 1 sheet of paper 1. Pick up the two spheres and identify which is heavier. 2. Predict which sphere will reach the fl oor fi rst if you release both of them simultaneously from the same height. Record your prediction and explain your reasoning. 3. Hold the two spheres at arm s length from your body at the same height. Release the two spheres. Record your observations. Use caution when dropping the spheres. Do not drop them near your feet or near other students. Pick up the spheres from the fl oor immediately. SKILLS HANDBOOK A Repeat Step 3 with one sphere and a fl at sheet of paper. Record your observations. 5. Crumple the piece of paper used in Step 4 into a ball of approximately the same size as one of the spheres. 6. Repeat Step 3 with the crumpled paper ball and the sphere. Record your observations. A. Did your observations in Step 3 support the prediction you made in Step 2? If not, provide reasons why the prediction was not supported. T/I B. Did the mass of each sphere in Step 3 affect the time it took the spheres to reach the fl oor? T/I C. Compare and contrast your observations for Step 3 with those for Step 4. T/I D. Compare and contrast your observations for Step 3 with those for Step 6. T/I Introduction 7

3 1.1 Distance, Position, and Displacement You see and interact with moving objects every day. Whether you are racing down a ski hill or running for a school bus, motion is part of your everyday life. Long jump athletes are very aware of distance, position, and displacement. Long jumpers run down a stretch of track to a foul line and then jump as far as possible into a sand pit (Figure 1). Their goal is to increase the distance of their jumps. To do this, they focus on their speed, strength, and technique. Successful long jumpers master this goal by applying the physics of motion. Learning Tip SI Metric Units The SI system (le Système international d unités) provides base units for measuring fundamental physical quantities. Examples of base units are metres (m) for distance and displacement and seconds (s) for time. Some physical quantities use derived units that are defined in terms of a combination of base units. An example of a derived unit is metres per second (m/s), which is the SI unit for speed and velocity. Figure 1 Long jumpers attempt to maximize the horizontal distance of their jumps. Describing the Motion of Objects kinematics the study of motion motion a change in an object s location as measured by a particular observer distance (d) the total length of the path travelled by an object in motion direction the line an object moves along from a particular starting point Career Link Many branches of engineering use principles of kinematics. To learn more about becoming an engineer, go to nelson science To understand the motion of objects, we must first be able to describe motion. Physicists use a number of specific terms and units to describe motion. You are likely familiar with many of these terms and units. Kinematics is the term used by physicists and engineers to describe the study of how objects move. What exactly is motion? Motion is a change in the location of an object, as measured by an observer. Distance, in physics terms, means the total length of the path travelled by an object in motion. The SI metric base unit for distance is the metre (m). To help you understand the terms that describe motion, imagine that you are at your home in Figure 2. You are at the location marked m. If you walk directly from home to your school in a straight line, you will travel a distance of 5 m. If you walk from your school to the library and then return home, you will travel an additional distance of 7 m 1 12 m 5 19 m. If your friend wants to know how to get to the library from your home, telling him to walk for 12 m is not very helpful. You also need to tell your friend which direction to go. Direction is the line an object moves along from a particular starting point, expressed in degrees on a compass or in terms of the compass points (north, west, east, and south). Directions can also be expressed as up, down, left, right, forward, and backwards. Directions are often expressed in brackets after the distance (or other value). For example, 5 m [E] indicates that the object is 5 m to the east. 8 Chapter 1 Motion in a Straight Line

4 Direction is important when describing motion. If the school in Figure 2 is your starting point, the library is in a different direction from your school than your home is. If the library is your starting point, then your school and home are in the same direction. W E school library home m 5 m 12 m Figure 2 Distance and direction along a straight line Scalar and vector Quantities A scalar quantity is a quantity that has magnitude (size) only. Distance is an example of a scalar quantity. Since direction is so important in describing motion, physicists frequently use terms that include direction in their definitions. A vector is a quantity that has magnitude (size) and also direction. An arrow is placed above the symbol for a variable when it represents a vector quantity. scalar a quantity that has only magnitude (size) vector a quantity that has magnitude (size) and direction Position and displacement Position is the distance and direction of an object from a particular reference point. Position is a vector quantity represented by the symbol d >. Notice the vector arrow above the symbol d. This arrow indicates that position is a vector: it has a direction as well as a magnitude. For example, if home is your reference point, the position of the school in Figure 2 is 5 m [E]. Note that the magnitude of the position is the same as the straight-line distance (5 m) from home to school, but the position also includes the direction (due east [E]). The position of the school from point m can be described by the equation WEB LInK To review scalar and vector quantities, go To SoN SCiENCE position (d > ) the distance and direction of an object from a reference point d > school 5 5 m 3E4 Now assume that the library is your reference point, or the point m. The position of the school from the reference point (library) can be described by the equation d > school 5 7 m 3W4 Once the position of an object has been described, you can describe what happens to the object when it moves from that position. This is displacement the change in an object s position. Displacement is represented by the symbol Dd >. Notice the vector arrow indicating that displacement is a vector quantity. The triangle symbol is the Greek letter delta. Delta is always read as change in, so Dd > is read as change in position. As with any change, displacement can be calculated by subtracting the initial position vector from the final position vector: displacement (Dd > ) the change in position of an object Dd > 5 d > final 2 d > initial When an object changes its position more than once (experiences two or more displacements), the total displacement Dd > T of the object can be calculated by adding the displacements using the following equation: Dd > T 5 Dd > 1 1 Dd > Distance, Position, and Displacement 9

5 Tutorial 1 Calculating Displacement for Motion in a Straight Line When you walk from one place to another, your position changes. This change in your position is displacement. The displacement can be calculated using your position at the beginning and the end of your journey with the equation Dd > 5 d > final 2 d > initial. Remember that position is a vector quantity, so you have to take direction into account. In the following Sample Problems, we will calculate displacements using a range of techniques. Refer to Figure 3 for the fi rst three Sample Problems. In Sample Problem 1, we will calculate the displacement of an object with an initial position of m. mall W home E school library Figure 3 1 m m 5 m 12 m Sample Problem 1: Calculating Displacement from a Zero Starting Point by Vector Subtraction Required: Dd > Imagine that you walk from home to school in a straight-line route. What is your displacement? Solution Figure 3 shows that home is the starting point for your journey. When you are at home, your position has not changed. Therefore, your initial position is zero. Your school has a position of 5 m [E] relative to your home. Given: d > school 5 5 m 3E4; d > home 5 m Sample Problem 2: Calculating Displacement by Vector Subtraction What is your displacement if you walk from your school to the library? Note that all positions are measured relative to your home. Given: d > school 5 5 m 3E4; d > library 5 12 m 3E4 Required: Dd > Analysis: Dd > 5 d > library 2 d > school Solution: Dd > 5 d > library 2 d > school 5 12 m 3E4 2 5 m 3E4 Dd > 5 7 m 3E4 Sample Problem 3: Calculating Total Displacement by Vector Addition One night after working at the library, you decide to go to the mall. What is your total displacement when walking from the library to the mall? Given: Dd > m 3W4; Dd > m 3W4 (from Figure 3) Required: Dd > T Analysis: In this problem, we are not simply calculating a change in position, we are fi nding the sum of two different displacements. The displacements are given in Figure 3. To calculate the total displacement, we will need to use vector addition. This is simple if both vectors have the same direction. Analysis: Dd > 5 d > school 2 d > home Solution: Dd > 5 d > school 2 d > home 5 5 m 3E4 2 m Dd > 5 5 m 3E4 Statement: Your displacement when walking from your home to school is 5 m [E]. Statement: Your displacement when walking from school to the library is 7 m [E]. Defi ning the initial starting position of your motion as m will often make displacement problems simpler. In Sample Problem 2, if we had defi ned m as being the location of the school, it would have been obvious from the diagram that the displacement from the school to the library is 7 m [E]. If the displacement from the library to your home is represented by Dd > 1 and the displacement from your home to the mall is represented by Dd > 2, then the total displacement, Dd > T, is given by vector addition of these two displacements: Dd > T 5 Dd > 1 1 Dd > 2 Solution: Dd > T 5 Dd > 1 1 Dd > m 3W4 1 1 m 3W4 Dd > T 5 22 m 3W4 Statement: When walking from the library to the mall, you experience a displacement of 22 m [W]. 1 Chapter 1 Motion in a Straight Line

6 Sample Problem 4: Calculating Total Displacement by Adding Displacements in Opposite Directions A dog is practising for her agility competition. She leaves her trainer and runs 8 m due west to pick up a ball. She then carries the ball 27 m due east and drops it into a bucket. What is the dog s total displacement? Solution In this problem, the given values are displacements. To calculate the total displacement, we will add these two displacement vectors. Given: Dd > m 3W4; Dd > m 3E4 Required: Dd > T Analysis: Dd > T 5 Dd > 1 1 Dd > 2 Solution: Dd > T 5 Dd > 1 1 Dd > m 3W m 3E4 At this point, it appears that we have a problem. We need to add a vector with a direction [W] to a vector with a direction [E]. We can transform this problem so that both vectors point in the same direction. To do so, consider the direction [E] to be the same as negative [W]. The vector 27 m [E] is the same as 27 m [W]. We can therefore rewrite the equation as follows. 5 8 m 3W m 3W4 Dd > T 5 53 m 3W4 Statement: The dog s total displacement is 53 m [W]. Practice 1. A golfer hits a ball from a golf tee at a position of 16.4 m [W] relative to the clubhouse. The ball comes to rest at a position of 64.9 m [W] relative to the clubhouse. Determine the displacement of the golf ball. T/I [ans: 48.5 m [W]] 2. A rabbit runs 3.8 m [N] and stops to nibble on some grass. The rabbit then hops 6.3 m [N] to scratch against a small tree. What is the rabbit s total displacement? T/I [ans: 1.1 m [N]] 3. A skateboarder slides 4.2 m up a ramp, stops, and then slides 2.7 m down the ramp before jumping off. What is his total displacement up the ramp? T/I [ans: 1.5 m [up]] Vector Scale Diagrams In Tutorial 1, you used algebra to determine the displacement of an object in a straight line. However, there is another method you can use to solve displacement problems: vector scale diagrams. Vector scale diagrams show the vectors associated with a displacement drawn to a particular scale. A vector can be represented by a directed line segment, which is a straight line between two points with a specific direction. Line segments have magnitude only (Figure 4(a)). A directed line segment is a line segment with an arrowhead pointing in a particular direction (Figure 4(b)). For example, AB > is a line segment in the direction from point A to point B. Line segment BA > is the same line segment but in the direction from point B to point A (Figure 4(c)). A directed line segment that represents a vector always has two ends. The end with the arrowhead is referred to as the tip. The other end is the tail. A vector scale diagram is a representation of motion using directed line segments drawn to scale with arrowheads to show their specific directions. Vector scale diagrams are very useful in measuring the total displacement of an object from its original position. B tail tip A A A (a) (b) (c) Figure 4 (a) A line segment (b) Directed line segment AB > (c) Directed line segment BA > B tip B tail vector scale diagram a vector diagram drawn using a specific scale directed line segment a straight line between two points with a specific direction Career Link Geomatics technicians collect data using GPS, surveying, and remote sensing. They also record data using geographic information systems (GIS). GIS analysts capture and manage spatial information for use in government and industry. To learn more about becoming a geomatics technician or GIS analyst, go to nelson science 1.1 Distance, Position, and Displacement 11

7 Consider two displacements, Dd > 1 = 7 m [W] and Dd > 2 = 5 m [W]. We can determine the total displacement that results from adding these vectors together by drawing a vector scale diagram. In general, when drawing a vector scale diagram, you should choose scales that produce diagrams approximately one-half to one full page in size. The larger the diagram, the more precise your results will be. Figure 5 shows a vector diagram drawn to a scale where 1 cm in the diagram represents 1 m in the real world. Note that each vector in Figure 5 has a tip (the end with an arrowhead) and a tail (the other end). Vectors can be added by joining them tip to tail. This is similar to using a number line in mathematics. Thus, after applying our chosen scale, Figure 5 shows Dd > 1 drawn as a vector 7. cm in length pointing due west. The tip of Dd > 1. is joined to the tail of Dd > 2. In other words, the displacement Dd > 2 is drawn as a directed line segment that is 5. cm long pointing due west, starting where the displacement Dd > 1 ends. The total displacement, Dd > T, is the displacement from the tail, or start, of the first vector to the tip, or end, of the second vector. In this case, Dd > T points due west and has a length of 12 cm. Converting this measurement by applying our scale gives a total displacement of 12 m [W]. For straight-line motion, vector scale diagrams are not very complex. We will look at more advanced vector scale diagrams in Chapter 2 when we consider motion in two dimensions. N d T 12 m [W] scale 1 cm : 1 m Figure 5 Vector scale diagram d 2 5 m [W] tip d 1 7 m [W] tail Tutorial 2 Determining Total Displacement for Two Motions in Opposite Directions Using Vector Scale Diagrams In the following Sample Problem, we will determine displacement by using vector scale diagrams. Consider an example in which motion occurs in two opposite directions. Sample Problem 1: Using a Vector Scale Diagram to Determine the Total Displacement for Two Motions in Opposite Directions Imagine that you are going to visit your friend. Before you get there, you decide to stop at the variety store. If you walk 2 m [N] from your home to the store, and then travel 6 m [S] to your store friend s house, what is your total displacement? Solution Let your initial displacement from your home to the store be Dd > 1 and your displacement from the store to your friend s house be Dd > 2. Given: Dd > 1 = 2 m [N]; Dd > 2 = 6 m [S] d 1 2 m [N] scale 1 cm : 1 m home N d 2 6 m [S] Required: Dd > T Analysis: Dd > T 5 Dd > 1 1 Dd > 2 Solution: Figure 6 shows the given vectors, with the tip of Dd > 1 joined to the tail of Dd > 2. The resultant vector Dd > T is drawn in red, from the tail of Dd > 1 to the tip of Dd > 2. The direction of Dd > T is [S]. Dd > T measures 4 cm in length in Figure 6, so using the scale of 1 cm : 1 m, the actual magnitude of Dd > T is 4 m. Statement: Relative to your starting point at your home, your total displacement is 4 m [S]. d T friend s house Figure 6 Solution scale diagram for adding vectors with a change in direction 12 Chapter 1 Motion in a Straight Line

8 Practice 1. A car drives 73 m [W] to a stop sign. It then continues on for a displacement of 46 m [W]. Use a vector scale diagram to determine the car s total displacement. T/I C [ans: 12 m [W]] 2. A robin flies 32 m [S] to catch a worm and then flies 59 m [N] back to its nest. Use a vector scale diagram to determine the robin s total displacement. T/I C [ans: 27 m [N]] 1.1 Summary Motion involves a change in the position of an object. Motion can be described using mathematical relationships. A scalar is a quantity that has magnitude (size) only. A vector is a quantity that has magnitude (size) and direction. You can determine the displacement of an object by subtracting the start position from the end position. You can determine total displacement by adding two or more displacements together algebraically or by using a vector scale diagram. Vectors can be added by joining them tip to tail. 1.1 Questions 1. Which of the following quantities are vectors, and which are scalars? Be sure to explain the reasoning for your answer. K/U C (a) A bird flies a distance of 2 m. (b) A train is travelling at 1 km/h due north. (c) It takes an athlete 1.37 s to run 1 m. 2. Explain the following in your own words: K/U C (a) the difference between position and displacement (b) the difference between distance and displacement 3. What is the displacement of a locomotive that changes its position from 25 m [W] to 76 m [W]? T/I 4. A car changes its position from 52 km [W] to 139 km [E]. What is the car s displacement? T/I 5. Determine the total displacement for each of the following motions by algebraic methods and by using scale diagrams. T/I C (a) Dd > m 3W4; Dd > m 3W4 (b) Dd > m 3W4; Dd > m 3E4 (c) Dd > m 3N4; Dd > m 3S4 (d) Dd > km 3W4; Dd > km 3E4; Dd > km 3W4 6. A person walks 1 paces forward followed by 3 paces forward, and finally 8 paces backwards. T/I C (a) Draw a vector scale diagram representing this person s motion. Use a scale of 1 cm 5 1 pace. (b) Check your answer by pacing out this motion yourself. How close is your experimental result to that predicted by your vector scale diagram? 1.1 Distance, Position, and Displacement 13

9 1.2 Figure 1 A laser speed device can accurately measure the speed of an oncoming vehicle. average speed (v av ) the total distance travelled divided by the total time taken to travel that distance Speed and Velocity If you have been a passenger in a car or are taking driving lessons, speed is something you have thought about. Knowing the speed at which a vehicle is moving is important for safety. Excessive speed is a contributing factor in many collisions. Speed can be measured in different ways. Police use laser speed devices to detect the speed of moving vehicles (Figure 1). In the laboratory, scientists and engineers can use electronic devices such as motion sensors to measure speed. Average Speed The average speed of a moving object is the total distance travelled divided by the total time elapsed. You are probably familiar with the speedometer of a passenger vehicle, which tells the speed of the vehicle in kilometres per hour (km/h). However, the SI unit for speed is metres per second (m/s). You do not need a special device like a police speed device to measure speed. If you know the distance travelled and the time it took an object to travel that distance, you can calculate the average speed of the object using the equation v av 5 Dd where v av is the average speed, Δd is the distance travelled, and Δt is the change in time. Like distance, speed is a scalar quantity. In the following Tutorial, we will determine the average speed of an object using this equation. Investigation Tutorial 1 Calculating Average Speed Watch Your Speed (p. 46) In this investigation you will use the average speed equation to determine average speed in a study of vehicles passing an observation point. The following Sample Problems will demonstrate how to use the equation for average speed. Sample Problem 1: Determining Average Speed Your dog runs in a straight line for a distance of 43 m in 28 s. What is your dog s average speed? Given: d = 43 m; Δt = 28 s Required: v av Analysis: v av 5 Dd Solution: v av 5 Dd 5 43 m 28 s v av m/s Statement: Your dog s average speed is 1.5 m/s. Sample Problem 2: Determining the Distance Travelled by a Ball Moving at Constant Speed A baseball rolls along a flat parking lot in a straight line at a constant speed of 3.8 m/s. How far will the baseball roll in 15 s? Given: v av = 3.8 m/s; Δt = 15 s Required: d 14 Chapter 1 Motion in a Straight Line

10 Analysis: v av 5 Dd Dd 5 v av Solution: Dd 5 v av 5 a3.8 m s b 115 s2 Dd 5 57 m Statement: The ball will roll 57 m in 15 s. Practice 1. A paper airplane fl ies 3.7 m in 1.8 s. What is the airplane s average speed? T/I [ans: 2.1 m/s] 2. A cheetah can run at a maximum speed of 3. km/h, or 8.33 m/s. How far can a cheetah run in 3.27 s? T/I [ans: 27.2 m] 3. How long does it take a rock to fall through 2.8 m of water if it falls at a constant speed of 1.2 m/s? T/I [ans: 2.3 s] LEARNING TIP Rounding in Calculations As a general rule, round final answers to the same number of significant digits as the given value with the fewest significant digits. Take extra care when rounding digits with multiple parts. You will see in this book that extra digits are carried in intermediate calculations. For more help with rounding, refer to the Skills Handbook. Research This Searching for Speeders Skills: Researching, Analyzing, Communicating, Identifying Alternatives, Defending a Decision SKILLS HANDBOOK A5 A laser speed device is used by police offi cers to measure the speed of moving vehicles. This device sends a pulse of infrared laser light at the speed of light ( m/s) toward a moving vehicle. The laser pulse refl ects off the vehicle and returns to a sensor on the speed device. A computer chip in the speed device determines the time it took for the pulse to travel to and from the moving vehicle. The speed device uses one half of this very short time and the speed of light to calculate the distance to the moving vehicle. The speed device s computer uses multiple distance readings to determine how the vehicle s distance is changing with time and then calculates the vehicle s speed. Modern speed devices send thousands of pulses of light each second, providing a high level of accuracy. 1. Conduct research to investigate how common laser speed devices are in the region where you live. 2. Investigate how speed affects the number of automobile collisions and fatalities in Canada. 3. Investigate alternative methods the police could use to determine the speed of a vehicle. A. Does the use of laser speed devices have an impact on the number of automobile collisions and fatalities in Canada? K/U B. Do you feel that the use of laser speed devices is the preferred way for police to monitor the speed of automobiles? C C. Laser speed devices and video recorders can now be used to capture the speed of a moving vehicle, the vehicle s licence plate number, the date, and the time in the same image. If these devices are set in a fi xed position, they can operate without the need for a police offi cer to be present. Data can be collected electronically and speeders can be sent a ticket through the mail. Do you support the use of devices like these in Ontario? Justify your decision. C A GO TO SON SCIENCE 1.2 Speed and Velocity 15

11 Average Velocity average velocity (v > av ) the total displacement, or change in position, divided by the total time for that displacement position time graph a graph describing the motion of an object, with position on the vertical axis and time on the horizontal axis slope (m) a measure of the steepness of a line rise vertical change between two points on a line run horizontal change between two points on a line Learning Tip Rates of Change Average speed and average velocity are examples of rates of change an important concept in science that describes how quickly a quantity is changing. Velocity is the rate of change of position, which means that the more rapidly an object s position is changing, the greater is the magnitude of its velocity. The average velocity of an object in motion is its total displacement, or change in position, divided by the total time taken for the motion. Velocity describes change in position over time. For instance, a cyclist travelling east at a constant speed of 11 m/s has a velocity of 11 m/s [E]. Since it has direction and magnitude, average velocity is a vector quantity. The SI unit for velocity is metres per second (m/s). The symbol for average velocity is v > av. A position time graph is a graph that describes the motion of an object, with position on the vertical axis and time on the horizontal axis. Figure 2 shows a position time graph for the motion of a rolling ball measured by students during an experiment. Notice that the points on the graph form a straight line that moves upward from left to right. Whenever an object is moving at a constant velocity, the position time graph of that motion is a straight line. You may recall from your mathematics studies that the slope of a line describes its steepness. The symbol for slope is m. Slope is determined, as shown in Figure 2, by comparing the magnitude of the rise (the change between points on the y-axis) and the magnitude of the run (the change between the same points on the x-axis). You can use this technique whether the graph passes through the origin or not; for example, in Figure 3 the motion begins at a position of 1 m [E] when t 5 s. For an object moving at a constant velocity, so that its position time graph is a straight line, the key relationship is this: The steeper the graph, the greater is the object s displacement in a given time interval, and the higher is its velocity. This can be confirmed using the information in Figure 2. Since the y-axis shows change in position, Dd >, and the x-axis shows change in time, t, the formula for the slope of this graph can be rewritten as follows: d (m [E]) slope 5 rise run m 5 d> 2 2 d > 1 t 2 2 t 1 or m 5 Dd> The slope of a position time graph gives the velocity of the object. Position v. Time for a Rolling Ball rise 4 m [E] 1 5 run 1 s Figure 2 Calculating the slope of a position time graph d (m [E]) Position v. Time for a Rolling Ball Figure 3 A position time graph with non-zero initial position 16 Chapter 1 Motion in a Straight Line

12 The (average) velocity of a moving object is given by the equation v > av 5 Dd> To determine the slope the average velocity from the zero point to the final data point for the x-axis and y-axis for the motion shown in Figure 2, substitute the initial and final displacement and time values into the equation we just derived: v > av 5 Dd> m 5 d> 2 2 d > 1 t 2 2 t 1 4 m 3E4 2 m 5 1 s 2 s m 5 4 m/s 3E4 The velocity of the rolling ball is 4 m/s [E]. Note that the slopes of the graphs shown in Figure 2 and Figure 3 are the same, 4 m/s [E]. The two motions are different in that the motion described by Figure 2 started m away from the observer, whereas the motion graphed in Figure 3 had an initial position of 1 m [E] from the observer. Calculating average velocity from the slope of a position time graph is a very useful technique because average velocity is often difficult to measure directly. However, position can be easily measured with equipment such as tape measures, motion sensors, and laser speed devices. Velocity (a vector quantity) is to speed (a scalar quantity) as displacement (a vector quantity) is to distance (a scalar quantity). The equation for average velocity should therefore look similar to the equation for average speed, except that velocity and displacement are vectors: v > av 5 Dd> where Dd > is the change in position and t is the change in time during the given time interval. This is the same equation as the one we just derived using the slope of a position time graph. Note that Δt can also be described by the equation Δt 5 t 2 2 t 1. Often, we can simplify this equation by considering t 1, the start time, to be s. In the following Tutorial, we will use the average velocity equation to determine unknown values. Learning Tip Calculations with Vectors In general, you cannot divide one vector by another, as dividing a direction has no meaning. However, if the directions of both vectors are the same, you can disregard the direction and divide one magnitude by the other. Tutorial 2 Solving Problems Using the Equation for Average Velocity The equation for average velocity can be used to solve for any of the three variables in the average velocity equation when the other two are known. In the following Sample Problems, we will review solving equations for an unknown variable using the equation for average velocity. Sample Problem 1: Calculating the Average Velocity of an Object On a windy day, the position of a balloon changes as it is blown 82 m [N] away from a child in 15 s. What is the average velocity of the balloon? Solution We are given the change in time and the change in position of the balloon, so we can solve for average velocity. Given: Dd > 5 82 m 3N4; 5 15 s Required: v > av Analysis: v > av 5 Dd> Solution: v > av 5 Dd> 82 m 3N s v > av m/s 3N4 Statement: The average velocity of the balloon is 5.5 m/s [N]. 1.2 Speed and Velocity 17

13 Sample Problem 2: Calculating the Time for a Displacement to Occur A subway train travels at an average velocity of 22.3 km/h [W]. How long will it take for the subway train to undergo a displacement of 241 m [W]? Given: v > av km/h 3W4; Dd > m 3W4 Required: t Analysis: v > av 5 Dd> Since we are given the average velocity and the displacement of the subway train, we can rearrange the average velocity equation to solve for the change in time. Solution: Before we solve this problem, we must first make sure that all of the given values are converted to SI metric units. This will require us to convert 22.3 km/h to metres per second. We will do this by multiplying 22.3 km/h by a series of ratios equal to 1. We will use these ratios so that the units that we do not want (kilometres and hours) will cancel, and we will be left with the units we do want (metres and seconds). Practice v > av 5 a22.3 km h 3W4 b a 1 h min b a1 6 min 6 s b a1 m 1 km b W4 3 1 m 5 a b s v > av m/s 3W4 1two extra digits carried2 Now that we have converted units, we can use the average velocity equation to determine the change in time. v > av 5 Dd> 5 Dd> v > av m 3W m/s 3W s 1. What is the average velocity of a soccer ball that is kicked and travels 2.17 m [E] in 1.36 s? T/I [ans: 1.6 m/s [E]] 2. How long will it take a cat to run 8.2 m [N] at an average velocity of 3.7 m/s [N]? T/I [ans: 2.2 s] Statement: It takes the subway train 38.9 s to be displaced 241 m [W] from its starting point. motion with uniform or constant velocity motion of an object at a constant speed in a straight line motion with non-uniform velocity (accelerated motion) motion in which the object s speed changes or the object does not travel in a straight line Motion with Uniform and Non-uniform Velocity Motion with uniform or constant velocity is motion at a constant speed in a straight line. It is the simplest type of motion that an object can undergo, except for being at rest. Note that both requirements (constant speed and straight line) must be met for an object s velocity to be uniform. In contrast, motion with non-uniform velocity is motion that is not at a constant speed or not in a straight line. Motion with non-uniform velocity may also be called accelerated motion. Table 1 shows some examples of motion with uniform velocity. You will learn more about accelerated motion in Section 1.3. Table 1 Examples of Uniform and Non-uniform Velocity Example Uniform velocity Non-uniform velocity Explanation A car travels down a straight highway at a steady 1 km/h. The car is travelling at a constant speed in a straight line. A passenger on an amusement park ride travels in a circle at a constant speed of 1.2 m/s. The passenger is travelling at a constant speed but not in a straight line. She is travelling in a circle. A parachutist jumps out of an aircraft. (after parachute opens) (before parachute opens) Before he opens the parachute, the speed of the parachutist will increase due to gravity. Once the parachute is opened, his speed will become constant due to air resistance. He will then fall at a constant speed in the same direction (downwards). 18 Chapter 1 Motion in a Straight Line

14 Determining Types of Motion from Position Time Graphs Recall that the slope of a position time graph gives the velocity of an object. A position time graph that describes constant velocity must be a straight line because motion with constant velocity is motion at a constant speed. Therefore, the slope of the position time graph must also be constant. Table 2 shows five position time graphs that represent commonly occurring types of motion. You will see these types of motion frequently in investigations. By the end of this unit, you should be able to identify the type of motion from the characteristics of its position time graph. Table 2 Interpreting Position Time Graphs Position time graph Type of motion Example Graph A d (m [E]) Graph B d (m [E]) Graph C d (m [E]) Graph D d (m [E]) Graph E d (m [E]) graph is a horizontal straight line the slope of a horizontal straight line is zero the object has a velocity of zero the object is at rest the object is at a constant positive position relative to the reference position the object is stationary at a location to the east of the reference position graph is a horizontal straight line the slope of a horizontal straight line is zero the object has a velocity of zero the object is at rest the object is at a constant negative position relative to the reference position the object is stationary at a location to the west of the reference position graph is a straight line with positive slope straight lines with non-zero slopes always represent constant (non-zero) velocity from the y-axis, we know the object is moving eastward the object s velocity can be determined from the slope of the graph (rise divided by run) graph is a straight line with positive slope, which represents constant (positive) velocity from the y-axis, we know the object is moving eastward the object s velocity can be determined from the slope of the graph since graph D has the steeper slope, we can conclude that this object has a greater velocity than the object described by graph C graph is a straight line, which represents constant velocity the slope of the graph is negative the object s velocity can be determined from the slope of the graph note that the direction for position on the y-axis is given by a vector with direction [E] the negative slope indicates that the object is moving westward 1.2 Speed and Velocity 19

15 Mini Investigation Bodies in Motion Skills: Performing, Observing, Analyzing Motion sensors are devices that send out ultrasonic wave pulses. When some of these waves reflect off an object, the waves return to the motion sensor. A computer can then analyze the data from the returning waves and generate real-time position time graphs. Using motion sensors can help you understand position time graphs. Equipment and Materials: motion sensor; computer or computer interface; graph paper; pencil 1. Connect the motion sensor to a computer or computer interface. 2. Place the motion sensor on a lab bench. Make sure that you have about 4 m of free space in front of the motion sensor. 3. You will be using your body to generate position time data with the motion sensor. Before you begin, sketch position time graphs to show the overall shape and features of the graphs that you predict the motion sensor will generate for each of the following scenarios: slow constant speed fast constant speed SKILLS HANDBOOK A6.2 speeding up slowing down 4. Using the motion sensor, generate real-time position time graphs for each scenario in Step 3 using body motions. A. Compare your predictions with the results. Explain any differences. T/I C B. If there is time, use the motion sensor to generate graphs that resemble as many letters of the alphabet as you can. T/I C 1.2 Summary Average speed is equal to the total distance travelled divided by the time taken for the motion. A position time graph describes motion graphically. The slope of the position time graph gives the velocity of an object. Average velocity is equal to the total displacement divided by the time taken for the motion. In other words, velocity describes change in position over time. Motion with uniform or constant velocity is motion at a constant speed in a straight line. Objects that are undergoing constant velocity have a position time graph that is a straight line. 1.2 Questions 1. When you are solving a problem, how do you know if you are given a speed value or a velocity value? K/U 2. Define motion with uniform velocity in your own words. K/U C 3. Give two real-life examples each of motion with uniform velocity and motion with non-uniform velocity. A 4. Determine the velocity for the motion described by the graph in Figure 4. T/I Figure 4 C1-P17-P11USB d (m [W]) 5. Copy and complete Table 3 in your notebook. T/I Table 3 v > av Dd > m/s [E] 25 m [E] 12.6 m [S] 16.3 s 4 m/s [N].25 s 6. What is the displacement of a horse that runs at a velocity of 3.2 m/s [S] for 12 s? T/I 7. How many seconds would it take a car travelling at 1. km/h to travel a distance of 16 m? T/I 8. What is the velocity (in metres per second) of a Canadian Forces CF-18 fi ghter jet that travels km [S] in.297 min? T/I 2 Chapter 1 Motion in a Straight Line

16 Acceleration Some theme parks have rides in which you are slowly carried up in a seat to the top of a tower, and then suddenly released (Figure 1). On the way down, your arms and hair may fly upward as the velocity of your seat increases. The thrill of this sudden change in motion can frighten and exhilarate you all at once. An even bigger thrill ride, however, is to be a pilot in a jet being launched from the deck of an aircraft carrier. These giant ships use catapults to move 35 kg jets from rest ( km/h) to 25 km/h in just 2.5 s. While it is true that objects sometimes move at constant velocity in everyday life, usually the velocities we observe are not constant. Objects that experience a change in velocity are said to be undergoing acceleration. Acceleration describes how quickly an object s velocity changes over time or the rate of change of velocity. We can study acceleration using a velocity time graph, which, similar to the position time graph, has time on the horizontal axis, but velocity rather than position on the vertical axis. Velocity time graphs are particularly useful when studying objects moving with uniform velocity (zero acceleration) or uniform acceleration (velocity changing, but at a constant rate). The velocity time graphs for both uniform velocity and uniform acceleration are always straight lines. By contrast, the position time graph of an accelerated motion is curved. Table 1 shows the position time graphs of objects moving with accelerated motion. 1.3 Figure 1 Sudden changes in velocity are part of the thrill of midway rides. acceleration (a > av) how quickly an object s velocity changes over time (rate of change of velocity) velocity time graph a graph describing the motion of an object, with velocity on the vertical axis and time on the horizontal axis Table 1 Interpreting Position Time Graphs Position time graph Type of motion Example Graph A d (m [E]) C1-F13-P11USB graph is a curve on any graph that curves, the slope or steepness of the graph changes from one point on the graph to another since the slope in graph A is constantly changing, the velocity is not constant since the graph lies above the x-axis and its slope is increasing, the velocity of the object is also increasing (speeding up) in a positive or eastward direction Graph B d (m [E]) graph is a curve on any graph that curves, the slope or steepness of the graph changes from one point on the graph to another since the slope in graph B is constantly changing, the velocity is not constant since the graph lies below the x-axis and its slope is negative but getting steeper, the object is speeding up as it moves in a negative or westward direction 1.3 Acceleration 21

17 Table 1 (continued) Position time graph Type of motion Example Graph C d (m [E]) graph is a curve on any graph that curves, the slope changes from one point on the curve to another since the slope in graph C is constantly changing, the velocity is not constant since the graph lies above the x-axis and its slope is decreasing, the velocity of the object is also decreasing (slowing down) in a positive or eastward direction Graph D d (m [E]) graph is a curve on any graph that curves, the slope or steepness of the graph changes from one point on the graph to another since the slope in graph D is constantly changing, the velocity is not constant since the graph lies below the x-axis and its slope is negative and getting shallower, the object is slowing down as it moves in a negative or westward direction Determining Acceleration from a Velocity Time Graph Figure 2 shows a velocity time graph for a skateboard rolling down a ramp. Notice that the line of the graph goes upward to the right and has x-intercept and y-intercept of zero. We can calculate the slope of the graph in Figure 2 using the equation slope 5 rise run m 5 Dv> 1m/s2 1s2 Since acceleration describes change in velocity over time, this suggests that the (average) acceleration of the skateboard is given by the equation Learning Tip Slope and Area of Velocity Time Graphs The slope of a velocity time graph gives the acceleration of the object. The area under a velocity time graph gives the displacement of the object. Why is displacement related to the area under a velocity time graph? One way to think about it is this: the greater the velocity during a given time interval, the greater the area under the graph, and the greater the displacement over that time interval. v (m/s [S]) a > av 5 Dv> Velocity v. Time for a Skateboard Rolling Down a Ramp Figure 2 Velocity time graph for a skateboard rolling down a ramp 22 Chapter 1 Motion in a Straight Line

18 In other words, The slope of a velocity time graph gives the average acceleration of an object. Acceleration over a time interval, that is, average acceleration, is given by the equation average acceleration 5 a > av 5 Dv> change in velocity change in time or a > av 5 v> f 2 v > i Recall that the SI unit for velocity is metres per second (m/s) and the SI unit for time is seconds (s). The SI unit for acceleration is a derived unit. A derived SI unit is a unit created by combining SI base units. We can derive the units for acceleration by dividing a velocity unit (m/s) by a time uni, as follows: units of acceleration 5 units of velocity per second 5 m s s 5 m s 3 1 s 5 m s 2 Learning Tip Square Seconds? What is a square second? Good question! When we write acceleration units as m/s 2, we are not implying that we have measured a square second. This is simply a shorter way of expressing the derived unit. You can also read the unit as metres per second, per second describing how many metres per second of velocity are gained or lost during each second of acceleration. Tutorial 1 Calculating Acceleration When you are given values for any three variables in the equation a > av 5 v> f 2 v > i acceleration, you can solve for the missing variable. for Sample Problem 1 What is the acceleration of the skateboard in Figure 2? Consider the motion between s and 1 s. Given: v > i 5 m/s; v > f 5 3 m/s 3S4; t i 5 s; t f 5 1 s Required: a > av Analysis: To calculate the average acceleration, which is the slope of the graph, we use the defining equation in the form that includes v > i, v > f, t i, and t f: a > av 5 Dv> 5 v> f 2 v > i t f 2 t i Statement: The skateboard is accelerating down the ramp at 3 m/s 2 [S]. What would the acceleration be if the same velocity change of 3 m/s [S] took place over a time interval of only 5 s? a > av 5 Dv> 3 m/s 3S4 5 5 s a > av 5 6 m/s 2 3S4 The time interval is shorter, so a more rapid acceleration occurs. Solution: a > av 5 v> f 2 v > i t f 2 t i 5 3 m/s 3S4 2 m/s 1 s 2 s a > av 5 3 m/s 2 3S4 1.3 Acceleration 23

19 Sample Problem 2 A bullet is found lodged deeply in a brick wall. As part of the investigation, a forensic scientist is experimenting with a rifle that was found nearby. She needs to determine the acceleration that a bullet from the rifle can achieve as a first step in linking the rifle to the bullet. During a test firing, she finds that the rifle bullet accelerates from rest to 12 m/s [E] in s as it travels down the rifle s barrel. What is the bullet s average acceleration? Given: v > i 5 m/s; v > f 5 12 m/s 3E4; s Required: a > av Sample Problem 3 When a hockey player hits a hockey puck with his stick, the velocity of the puck changes from 8. m/s [N] to 1. m/s [S] over a time interval of.5 s. What is the acceleration of the puck? Given: v > i 5 8. m/s 3N4; v > f 5 1. m/s 3S4; 5.5 s Required: a > av Analysis: a > av 5 v> f 2 v > i Solution: a > av 5 v> f 2 v > i a > 1. m/s 3S m/s 3N4 av 5.5 s At this point, it would appear that we have a dilemma. In the numerator of the fraction, we must subtract a vector with a direction [N] from a vector with a direction [S]. Analysis: a > av 5 v> f 2 v > i Solution: a > av 5 v> f 2 v > i To solve this dilemma, we can use a technique from Section 1.2. We will change a vector subtraction problem into a vector addition problem. Recall that negative [N] is the same as positive [S]. a > 1. m/s 3S m/s 3S4 av 5.5 s 5 12 m s 3E4 2 m s s a > av m/s 2 3E4 Statement: The acceleration of the bullet is m/s 2 [E]. 18. m/s 3S4.5 s a > av m/s 2 3S4 Statement: The hockey puck s acceleration is m/s 2 [S]. Notice that the initial velocity of the puck is north, while its final velocity is south. The acceleration is in the opposite direction to the initial motion, so the puck slows down and comes to rest. It then continues accelerating south, increasing its velocity to 1 m/s [S]. This is why the final velocity is due south. Sample Problem 3 can also be solved by using a vector scale diagram. Practice 1. A catapult accelerates a rock from rest to a velocity of 15. m/s [S] over a time interval of 12.5 s. What is the rock s average acceleration? T/I [ans: 1.2 m/s 2 [S]] 2. As a car approaches a highway on-ramp, it increases its velocity from 17 m/s [N] to 25 m/s [N] over 12 s. What is the car s average acceleration? T/I [ans:.67 m/s 2 [N]] 3. A squash ball with an initial velocity of 25 m/s [W] is hit by a squash racket, changing its velocity to 29 m/s [E] in.25 s. What is the squash ball s average acceleration? T/I [ans: m/s 2 [E]] You can use both the defining equation for average acceleration and velocity time graphs to determine other information about the motion of an object, besides acceleration itself. In Tutorial 2, you will use the equation in a different form to determine a different quantity. Tutorial 3 introduces the idea of determining displacement via the area under a velocity time graph. 24 Chapter 1 Motion in a Straight Line

20 Tutorial 2 Solving the Acceleration Equation for Other Variables In the following Sample Problem, we will explore how to solve the defining acceleration equation for other variables. Sample Problem 1: Solving the Acceleration Equation for Final Velocity A racehorse takes 2.7 s to accelerate from a trot to a gallop. If a > av 5 v > f 2 v > i the horse s initial velocity is 3.61 m/s [W] and it experiences an acceleration of 2.77 m/s 2 [W], what is the racehorse s velocity v > f 5 a > av 1 v > i when it gallops? Given: s; v > i m/s 3W4; a > 5 a2.77 m av m/s 2 3W4 s 3W4 b 12.7 s m 2 s 3W4 Required: v > m/s 3W m/s 3W4 f Analysis: a > av 5 v> f 2 v > v > f m/s 3W4 i Solution: Rearrange the equation for acceleration to solve for the final velocity. Practice 1. How long does it take a radio-controlled car to accelerate from 3.2 m/s [W] to 5.8 m/s [W] if it experiences an average acceleration of 1.23 m/s 2 [W]? T/I [ans: 2.1 s] 2. A speedboat experiences an average acceleration of 2.4 m/s 2 [W]. If the boat accelerates for 6.2 s and has a final velocity of 17 m/s [W], what was the initial velocity of the speedboat? T/I [ans: 2.1 m/s [W]] Statement: When it gallops, the racehorse has a velocity of 11.1 m/s [W]. Tutorial 3 Determining Displacement from a Velocity Time Graph By further analyzing the velocity time graph shown in Figure 2 on page 22, we can determine even more information about the motion of the skateboard. Figure 3 has the same data plotted as Figure 2, except that the area under the line is shaded. The area under a velocity time graph gives the displacement of the object. Note how the area under the straight line in Figure 3 forms a triangle. The area (A) of a triangle is determined by the equation A bh Dd > bh where b is the length of the base of the triangle and h is the height of the triangle. We can determine the displacement of the skateboard from Figure 3 by calculating the area of this triangle. v (m/s [S]) Velocity v. Time for a Skateboard Rolling Down a Ramp height: h 3. m/s [S] 5. base: b 1. s Figure 3 Velocity time graph showing the area underneath the line 1.3 Acceleration 25

21 Sample Problem 1: Determining Displacement from a Velocity Time Graph What is the displacement represented by the graph in Figure 3 on the previous page? Solution: Dd > s2 a3. m s 3S4 b Given: b = 1. s; h = 3. m/s [S] Required: Dd > Dd > 5 15 m 3S4 Statement: The object was displaced 15 m [S] in 1 s. Analysis: Dd > bh Sample Problem 2: Determining Displacement from a More Complex Velocity Time Graph What is the displacement represented by the graph in Figure 4 over the time interval from s to 1. s? v (m/s [S]) Velocity v. Time for a Complex Motion h w 1. m/s [S] b 5. s l 5. s Figure 4 Velocity time graph showing more complex motion The graph in Figure 4 is a more complex velocity time graph than we worked with in Sample Problem 1. However, the displacement of the object can still be determined by calculating the area under the velocity time graph. To calculate this displacement, we will need to break the area under the line into a rectangle and a triangle, and add these two areas together. We know the area (A) of a triangle is given by A triangle 5 1 bh. The area of a rectangle is its length (l ) multiplied 2 by its width (w ), or A rectangle = lw. Given: b = 5. s; h = 1. m/s [S]; l = 5. s; w = 1. m/s [S] Required: Dd > Analysis: Dd > 5 A triangle 1 A rectangle Solution: Dd > 5 A triangle 1 A rectangle 5 1 bh 1 lw s2 a1. m s 3S4 b s2 a1. m s 3S4 b 5 25 m 3S m 3S4 Dd > 5 75 m 3S4 Statement: The object has travelled 75 m [S] after 1. s. Practice 1. Determine the displacement represented by the graph in Figure 4 over the following time intervals: T/I (a) from s to 4. s [ans: 16 m [S]] (b) from s to 7.5 s [ans: 5 m [S]] 26 Chapter 1 Motion in a Straight Line

22 Mini Investigation Motion Simulations Skills: Predicting, Performing, Observing, Analyzing, Communicating SKILLS HANDBOOK A2.1 In this investigation, you will use a computer simulation for four different motion scenarios and then analyze the scenarios by graphing. Equipment and Materials: computer access; graphing paper 1. Go to the Nelson website and fi nd the link for this Mini Investigation. 2. Go to the simulation, and take a few minutes to familiarize yourself with how it operates. You will be asked to run this simulation for each of the following scenarios: a positive velocity value a negative velocity value a negative initial position and a positive velocity value a negative initial position and a negative velocity value A. Before you run these scenarios, write a brief statement and draw a sketch to predict how each graph will appear. T/I C B. After you have run each scenario, sketch the resulting position time and velocity time graphs from the actual data you collected. T/I C C. Compare your sketches from Question A to the graphs that were generated when you ran each scenario. Explain any discrepancies or misconceptions that you may have had. T/I go To SoN SCiENCE Figure 5 Launch of NASA s Mars Pathfi nder mission Instantaneous velocity and Average velocity The velocity of any object that is accelerating is changing over time. In motion with uniform acceleration, the velocity of an object changes at a constant (uniform) rate. During a launch (Figure 5), a spacecraft accelerates upward at a rapid rate. NASA personnel may need to determine the velocity of the spacecraft at specific points in time. They can do this by plotting position and time data on a graph and determining the spacecraft s instantaneous velocity. Instantaneous velocity, or v > inst, is the velocity of an object at a specific instant in time. By comparison, average velocity (v > av) is determined over a time interval. Both types of velocity are rates of change of position, but they tell us different things about the motion of an object. motion with uniform acceleration motion in which velocity changes at a constant rate instantaneous velocity ( v > inst ) the velocity of an object at a specifi c instant in time Tutorial 4 Determining Instantaneous and Average Velocity Figure 6 on the next page shows a position time graph of an object that is undergoing uniform acceleration. Moving along the curve, the slope of the curve progressively increases. From this, we know that the velocity of the object is constantly increasing. To determine the instantaneous velocity of the object at a specifi c time, we must calculate the slope of the tangent of the line on the position time graph at that time. A tangent is a straight line that contacts a curve at a single point and extends in the same direction as the slope of the curve at the point. A plane mirror can be used to draw a tangent to a curved line. Place the mirror as perpendicular as possible to the line at the point desired. Adjust the angle of the mirror so that the real curve merges smoothly with its image in the mirror, which will occur when the mirror is perpendicular to the curved line at that point. Draw a line perpendicular to the mirror to obtain the tangent to the curve. 1.3 Acceleration 27

23 d (m [E]) Position v. Time for Motion with Uniform Acceleration rise run Figure 6 Position time graph with non-constant velocity We can use Figure 6 to determine the instantaneous velocity of the object at any specific point in time. To determine the instantaneous velocity, we must determine the slope of the tangent of the line on the position time graph at that specific time. Sample Problem 1: Determining Instantaneous Velocity Consider the point on the curve in Figure 6 at 2. s on the x-axis. What is the instantaneous velocity of the object at this time? Given: t 5 2. s; position time graph Required: v > inst Analysis: v > inst is equal to the slope, m, of the tangent to the curve at t 5 2. s, so m 5 Dd> In Figure 6, the tangent to the point on the curve at t 5 2. s has been extended until it crosses a convenient grid line. Calculate the slope of the tangent to determine the instantaneous velocity. Solution: m 5 8. m 3E4 2. s v > inst 5 4. m/s 3E4 Statement: The instantaneous velocity of the object at 2. s is 4. m/s [E]. Since the object is accelerating, if we had calculated the slope of the tangent at time t 5 1. s, the velocity would have been smaller in magnitude, and if we had calculated the slope of the tangent at time t 5 3. s, the velocity would have been greater. We can also use Figure 6 to determine the average velocity of the object. Sample Problem 2: Determining Average Velocity from a Position Time Graph What is the average velocity of the object in Figure 6 over the time interval from. s to 2. s? Given: d > 1 5. m; d > m 3E4; t 1 5. s; t s Required: v > av Analysis: Recall that average velocity is the total displacement of an object divided by the total time taken. v > av 5 Dd> v > av 5 d> 2 2 d > 1 t 2 2 t 1 Solution: v > av 5 d> 2 2 d > 1 t 2 2 t m 3E4 2. m 2. s 2. s v > av 5 2. m/s [E] Statement: The average velocity of the object over the time interval from. s to 2. s is 2. m/s [E]. 28 Chapter 1 Motion in a Straight Line

24 Practice 1. (a) Determine the instantaneous velocity at t 5 1. s for the graph shown in Figure 6. [ans: 2. m/s [E]] (b) Determine the instantaneous velocity at t 5 3. s for the graph shown in Figure 6. [ans: 6. m/s [E]] (c) Compare your answers to (a) and (b) with the solution to Sample Problem 1 above. Is it possible that the object is moving with constant acceleration? Explain. T/I 2. (a) Determine the instantaneous velocity at t 5 5. s for the graph shown in Figure 7. [ans: 3 m/s [E]] (b) Determine the average velocity from t 5 s to t 5 1. s for the graph Figure 7. [ans: 3 m/s [E]] (c) Notice that for the motion described in Figure 7, t 5 5. s is the midpoint in time. Write a statement describing the relationship that exists between the average velocity and the instantaneous velocity at the midpoint in time when an object is accelerating uniformly. T/I C d (m [E]) Figure 7 Position v. Time for Accelerated Motion Note that in Tutorial 4, the average velocity over the first 2. s of the motion was less than the instantaneous velocity at 2. s. For motion with non-uniform velocity but uniform acceleration, average and instantaneous velocities are not necessarily equal. The only situation where the average velocity and the instantaneous velocity are the same is at the midpoint in the time interval. 1.3 Summary Acceleration describes change in velocity over time. The slope of a velocity time graph gives the acceleration of the object whose motion it describes. The area under a velocity time graph gives the displacement of the object whose motion it describes. The instantaneous velocity of an object is its velocity at a specific instant in time. It is equal to the slope of the tangent to the position time graph at that instant in time. For motion with non-uniform velocity, average and instantaneous velocities are not necessarily equal. 1.3 Acceleration 29

25 1.3 Questions 1. Describe three characteristics that an accelerating object may exhibit. Give a real-world example of each characteristic. K/U C 2. Describe, in your own words, how you would determine the acceleration of an object from a velocity time graph. K/U C 3. Describe, in your own words, how you would determine the displacement of an object from a velocity time graph. K/U C 4. Determine the average acceleration described by each of the following graphs. T/I (a) (b) (c) v (m/s [E]) v (m/s [E]) v (m/s [E]) Velocity v. Time for Accelerated Motion Velocity v. Time for Accelerated Motion Velocity v. Time for Accelerated Motion One of your classmates makes the following statement: If an object has an initial velocity of 1 m/s [N] and a final velocity of 1 m/s [S], this object has clearly not accelerated, as it is travelling at a constant speed. Write an to this student explaining why this statement is incorrect. K/U C 6. (a) Describe the motion of the object in all three segments of the graph shown in Figure 8. (b) Calculate the average acceleration of the object in all three segments of the graph in Figure 8. (c) Calculate the total displacement of the object from s to 4. s, from 4. s to 7. s, and from 7. s to 1. s. T/I C Velocity v. Time for Complex Motion Figure 8 v (m/s [W]) 7. What is the average acceleration of a sports car that increases its velocity from 2. m/s [W] to 4.5 m/s [W] in 1.9 s? T/I 8. If a child on a bicycle can accelerate at an average rate of.53 m/s 2, how long would it take to increase the bicycle s velocity from.68 m/s [N] to.89 m/s [N]? T/I 9. (a) While approaching a red light, a student driver begins to apply the brakes. If the car s brakes can cause an average acceleration of 2.9 m/s 2 [S] and it takes 5.72 s for the car to come to rest, what was the car s initial velocity? (b) What is the significance of the direction of the initial velocity and that of the acceleration? T/I 1. What is the average acceleration of a tennis ball that has an initial velocity of 6. m/s [E] and a final velocity of 7.3 m/s [W], if it is in contact with a tennis racket for.94 s? T/I 11. (a) Determine the instantaneous velocity at t = 6. s in Figure 9. (b) Determine the average velocity of the motion depicted in Figure 9. T/I d (m [W]) Figure 9 Position v. Time for Accelerated Motion Chapter 1 Motion in a Straight Line

26 Comparing Graphs of Linear Motion Cheetahs are adapted for speed they are the fastest land animals. They can accelerate at faster rates than most sports cars (Figure 1). Cheetahs have been measured accelerating at rates greater than 1 m/s 2. To put this in perspective, a sports car can accelerate at approximately 7.2 m/s 2. In fact, cheetahs are capable of accelerating from rest to 1 m/s in only three strides. You have already seen how position time and velocity time graphs can be used to analyze the linear motion of objects. In this section, we will introduce acceleration time graphs and use all three types of graphs to analyze motion in more detail. 1.4 Figure 1 Cheetahs have the greatest acceleration of any animal. Acceleration Time Graphs Earlier in this chapter, you learned how to find the displacement, or change in position, of an object by determining the area under a velocity time graph. In a similar way, we can determine the change in velocity of an object from the area under an acceleration time graph, which has acceleration on the vertical axis and time on the horizontal axis. Consider the acceleration time graph in Figure 2, which shows the motion of a cheetah. The points plotted on this graph lie along a horizontal straight line with a non-zero y-intercept. The acceleration is a constant 4. m/s 2, so this graph represents uniform acceleration. acceleration time graph a graph describing motion of an object, with acceleration on the vertical axis and time on the horizontal axis a (m/s 2 [W]) Acceleration v. Time for Motion with Uniform Acceleration Figure 2 Acceleration time graph showing motion with uniform acceleration 1.4 Comparing Graphs of Linear Motion 31

27 Investigation Uniform Velocity (p. 47) In this investigation, you will use a motion sensor to generate different types of motion graphs for an object moving with uniform velocity, and analyze these graphs. If we calculate the area under the acceleration time graph in Figure 2 from s to 5. s, we will be determining the change in velocity of the object from t 5 s to t 5 5. s: A 5 lw s2 a4. m s 2 3W4 b A 5 2 m/s 3W4 Since the units are metres per second, the area we calculated represents a change in velocity. The area under an acceleration time graph represents the change in velocity of an object. If the initial velocity of the cheetah is zero (the object is at rest), the final velocity is equal to the change in velocity, 2 m/s [W]. If the initial velocity is 5 m/s [W], however, then the graph tells us that the final velocity is 5 m/s [W] 1 2 m/s [W] 5 25 m/s [W] The graph does not tell us what the initial and final velocities are; it just tells us the change in velocity that occurs in the time interval. Relationships among Linear Motion Graphs Investigation Motion Down a Ramp (p. 48) In this investigation, you will use a motion sensor and different types of motion graphs to analyze the motion of an object rolling down a ramp. Graphical analysis is one of the most powerful analytical tools available to physicists. In studying the motion of objects, analyzing position time, velocity time, and acceleration time graphs can help us gain insight into real-life events such as the motion of the cheetah shown in Figure 1. This is particularly important because most objects in nature do not come equipped with a speedometer. Figure 3 compares the three types of graphs of linear motion. All three graphs represent the same type of motion: uniform acceleration. Nevertheless, the three graphs look very different. When analyzing a motion graph, you may read information directly from the graph or determine further information by calculating the slope or the area of the graph. d (m) area slope v (m/s) area slope a (m/s 2 ) Figure 3 Position time, velocity time, and acceleration time graphs of the same motion 32 Chapter 1 Motion in a Straight Line

28 Tutorial 1 Creating One Type of Motion Graph from Another By using the information in Figure 4, we can analyze an acceleration time graph further and get more information about the motion it describes. Sample Problem 1: Creating a Velocity Time Graph from an Acceleration Time Graph Use the acceleration time graph in Figure 4 to generate velocity and time data for the object. Then use these data to plot a velocity time graph. Step 1. To generate the velocity time data, first calculate the area under the graph for several time points in Figure 4. Since the line is horizontal, we use the formula for the area of a rectangle, A 5 lw. a (m/s 2 [W]) For Figure 4, l 5 t 1s2; w 5 a > 1m/s 2 3W42; and A 5 v > 1m/s 3W42, so in calculating A = lw, we are actually calculating v > 5 1Da > Acceleration v. Time for Motion with Uniform Acceleration Figure 4 Using an acceleration time graph to create other motion graphs Step 2. Table 1 shows the calculations for the area under the graph at 1 s intervals from t = s to t = 5. s. Table 1 Calculating the Velocity at Various Time Points in Figure 4 Step 3. Plot the data to create a velocity time graph (Figure 5). v (m/s) [W]) Velocity v. Time for Motion with Uniform Acceleration Figure 5 Velocity time solution graph Figure 5 shows the resulting graph. It is an increasing straight line with a zero intercept. It describes precisely the same motion that was described by the acceleration time graph in Figure 4. Both graphs describe uniform acceleration. Time Acceleration a > 1m/s 2 3W42 Equation v > 5 1Da > Velocity v > 1m/s2 3W v > 5 a4. m s 3W4b 1 s2 2 v > 5 a4. m s 3W4b 11. s v > 5 a4. m s 3W4b 12. s v > 5 a4. m s 3W4b 13. s v > 5 a4. m s 3W4b 14. s v > 5 a4. m s 3W4b 15. s Comparing Graphs of Linear Motion 33

29 Sample Problem 2: Creating an Acceleration Time Graph from a Velocity Time Graph Use the velocity time graph shown in Figure 6 to plot the corresponding acceleration time graph. v (m/s [W]) Velocity v. Time for Motion with Uniform Acceleration Figure 6 Given velocity time graph Step 1. The data plotted on the velocity time graph in Figure 6 form an increasing straight-line graph with a zero intercept. You can determine acceleration from a velocity time graph by calculating its slope. Since the velocity time graph in Figure 6 is a straight line, its slope does not change. So we can calculate the slope or acceleration over any time interval. slope 5 rise run a > av 5 Dv> m/s 3W4 3. s a > av m/s 2 [W] Step 2. Use this value to create an acceleration time graph (Figure 7). a (m/s 2 [W]) Acceleration v. Time for Motion with Uniform Acceleration Figure 7 Acceleration time solution graph Figure 7 shows the corresponding acceleration time graph. This graph shows a horizontal straight line with a y-intercept of 2.6 m/s 2 [W]. Note: If a velocity time graph is not a straight line, you will need to determine the slope of the tangent for each time data point, and then use these data to plot the corresponding acceleration time graph. Practice 1. Generate position time and acceleration time data representing the motion of the object shown in Figure 8. Use the data to plot the corresponding position time and acceleration time graphs. T/I C v (m/s) [N] Figure 8 Velocity v. Time for Motion with Uniform Acceleration Chapter 1 Motion in a Straight Line

30 1.4 Summary The area under an acceleration time graph gives the velocity of the object. Given one type of motion graph, you can read or calculate data from it in order to construct a different type of graph. 1.4 Questions 1. Copy and complete Table 2 in your notebook by adding a check mark in each column that applies. K/U Table 2 How do you determine position velocity velocity velocity acceleration acceleration Given a position time graph position time graph velocity time graph acceleration time graph velocity time graph acceleration time graph Read information from graph Take the slope Find the area 2. From the velocity time graph in Figure 9, generate position time data and then plot the corresponding position time graph, assuming the initial position is m. T/I C v (m/s) [S] Velocity v. Time for Complex Motion Figure 9 d (m [S]) Position v. Time for Accelerated Motion Figure 1 4. Use the data in the velocity time graph shown in Figure 11 to plot the corresponding acceleration time graph. T/I C 3. Consider the position time graph shown in Figure 1. T/I (a) What is the position of the object at t 5 5. s? (b) What is the instantaneous velocity of the object at t 5 3. s? (c) What is the average velocity for the object s motion from s to 6. s? v (m/s [S]) Figure 11 Velocity v. Time for Accelerated Motion Comparing Graphs of Linear Motion 35

31 1.5 Five Key Equations for Motion with Uniform Acceleration Graphical analysis is an important tool for physicists to use to solve problems. Sometimes, however, we have enough information to allow us to solve problems algebraically. Algebraic methods tend to be quicker and more convenient than graphical analysis. For example, if you want to determine how far a passing vehicle would travel in a given amount of time, you could perform an experiment using a motion sensor. You would collect position time data with the motion sensor and then plot the data on a graph. From the graph, you could then measure how far the vehicle has gone in a given amount of time. However, if you were in the vehicle, you would simply use the vehicle s speedometer to determine the speed of the vehicle. Knowing the speed of your vehicle, you could easily determine how far it would travel in a given time interval using the equation v av 5 Dd. As you can see, the best way to solve a problem is usually determined by the information that is available to you. To be able to solve problems related to motion with uniform acceleration, in which the velocity may change but the acceleration is constant, we need to derive algebraic equations that describe this type of motion. We will start with equations that we have already used in previous sections. v (m/s [E]) v f v i area 2 area 1 A Displacement Equation for Uniformly Accelerated Motion The velocity time graph in Figure 1 shows a straight line with a non-zero intercept. This graph is a non-horizontal straight line, showing that the object is undergoing uniform, or constant, acceleration. In other words, the velocity is increasing at a uniform, or constant, rate. We know that to determine the displacement of this object from the velocity time graph, we must determine the area under the line. For the graph in Figure 1, we must determine the area of a rectangle and a triangle: Figure 1 A velocity time graph for an object undergoing uniform acceleration Dd > 5 A triangle 1 A rectangle 5 1 bh 1 lw v> f 2 v > i 2 1 v> i Learning Tip Interpreting Areas Under a Motion Graph Notice that the rectangular area (green) in Figure 1 represents the displacement the object would have undergone had it continued at constant velocity v > i. The triangular area (blue) represents the extra displacement the object experienced due to its acceleration v> f v> i 1 v > i v> f v> i Dd > 5 a v> f 1 v > i b (Equation 1) 2 We can use Equation 1 to determine the displacement of an object that is undergoing uniform acceleration. Equation 1 is very similar to an equation that we previously developed from the defining equation for average velocity: Dd > 5 v > av. We can relate the average velocity to the initial and final velocities by the equation v > av 5 a v> f 1 v > i b 2 Then substituting v > av in place of a v> f 1 v > i b in Equation 1 gives Dd > 5 v > 2 av directly. 36 Chapter 1 Motion in a Straight Line

32 As you will see, Equation 1 can help us to solve many motion problems. However, in some situations we will not know the initial velocity, the final velocity, or the time interval for a given scenario. We could use the defining equation for acceleration in a two-step process, but this tends to be difficult. To simplify things, we can derive a number of other motion equations that will allow us to solve problems in one step. Additional Motion Equations Consider the defining equation for acceleration: a > av 5 v> f 2 v > 1 If we rearrange this equation to solve for final velocity (v > f), we get Equation 2: v > f 5 v > i 1 a > av (Equation 2) You may use Equation 2 in problems that do not directly involve displacement. If we substitute the expression v i 1 a > av from Equation 2 into Equation 1, we get v > f 5 v > i 1 a > av (Equation 2) Dd > 5 a v> f 1 v > i b (Equation 1) v> i 1 a > av 1 v > i v> i 1 a > av2 Dd > 5 v > i a> av 2 (Equation 3) This is Equation 3, which allows you to determine the displacement of an object moving with uniform acceleration given a value for acceleration rather than a final velocity. The Five Key Equations of Accelerated Motion Table 1 shows the five key equations of accelerated motion. You should be able to solve any kinematics question by correctly choosing one of these five equations. You have seen how the first three are developed. We will leave the others to be developed as an exercise. Table 1 The Five Key Equations of Accelerated Motion Equation Variables found in equation Variables not in equation Equation 1 Dd > 5 a v> f 1 v > i b 2 Dd >,, v > f, v > i Equation 2 v > f 5 v > i 1 a > av a > av,, v > f, v > i Dd > a > av Equation 3 Dd > 5 v > i a> av 2 Dd >, a > av,, v > i v > f Equation 4 v 2 f 5 v 2 i 1 2a av Dd Dd, a av, v f, v i Equation 5 Dd > 5 v > f a> av 2 Dd >, a > av,, v > f v > i 1.5 Five Key Equations for Motion with Uniform Acceleration 37

33 Tutorial 1 Using the Five Key Equations of Accelerated Motion The following Sample Problems will demonstrate how to choose equations and solve problems involving the five key motion equations. Sample Problem 1 A sports car approaches a highway on-ramp at a velocity of 2. m/s [E]. If the car accelerates at a rate of 3.2 m/s 2 [E] for 5. s, what is the displacement of the car? Given: v > i 5 2. m/s 3E4; a > av m/s 2 3E4; 5 5. s Required: Dd > C3-F54-OP11USB.ai Analysis: Our first task is to determine which of the five equations of accelerated motion to use. Usually, you can solve a problem using only one of the five equations. We simply identify which equation contains all the variables for which we have given values and the unknown variable that we are asked to calculate. In Table 1, we see that Equation 3 has all the Sample Problem 2 A sailboat accelerates uniformly from 6. m/s [N] to 8. m/s [N] at a rate of.5 m/s 2 [N]. What distance does the boat travel? Given: v i 5 6. m/s; v f 5 8. m/s; a av 5.5 m/s 2 Required: Δd Analysis: In Table 1, we see that Equation 4 will allow us to solve for the unknown variable. First, we rearrange the equation to solve for d: v f 2 5 v i 2 1 2a av Dd v f 2 2 v i 2 5 2a av Dd Dd 5 v f 2 2 v i 2 2a av Sample Problem 3 A dart is thrown at a target that is supported by a wooden backstop. It strikes the backstop with an initial velocity of 35 m/s [E]. The dart comes to rest in.5 s. (a) What is the acceleration of the dart? (b) How far does the dart penetrate into the backstop? For (a): Given: v > i 5 35 m/s 3E4; v > f 5 m/s; 5.5 s Required: a > av Analysis: We may use the defining equation for acceleration. a > av 5 v> f 2 v > i given variables and will allow us to solve for the unknown variable. Dd > 5 v > i a> av 2 Solution: Dd > 5 v > i a> av 2 5 a2. m s 3E4 b 15. s a3.2 m s 2 3E4 b 15. s m 3E4 1 4 m 3E4 Dd > m 3E4 Statement: During the 5. s time interval, the car is displaced m [E]. Solution: Dd 5 v f 2 2 v i 2 2a av a8. m 2 s b 2 a6. m 2 s b 5 2a.5 m s 2b 5 Δd 5 28 m m2 s 2 1 m s 2 Statement: The boat travels a distance of 28 m. Solution: a > av 5 v> f 2 v > i m s 2 35 m s 3E4 5.5 s 5 27 m/s 2 3E4 a > av m/s 2 3W4 Notice that the acceleration is in the opposite direction to the initial motion. This must be true in order for the velocity of the dart to decrease to zero as it comes to rest. If the acceleration were in the same direction as the initial velocity, the final velocity would be greater than the initial velocity. Statement: The acceleration of the dart is m/s 2 [W]. 38 Chapter 1 Motion in a Straight Line

34 To solve (b), we have sufficient information to solve the problem using any equation with displacement in it. Generally speaking, in a two-part problem like this, it is a good idea to try to find an equation that uses only given information. Then, if we have made an error in calculating the first part (acceleration), our next calculation would be unaffected by the error. Therefore, we will use Equation 1 to solve (b), since it can be solved using only given information. Given: v > i 5 35 m/s 3E4; v > f 5 m/s; 5.5 s Required: Dd > Solution: Dd > 5 a v> f 1 v > i b 2 m s 1 35 m s 3E s2 2 Dd > 5.88 m 3E4 Statement: The displacement of the dart into the backstop is.88 m [E]. Analysis: Dd > 5 a v> f 1 v > i b 2 Practice 1. A football player initially at rest accelerates uniformly as she runs down the field, travelling 17 m [E] in 3.8 s. What is her final velocity? T/I [ans: 8.9 m/s [E]] 2. A child on a toboggan sits at rest on the top of a tobogganing hill. If the child travels 7. m [downhill] in 5.3 s while accelerating uniformly, what acceleration does the child experience? T/I [ans: 5. m/s 2 [downhill]] 1.5 Summary The five key equations of accelerated motion, listed in Table 1 on page 37, apply to motion with uniform (constant) acceleration. They involve the variables for displacement, initial velocity, final velocity, acceleration, and time interval. When solving uniform acceleration problems, choose which equation(s) to use based on the given and required variables of the problem. 1.5 Questions 1. A car accelerates from rest at a rate of 2. m/s 2 [N]. What is the displacement of the car at t = 15 s? T/I 2. An astronaut is piloting her spacecraft toward the International Space Station. To stop the spacecraft, she fires the retro-rockets, which cause the spacecraft to slow down from 2. m/s [E] to. m/s in 12 s. T/I (a) What is the acceleration of the spacecraft? (b) What is the displacement of the spacecraft when it comes to rest? 3. A helicopter travelling at a velocity of 15 m/s [W] accelerates uniformly at a rate of 7. m/s 2 [E] for 4. s. What is the helicopter s final velocity? T/I 4. Two go-carts, A and B, race each other around a 1. km track. Go-cart A travels at a constant speed of 2. m/s. Go-cart B accelerates uniformly from rest at a rate of.333 m/s 2. Which go-cart wins the race and by how much time? T/I 5. A boat increases its speed from 5. m/s to 7.5 m/s over a distance of 5. m. What is the boat s acceleration? T/I 6. Within 4. s of liftoff, a spacecraft that is uniformly accelerating straight upward from rest reaches an altitude of m [up]. T/I (a) What is the spacecraft s acceleration? (b) At what velocity is the spacecraft travelling when it reaches this altitude? 7. Derive Equation 4 and Equation 5 in Table 1 on page 37 by substituting other expressions. T/I 1.5 Five Key Equations for Motion with Uniform Acceleration 39

35 1.6 Acceleration Near Earth s Surface During a basketball game, the player takes the ball and shoots it toward the basket (Figure 1). The ball briefly skims the rim, then drops through the net to the floor. The player makes the basket because of her skill, with a little bit of help from the force of gravity. Gravity causes all objects to accelerate toward Earth s centre. If you have accidentally dropped an object such as a glass, you have directly experienced how significant the effect of gravity is. Learning Tip Describing Vertical Motion Directions Vertical motion problems are vector questions that require directions to be indicated. Directions can be simplified by defining up as positive and down as negative. This means that the acceleration due to gravity should be negative in your calculations. Figure 1 Earth s gravity plays a key role in basketball and most other sports. Acceleration Due to Gravity acceleration due to gravity (g) the acceleration that occurs when an object is allowed to fall freely; close to Earth s surface, g has a value of 9.8 m/s 2 free fall the acceleration due to gravity of an object in the absence of air resistance Acceleration due to gravity is the acceleration that occurs when an object is allowed to fall freely. The symbol for acceleration due to gravity is g. Physicists have determined that the average value of g measured very close to Earth s surface is 9.8 m/s 2. Different places on Earth have different values for g. For example, the value of g in Mexico City is slightly but measurably lower than the average of 9.8 m/s 2 because the city has a very high elevation and is therefore farther from Earth s centre. This had an interesting effect on the 1968 Summer Olympic Games, which were held in Mexico City. Many high jump, long jump, and pole-vaulting records were broken during these Olympic Games, attributed in part to the lower value of g. The value of g is different on different planets and other celestial objects. The acceleration due to gravity of an object near Earth s surface will be about 9.8 m/s 2 only if it is dropped in a vacuum. This type of motion is referred to as free fall, which is acceleration that occurs when there is no air resistance or other force affecting the motion of the object besides gravity. Uniform Vertical Acceleration All objects that move freely in the vertical direction experience acceleration due to gravity (g). In the previous section, you worked with the five key motion equations. Since we know the average value of g close to Earth s surface, we can use the motion equations to explore how gravity affects objects that are moving vertically. Tutorial 1 Motion of an Object Falling Straight Down C3-F54-OP11USB.ai The following Sample Problems will demonstrate, using the five key equations of motion, how to solve problems involving vertical motion. You might want to review the equations in Table 1 on page 37. Sample Problem 1: Determining the Time It Takes for an Object to Fall to the Ground A flowerpot is knocked off a window ledge and accelerates Solution uniformly to the ground. If the window ledge is 1. m above the We know that the motion of the flowerpot is straight down. ground and there is no air resistance, how long does it take the However, we cannot describe the direction of the vector as we flowerpot to reach the ground? have previously, using [E], [W], [N], or [S]. So we will let vectors 4 Chapter 1 Motion in a Straight Line

36 with directions up and to the right be indicated by positive values. Vectors with directions down and to the left will be indicated by negative values. Since the flowerpot is not thrown upward or downward, we can assume that it was initially at rest. Therefore, the initial velocity is m/s. Given: Dd > m; a > 5 g > m/s 2 ; v > i 5 m/s Required: Δt Analysis: Since we are given the displacement, acceleration, and initial velocity of the flowerpot, we can use Equation 3 to solve for time. Dd > 5 v > i a> 12 2 Solution: Notice that the given displacement and acceleration values are both negative, because both vectors point downward. In general, it is not valid to divide one vector by another. However, since the motion is in a straight line, and the directions are given by the minus signs, we are able to divide the vector values. This equation can be rewritten as follows. Dd > a> Å 2Dd a 5 Å m m/s s We know we should take the positive root because time intervals are always positive. Statement: The flowerpot will take 1.4 s to reach the ground. Sample Problem 2: Determining the Final Velocity for a Falling Object What is the final velocity of the flowerpot in Sample Problem 1 just before it hits the ground? Solution To solve this problem, we need to find the final velocity of the flowerpot just before it hits the ground. If the question asked for its velocity after it had hit the ground and came to rest, then the final velocity would be m/s. Here, we are determining the velocity while the flowerpot is still in motion, just before it hits the ground. We will again write vectors with directions that are up and to the right as having positive values and those with directions down and to the left as having negative values. Given: Dd > m; a > 5 g > m/s 2 ; v > i 5 m/s Required: v > f Analysis: We can choose any of the five key equations of motion that has v > f as a variable. It is always wise to choose the equation that only requires you to use given information. We will use the vector directions for the given variables but will only calculate the magnitude of the final velocity. We know that the direction of the final velocity must be downward. v f 2 5 v i 2 1 2aDd v f 5 "v 2 i 1 2aDd Solution: Since the initial velocity is zero, v f 5 "2aDd 5 Å 2a29.8 m s 2b 121. m m/s Statement: The flowerpot is travelling at 14 m/s downward just before it hits the ground. Practice 1. A ball is dropped from the roof of a building. If it takes the ball 2.6 s to reach the ground, how tall is the building? T/I [ans: 33 m] 2. A hot air balloon is hovering at a height of 52 m above the ground. A penny is dropped from the balloon. Assume no air resistance. T/I (a) How long does it take the penny to hit the ground? [ans: 3.3 s] (b) What is the final velocity of the penny just before it hits the ground? [ans: 32 m/s] Tutorial 2 Motion of an Object Thrown Straight Up The following Sample Problems involve analyzing the motion of an object that is first thrown straight up and then falls to Earth. Sample Problem 1: Determining the Height Reached by a Ball Thrown Straight Up in the Air A tennis ball is thrown straight up in the air, leaving the person s hand with an initial velocity of 3. m/s, as shown in Figure 2 on the next page. How high, from where it was thrown, does the ball go? 1.6 Acceleration Near Earth s Surface 41

37 v f = v i a upward vector will be made positive and the downward vector negative. At its maximum height, the ball comes to rest momentarily, so the fi nal velocity, v > f, is zero. Then we can use Equation 4. Given: v > i 5 3. m/s; a > m/s 2 ; v > f 5 m/s Required: Dd Analysis: v f 2 5 v i 2 1 2aDd 5 v i 2 1 2aDd 2v 2 i 5 2aDd Dd 5 2v i 2 2a Solution: Dd 5 2v i 2 2a 2a3. m 2 s b 5 2a29.8 m s 2b Figure 2 The motion of a ball thrown straight upward Figure 2 shows the path of the tennis ball. Notice that the vectors for the initial velocity, v > i, and the acceleration, a >, are in opposite directions to each other. Using our previous convention, the 29. m2 s m s 2 Dd 5.46 m Statement: The maximum height attained by the tennis ball is.46 m (46 cm). Sample Problem 2: Determining the Time for a Ball Thrown Upward to Attain Its Maximum Height How long will it take the ball shown in Figure 2 to reach its maximum height? Given: v > i 5 3. m/s; a > m/s 2 ; v > f 5 m/s Required: Δt Analysis: To determine the time that this motion takes, we will identify an equation that uses only given information. Since the fi nal velocity is equal to zero, 5 2v> i a > Solution: 5 2v> i a > a > 5 v> f 2 v > i 5 v> f 2 v > i a > 2a3. m s b m s s Statement: It will take the tennis ball.31 s to reach its maximum height. Practice 1. A golf ball is thrown straight up in the air at a velocity of 8.3 m/s. T/I (a) Determine the maximum height of the golf ball. [ans: 3.5 m] (b) How long will it take the ball to reach its maximum height? [ans:.85 s] (c) How long will it take the ball to fall from its maximum height to the height from which it was initially launched? [ans:.85 s] 2. A rock is thrown downward from a bridge that is 12 m above a small creek. The rock has an initial velocity of 3. m/s downward. What is the velocity of the rock just before it hits the water? T/I [ans: 16 m/s down] 42 Chapter 1 Motion in a Straight Line

38 Figure 3 The value of 9.8 m/s 2 for acceleration near Earth assumes that there are no other forces acting on an object, such as air resistance. Free Fall and Terminal Velocity In real-life situations, there will always be some air resistance. Sometimes air resistance can be enough to have a significant effect on the motion of a falling object. For example, when a parachutist jumps out of an aircraft, he can control the amount of air resistance based on how he positions his body. If the parachutist dives out of the aircraft head first, he will experience very little air resistance. Most parachutists will try to fall so that as much of the surface area of their body is in contact with the air as possible. In other words, most parachutists will fall in a belly flop (Figure 3). When the air resistance on the parachutist is equal to the force due to gravity acting on the parachutist, the parachutist will stop accelerating and stay at a constant velocity, called the terminal velocity. You will learn more about air resistance and free fall in Unit Summary terminal velocity the velocity of an object when the force due to air resistance equals the force due to gravity on the object The symbol g is used to represent the acceleration due to gravity. All objects in free fall close to Earth s surface will accelerate at 9.8 m/s 2 toward Earth s centre. Air resistance can cause objects to accelerate at values less than g. When an object reaches terminal velocity, it will fall at a constant velocity. 1.6 Questions 1. Describe the motion of an object that is dropped close to Earth s surface. K/U C 2. A basketball player jumps up to make a basket and appears to hang in mid-air. Write a brief description explaining to a Grade 9 student what is occurring and why. C 3. A baseball is thrown straight up in the air, reaches its maximum height, and falls back down to the height from which it was originally thrown. What is the acceleration of the ball (a) halfway up to its maximum height? (b) at its maximum height? (c) halfway back down to the initial height from which it was thrown? T/I 4. A rubber ball is dropped from a height of 1.5 m. T/I (a) How long does it take to hit the ground? (b) What is the velocity of the ball when it has travelled a distance halfway to the ground? 5. An arrow is shot straight up into the air at 8. m/s. T/I (a) What is the arrow s maximum height? (b) How long does the arrow take to reach its maximum height? (c) Determine the total amount of time that the arrow is in the air. 6. A rock is thrown down from the top of a cliff with a velocity of 3.61 m/s [down]. The cliff is 28.4 m above the ground. Determine the velocity of the rock just before it hits the ground. T/I 7. Describe the motion of the object represented by the velocity time graph in Figure 4. Give an example of an object that might undergo this type of motion. K/U v (m/s [up]) Figure 4 Velocity v. Time for Accelerated Motion Research and describe a real-life situation where an object or person experiences an acceleration greater than g (9.8 m/s 2 ). A go to nelson science 1.6 Acceleration Near Earth s Surface 43

39 1.7 Explore an Issue in Vehicle Safety Defining the Issue Researching Identifying Alternatives Skills Menu Analyzing Defending a Decision Communicating Evaluating Figure 1 Many truck drivers object to the use of devices to limit speeds, arguing that they will cause more frustration on the road. Electronic Speed Limiters for Teen Drivers Today s automobiles are the safest vehicles ever. Advances in automotive technology have produced many safety devices, including seat belts, airbags, electronic stability control, electronic brake force distribution, and antilock brakes. All of these technologies are known to save lives when used correctly. Transport Canada indicates that, since 1987, the annual number of automobile accident fatalities in Canada has dropped by 33 %. If we are to decrease the number of automobile accidents even further, drivers will need to improve their safe driving practices. Statistics from the U.S. National Highway Traffic Safety Administration (NHTSA) show that teenage drivers are more likely to take driving risks, such as speeding, than older drivers. Speeding is a contributing factor in 3 % of all fatal crashes. Statistics also show that teens are less likely to wear seat belts than other drivers. In addition to contributing to driving accidents, speed can affect the emissions released by vehicles as well as their efficiency. Automobile emissions make up a substantial portion of greenhouse gas emissions. Vehicle manufacturers have embraced advances in automobile technology to produce substantially more fuel-efficient vehicles. Even a highly fuel-efficient vehicle is at its most efficient when driven at a reduced speed. Transport Canada indicates that by reducing a vehicle s speed from 12 km/h to 1 km/h, fuel consumption could drop by as much as 2 %. In Ontario and Québec, legislation has recently been passed to limit the speed of transport trucks to 15 km/h (Figure 1). A speed limiter is an electronic device that prevents a vehicle from travelling above a certain speed. This has been suggested as a way to increase highway safety and reduce the environmental impact of truck emissions. Some opponents of speed limiters say that the inability to go over a particular speed may prevent drivers from avoiding accidents. Others say that there is not enough evidence to prove that speed limiters actually prevent accidents, and that speed limiters can be tampered with. The Issue Suppose you are a member of the provincial student council association. The provincial government has tabled a bill that will call for the implementation of mandatory speed limiters for teenage drivers. The government is proposing that all automobiles sold in Ontario have an electronic device installed that will electronically limit teen drivers to a maximum speed of 9 km/h sound an alarm when the vehicle reaches speeds of 6 km/h, 7 km/h, and 8 km/h The provincial government has argued that installing these devices will increase highway safety, reduce automobile emissions, and protect new drivers. The provincial student council association will be making a presentation during a town hall meeting in which the proposed legislation is being discussed. Members of the Ontario Ministry of Transportation will be present. Role You will be acting as a member of the provincial student council association representing young drivers in Ontario. 44 Chapter 1 Motion in a Straight Line

40 AUDIEnCE Your audience will be the Ministry of Transportation but you will present your views at a town hall meeting where the speed limiter legislation is being discussed. Goal To convince the Ministry of Transportation to either support this bill in its current format recommend modifications to the bill not support the bill SKILLS Research HANDBOOK A5 Conduct library and Internet research about speed limiters and their effectiveness. Collect information about teen driving habits and their effect on road safety. You might conduct a survey among your fellow students for their opinions. You may wish to research the following: What positive effects have speed limiters had on safety among truck drivers? What negative effects have speed limiters had on safety? Would those who want to break the law be able to find a way to disable a speed limiter? WEB LInK To learn more about speed limiters, go TO SON SCiENCE Identify Solutions After reviewing your research, consider the impact the speed limiter bill could have on you and on the citizens of Ontario. What are the benefits and drawbacks of using speed limiters, and whom would they harm or benefit? What modifications to the bill might you recommend? Make a Decision Based on your research, decide which of the three possible solutions you will support. Alternatively, come up with a solution of your own. Communicate Prepare a position paper, photo essay, or other presentation format supporting your chosen decision. Be sure to support your decisions with research that will convince your audience (the Ministry of Transportation) to support your choice. Plan for Action This issue has the potential to affect you and your classmates for many years. Make a plan to make students aware of the potential for speed limiters that will electronically control driving behaviour among young drivers. Your plan should include an opportunity for young drivers to voice their opinions on this issue. For example, you might plan to start a group on a social networking site, or write an article for your school paper. 1.7 Explore an issue in Vehicle Safety 45

41 CHAPTER 1 Investigations Investigation OBSERVATIONAL STUDY Watch Your Speed In this observational study, you will use very simple equipment to measure the speed of real-life objects. purpose In this study, you will use a stopwatch and a tape measure to determine the average speed of vehicles driving past your school. You will then determine the percentage of vehicles exceeding the posted speed limit. Equipment and Materials 25 m 5 m tape measure stake or other marker stopwatch SKILLS HANDBOOK A2.4 procedure 1. Choose a location near your school where vehicles will be travelling at what appears to be a constant speed. For example, be sure not to be close to a stop sign or stoplight. 2. Choose an obvious landmark such as a telephone pole as a starting point. Making sure that you stay off the road, measure a distance from your starting point for 1 m to 2 m along the curb. Mark the end of your known distance with a stake or other obvious object. 3. Stand well back from the curb before taking time measurements. Do not stand on the road or too close to the curb for this study. Be aware of your surroundings and stay well away from the road at all times. Ensure you have your teacher s permission before performing this investigation. Questioning Researching Hypothesizing Predicting Planning Controlling Variables Performing 4. Start your stopwatch as each vehicle passes the starting point, and stop the stopwatch as the vehicle passes the stake representing the end of your measured distance. Record your observations. 5. Use the defining equation for average speed, v av 5 Dd, to determine the average speed of the vehicle. 6. Repeat this procedure for 5 vehicles. 7. Determine the speed limit in front of your school. 8. Calculate the percentage of vehicles passing your school that are exceeding the posted speed limit. Analyze and Evaluate SKILLS MEnU Observing Analyzing Evaluating Communicating SKILLS HANDBOOK A6.2 (a) Show one sample calculation determining the average speed of a vehicle. T/I C (b) Plot the data you have gathered on a graph and analyze your results. Summarize your findings in a brief paragraph. Be sure to explain whether your data suggest that the speed limit near your school is being observed. T/I (c) List any sources of uncertainty that may have affected your results. T/I (d) If you were to perform this study again, what would you do to reduce the uncertainty? T/I Apply and Extend (e) Using the evidence gathered from this investigation, develop a road safety action plan that will help improve the level of road safety at your school. C 46 Chapter 1 Motion in a Straight Line

42 Investigation OBSERVATIONAL STUDY Uniform Velocity There are a number of devices and methods for collecting motion data in the laboratory, including tickertape devices, air tables, spark-gap timers, photogates, video analysis, and motion sensors. Tickertape devices, air tables, and spark-gap timers rely on generating a series of dots on either a paper tape or a paper sheet. These dots are generated at a constant rate and can be used to gather position time data. Photogates, video analysis, and motion sensors use computer technology to collect and analyze real-time data. A motion sensor is an ultrasonic transceiver. It transmits invisible ultrasonic waves into space. These waves are reflected from objects in their path back to the motion sensor s receiver. Computer software running on the interface, or on a computer to which the motion sensor is connected, is able to use these ultrasonic pulses to collect position time data. Motion sensors and their associated software can be used to generate position time, speed time, and acceleration time data and graphs. As an experimenter, you can choose which type of data to collect. In this investigation you will use a motion sensor to collect position time, velocity time, and acceleration time data for a motorized toy car as it moves across a horizontal lab bench. SKILLS purpose HANDBOOK A2.4 To determine which type of motion (constant velocity, uniform acceleration, non-uniform acceleration) a motorized electric toy undergoes as it moves across a horizontal lab bench. Equipment and Materials motion sensor computer or computer interface motorized toy car metre stick data storage device such as a USB drive (memory stick) graph paper (optional) procedure 1. Connect the motion sensor to a computer or computer interface. 2. Set the software or computer interface to collect position time, velocity time, and acceleration time data (all three will be collected at once). 3. Turn on the toy car and hold it approximately 4 cm above the motion sensor. Questioning Researching Hypothesizing Predicting Planning Controlling Variables Performing SKILLS MEnU Observing Analyzing Evaluating Communicating 4. Start collecting data with the motion sensor. You will hear a clicking sound and see a blinking light on the motion sensor. 5. Quickly place the toy car on the lab bench so that it is 15 cm in front of the motion sensor (Figure 1). (Most motion sensors cannot collect data when objects are placed less than 15 cm away from them.) Figure 1 Positioning the car in front of the sensor 6. Allow the car to run approximately 2 m in a straight line away from the motion sensor. 7. Save your data. If necessary, export the data to a spreadsheet program. Analyze and Evaluate SKILLS HANDBOOK A6.5 (a) Generate full-page position time, velocity time, and acceleration time graphs using graph paper or computer software. Describe the characteristics of each graph (for example, straight line, curve, increasing, decreasing, horizontal, zero or non-zero intercept value). T/I C (b) What type of relationship is represented by each graph, linear or non-linear? T/I (c) What type of motion is described by each graph? K/U (d) Determine the slope of the position time graph. How does this value compare to the data shown in the velocity time graph? T/I C (e) Determine the slope of the velocity time graph. How does this value compare to the data shown in the acceleration time graph? T/I C Chapter 1 Investigations 47

43 (f) Calculate the area under the velocity time graph. How does this value compare to the final position of your toy car? T/I C (g) List any sources of uncertainty that may have affected your results. T/I (h) If you performed this experiment again, what would you do to reduce the uncertainty in your measurements? T/I Apply and Extend (i) In this investigation you used graphical methods to analyze your data. Is it possible to use algebraic methods to determine the toy car s average speed? Explain your reasoning. T/I (j) Perform research to determine how the technology you used in this investigation compares to sensors used in point-and-shoot digital cameras (Figure 2). How is it similar? How is it different? A Figure 2 Many digital cameras use motion sensors. go TO SON SCiENCE Investigation OBSERVATIONAL STUDY Motion Down a Ramp If you have ever been skiing or tobogganing, you know how thrilling motion can be when you are going down a hill or an incline. In this study you will use a motion sensor to collect position time, velocity time, and acceleration time data for a PVC pipe as it rolls down a ramp. While the procedure you will use in this investigation is similar to that used in Investigation 1.4.1, you will find that your results are significantly different. You will find comparing results of the two investigations to be helpful for your analysis and evaluation. purpose To determine which type of motion (constant velocity, uniform acceleration, non-uniform acceleration) a PVC pipe will undergo when it rolls down an inclined ramp. SKILLS HANDBOOK A2.4 Equipment and Materials motion sensor computer or computer interface data storage device, such as a USB drive (memory stick) PVC pipe ramp, 1.5 m or longer textbooks or wood blocks metre stick masking tape graph paper (optional) Questioning Researching Hypothesizing Predicting Planning Controlling Variables Performing SKILLS MEnU Observing Analyzing Evaluating Communicating procedure 1. Connect the motion sensor to a computer or computer interface. 2. Set the computer software or computer interface to collect position time, velocity time, and acceleration time data (all three will be collected at once). 3. Use the books or wood blocks to incline the ramp to approximately 3. Secure the motion sensor at the top of the inclined ramp with masking tape. Use a piece of masking tape to mark a point 15 cm in front of the motion sensor. Be sure to affix the masking tape in such a way that it will not interfere with the motion of your PVC pipe (Figure 1). Figure 1 Positioning the PVC pipe on the ramp 48 Chapter 1 Motion in a Straight Line

44 4. Hold the PVC pipe on the ramp just below the masking tape 15 cm in front of the motion sensor. Be sure to hold the PVC pipe at rest. 5. Start collecting data with the motion sensor. You will hear a clicking sound and see a blinking light on the motion sensor. 6. Quickly release the PVC pipe and allow it to roll straight down the ramp. Catch the PVC pipe at the bottom of the ramp. 7. Save your data. If necessary, export the data to a spreadsheet before your analysis. Analyze and Evaluate SKILLS HANDBOOK A6.5 (a) Generate full-page graphs of the position time, velocity time, and acceleration time data. Describe the characteristics of each graph (for example, straight line, curve, increasing, decreasing, horizontal, zero or non-zero intercept value). T/I C (b) What type of relationship is represented in each graph, linear or non-linear? T/I (c) What type of motion is described by each graph? K/U (d) Determine the slope of the velocity time graph. How does this calculated value compare to the data shown in your acceleration time graph? T/I C (e) Calculate the area under the acceleration time graph. How does this calculated value compare to the final velocity of the PVC pipe? T/I C (f) List any sources of uncertainty that may have affected your results. T/I (g) If you were able to perform this study again, what could you do to reduce the uncertainty? T/I Apply and Extend (h) Repeat this investigation using remote control vehicles. Generate data that show constant velocity and uniform acceleration. Compare the acceleration of different brands and models of remote control vehicles. T/I C (i) Repeat this experiment by rolling the PVC pipe up the inclined ramp. Be sure not to push the pipe hard enough to physically hit the motion sensor. How do the data collected for this investigation compare to the data collected in the original investigation? T/I C (j) Roller coasters are some of the most exciting thrill rides available. Modern roller coasters twist, turn, and spiral in a complex and thrilling way. Old-fashioned roller coasters like the Mighty Canadian Minebuster at Canada s Wonderland are wooden and do not spiral but can still make for an exciting ride (Figure 2). Use the knowledge you gained from this investigation, and any personal life experiences, to help you write a letter to your younger sister explaining which part of an old-fashioned roller coaster ride is the most exciting, which part is the least exciting, and why. T/I C Figure 2 What makes a roller coaster ride like this exciting? Chapter 1 Investigations 49