Motion in 1 Dimension

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Mitchell Miles
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1 Motion in 1 Dimension Physics is all about describing motion. For now we are going to discuss motion in 1 dimension, which means either along the x axis or the y axis. To describe an object s motion, we must be able to specify its position (x) at all times. An object s position is the x or y coordinate of its location. Since we are discussing motion along the x-axis first, we will assign an object s initial position the label x i and its final position the label x f. When an object changes its position, it experiences a displacement. This is a vector that points from an initial position x i to a final position x f. The magnitude is the shortest distance between the two positions. Mathematically, displacement is given by x x x f i x is positive if x ( object moves right f > ) x i Positive displacement The 3 bar equal sign means mathematical definition. x is negative ( object moves left Negative displacement if x ) f < x i

2 We will use the meter as our standard S.I. unit of measurement. Do you remember your conversions? 1 meter = 100 centimeters 1 meter = 1000 millimeters 1 kilometer = 1000 meters FROM THIS POINT ON, YOU REALLY NEED TO KNOW THOSE METRIC CONVERSIONS!!! Now it s time to realize that there is a HUGE difference between distance traveled and displacement!! As an example A baseball player hitting a home run travels around the bases. Questions What is his initial position? What is his final position? What distance did he run? What is his displacement?

98 = -0.11 m SAMPLE PROBLEM B A lazy susan is spun 3 times.

3 SAMPLE PROBLEM A A marble rolls alongside a meter stick. It begins moving at the 98- cm mark and stops on the 87-cm mark. Calculate its displacement. x = x f x i = = m SAMPLE PROBLEM B A lazy susan is spun 3 times. A container of parsley flakes travels around a circular path having a radius of 6.0 cm. (a) What was the distance traveled by the parsley? (b) What was the parsley s displacement? a) d = 3 (2πr) = 6π(0.06) = m b) x = 0 m

4 Graphing Position vs. Time (x vs. t) Important relationships can be discovered from graphs, so let s plot a sample position vs. time (x vs. t) graph to analyze. 4 Steps to Graphing Position vs. Time Step 1: Label the x-axis with Time (units) Label the y-axis with Position (units) Entitle the graph Position vs. Time Step 2: Determine an appropriate scale for each axis. The idea is to make the graph as large as possible! Step 3: Plot the data neatly and connect the points with a smooth line. Step 4: Calculate the slope for each segment of the graph using the equation below. Don t forget units as this slope will now be a meaningful quantity!!! rise y m = = or for our graph : run x x t

5 SAMPLE (x vs. t) GRAPH Position (m) Time (s) Step 1: Label, label, label! Step 2: Determine an appropriate scale for each axis. Step 3: Plot the data neatly and connect the points with a smooth line. Step 4: Calculate the slope for each segment of the graph. Slope 1 = x / t = 1200 m / 600 s = 2 m/s Slope 2 = x / t = 0 m / 400 s = 0 m/s Slope 3 = x / t = -800 m / 800 s = -1 m/s What is significant about the slope s units? m/s. Indicates that the slope is telling you velocity.

6 v Average Velocity This slope is so important that it has been given a special name; Average Velocity. Average velocity is the change in position with time. Since velocity depends on displacement, velocity is also a vector quantity. meters per second (m/s) is the standard S.I. unit of measurement We can find between any two points by drawing a straight line between them and calculating the slope, regardless of the graph s shape. Average Velocity is defined as: x v x = slope of x vs. t v x is positive if x is positive or slope of x vs. t is positive t graph v x is negative if x is negative or slope of x vs. t is negative

7 PRACTICE PROBLEM A NAME: A farmer collected the following data after lopping the head off one of his chickens. Set up a position vs. time graph to help the farmer understand the motion of the bird. Label each straight-line segment A J. Position (m) Time (s) Calculate V AVG for the following time intervals: (0 2 sec): x/ t = 8m/2s = 4 m/s (2 4 sec): x/ t = -6m/2s = -3 m/s (1 6 sec): x/ t = -2m/5s = -0.4 m/s (4 9 sec): x/ t = 4m/5s = 0.8 m/s (4 6 sec): x/ t = 0m/2s = 0 m/s (0 17 sec): x/ t = 6m/17s = m/s Describe the motion of the bird for each segment: A: B: C: D: E: F: G: H: I: J:

8 Velocity vs. Time Graphs Instead of graphing just position vs. time, we are also able to graph velocity vs. time. This graph will show you how velocity changes with time. Therefore, the rules that applied to a position vs. time graph do not necessarily apply to a velocity vs. time graph. velocity (m/s) time (s) x = positive v = increasing a = positive time (s) x = positive v = increasing a = positive velocity (m/s) velocity (m/s) velocity (m/s) time (s) x = positive v = constant a = zero time (s) x = positive v = decreasing a = negative

9 Instantaneous Velocity v Why would not be useful if you wanted to know your velocity the instant you noticed the flashing lights of a police car behind you? We need to be able to specify our velocity just as precisely as we can our position by knowing what is happening at a specific time. We could not actually do this mathematically until the late 1600 s when Calculus gave us this ability. We can do this in Physics by taking smaller and smaller intervals of time on our x vs. t graph. The limiting value of the ratio as approaches zero is called Instantaneous Velocity. x t t Notice that this is just the limit of the average velocity as time approaches zero. We can interpret V graphically by drawing a line tangent to the curve at a specific time. We can then calculate the slope of the tangent line.

10 Instantaneous Velocity is defined as: v x lim = t t 0 slope of tangent line From here on, we use the term velocity to designate instantaneous velocity. SAMPLE PROBLEM An object moves along the x-axis as shown in the position vs. time graph to the right. (a) Determine the object s displacement during the time intervals t = 1 s to t = 3 s and t = 4 s to t = 6 s. (b) Calculate the average velocity during these two time intervals. (c) Find the object s instantaneous velocity at t = 2.0 s and at 6.0 s. x = x f x i = 3m 11m = -8 m x = x f x i = 5m 2m = 3 m v = x / t = -8 / 2 = -4 m/s v = x / t = 3 / 2 = 1.5 m/s v = x / t = -13 / 3.5 = m/s

11 Instantaneous Velocity (Using Calculus) SAMPLE PROBLEM A person moves along the x-axis. Her position varies with time according to the expression x = 2t 2-4t where x is her instantaneous position in meters while t represents the time at that position. The graph for this motion is shown below. Find the instantaneous velocity at t = 0s, t = 1s, and t = 3s. Step 1 = Take the derivative of the function: x = 2t 2 4t v = 4t 4 Step 2 = Plug in your value of time v(0) = 4(0) 4 = -4 m/s v(1) = 4(1) 4 = 0 m/s v(3) = 4(3) 4 = 8 m/s

In everyday usage, the words speed and velocity are used interchangeably. However, there is a HUGE difference!!! EXAMPLE During the 2008 Summer Olympics, Michael Phelps swam four 50-m laps in 1.

12 In everyday usage, the words speed and velocity are used interchangeably. However, there is a HUGE difference!!! EXAMPLE During the 2008 Summer Olympics, Michael Phelps swam four 50-m laps in 1.52 minutes to win the men s 200-m Butterfly and set a new world record. QUESTIONS: What distance was covered? 4 x 50 m = 200 m What was the swimmer s displacement? x = 0m What was the swimmer s average velocity? V = x/ t = 0m / 1.52 min = 0 m/s This is not very useful, so we still need to be able to quantify how fast she swam.

13 This is where the term Average Speed comes in. Speed is a scalar quantity so the direction is meaningless in describing it. Average Speed is given by: s = total distance total time Instantaneous Speed is defined as the magnitude of the instantaneous velocity. Your car s speedometer measures instantaneous speed!! You need to very careful here! Average Speed IS NOT the magnitude of Average Velocity because average speed is dependent on distance and average velocity depends on displacement!

Acceleration We have already seen that velocity can change while a particle moves. We can quantify changes in velocity as a function of time just as we could with x(t).

14 Acceleration We have already seen that velocity can change while a particle moves. We can quantify changes in velocity as a function of time just as we could with x(t). When velocity changes with time, a particle is undergoing Acceleration. Mathematically, acceleration is defined in the following way: v a = slope of v vs. t t graph As with velocity, when the motion is 1-Dimensional, we can simply use +/- to indicate direction. meters per second 2 (m/s 2 ) is the standard S.I. unit of measurement.

SAMPLE PROBLEM A Suppose a plane starts from rest (v i = 0 m/s) when t i = 0 s.

Determine the acceleration of the plane. 260 km x 1000 m x 1 hr = 72.222 m/s hr 1 km 3600s a = (V f V i ) / t = (72.222 0) / 29 = 2.

The driver begins slowing down when t = 9.0 s and the car s velocity is v = 28 m/s. When t = 12.0 s, the velocity has been reduced to v = 13 m/s.

15 SAMPLE PROBLEM A Suppose a plane starts from rest (v i = 0 m/s) when t i = 0 s. The plane accelerates down the runway and at t f = 29 s attains a velocity of v f = km/h, where the plus sign indicates the velocity points to the right. Determine the acceleration of the plane. 260 km x 1000 m x 1 hr = m/s hr 1 km 3600s a = (V f V i ) / t = ( ) / 29 = m/s 2 SAMPLE PROBLEM B A drag racer crosses the finish line, and the driver deploys a parachute and applies the brakes to slow down. The driver begins slowing down when t = 9.0 s and the car s velocity is v = 28 m/s. When t = 12.0 s, the velocity has been reduced to v = 13 m/s. What is the acceleration of the dragster? a = (V f V i ) / t = (13 28) / (12-9) = -5 m/s 2 SAMPLE PROBLEM C A glider s velocity on a tilted air track increases from cm/s at the time t = 0.33 s to cm/s at a time of t = 2.33 s. What is the acceleration of the glider? -17 cm/s = m/s, -55 cm/s = m/s a = (V f V i ) / t = (-0.55 (-0.17)) / 2 = 0.19 m/s 2

16 1-D Motion w/ Constant Acceleration The following kinematic equations only work for situations involving constant acceleration!!! v f = vi + at 2 v f 2 i = v + 2a x 1 x = ( vi + v f ) t 2 1 x v t at 2 = i + 2 You need to have these memorized by next class!

Physics Kinematics SAMPLE PROBLEMS Name: Date: Period: Directions: Solve the following problems in the space provided. Use the tables to organize your data.

17 Physics Kinematics SAMPLE PROBLEMS Name: Date: Period: Directions: Solve the following problems in the space provided. Use the tables to organize your data. If a variable is not given, record a in the block. If a variable is unknown, record a? in the block. Convert all measurements into units of meters, seconds and meters per second. Be sure to show all work! SAMPLE PROBLEM A The doors open for a back to school sale and a crazed teen sprints from her mom s car. If the girl reaches a velocity of +4.6 m/s in 1.9 s, determine how far she ran to the store. DATA TABLE x v i v f a t? x = ½ (V i + V f )t = 0.5 ( )1.9 = 4.37 m SAMPLE PROBLEM B Starting from rest, a cow reaches a velocity of +2.5 m/s in a time of 5.0 s. A Calculate the cow s average acceleration. B By how much does the cow s speed change each second? C Calculate the bovine s displacement during this time interval. DATA TABLE x v i v f a t? A. a = (V f V i ) / t = (2.5 0) / 5 = 0.5 m/s 2 B. Increases m/s for every second of movement C. x = V i t + ½ at 2 = 0.5(0.5)5 2 = 6.25 m

SAMPLE PROBLEM C A man s 2010 Porsche 911 reaches 60 mi/h with a constant acceleration of + 8 m/s 2. Assuming the car starts from rest, how long does it take the car to reach this speed?

352 s SAMPLE PROBLEM D Billy Bob is going to fetch some dinner. He catches sight of a squirrel that he fancies and takes off a runnin. Billy Bob starts from rest, accelerates at the rate of + 1.

5 - V f 2 = V i 2 + 2a x x = (V f 2 V i 2 ) / 2a = (4.7 2 0 2 ) / (2(1.5)) = 7.363 m SAMPLE PROBLEM E A motorcycle accelerates from 90 mi/h to 65 mi/h in 2.3 s.

18 SAMPLE PROBLEM C A man s 2010 Porsche 911 reaches 60 mi/h with a constant acceleration of + 8 m/s 2. Assuming the car starts from rest, how long does it take the car to reach this speed? (Hint: 1 mile = 1609 m) DATA TABLE x v i v f a t ? 60 mi x 1609 m x 1 hr = m/s hr 1 mi 3600s t = (V f V i ) / a = ( ) / 8 = s SAMPLE PROBLEM D Billy Bob is going to fetch some dinner. He catches sight of a squirrel that he fancies and takes off a runnin. Billy Bob starts from rest, accelerates at the rate of m/s 2 and reaches a velocity of m/s before catching the critter. Calculate Billy s displacement for the hunt. DATA TABLE x v i v f a t? V f 2 = V i 2 + 2a x x = (V f 2 V i 2 ) / 2a = ( ) / (2(1.5)) = m SAMPLE PROBLEM E A motorcycle accelerates from 90 mi/h to 65 mi/h in 2.3 s. Find the motorcycle s acceleration and displacement during this time interval. DATA TABLE x v i v f a t? ? mi x 1609 m x 1 hr = m/s hr 1 mi 3600s 65 mi x 1609 m x 1 hr = m/s hr 1 mi 3600s a = (V f V i ) / t = ( ) / 2.3 = m/s 2 x = V i t + ½ at 2 = (2.3) + 0.5(-4.858)2.3 2 = m

SAMPLE PROBLEM F A car, initially traveling at 70.0 km/h, accelerates at the constant rate of -1.50 m/s 2. How far will the car travel in 10.0 s? DATA TABLE x v i v f a t? 19.444-1.

A car traveling at a constant speed of 45.0 m/s passes a trooper hidden behind a billboard.

19 SAMPLE PROBLEM F A car, initially traveling at 70.0 km/h, accelerates at the constant rate of m/s 2. How far will the car travel in 10.0 s? DATA TABLE x v i v f a t? km x 1000 m x 1 hr = m/s hr 1 km 3600s x = V i t + ½ at 2 = (10) + 0.5(-1.5)10 2 = m CHALLENGE PROBLEM Watch Out for the Speed Limit! A car traveling at a constant speed of 45.0 m/s passes a trooper hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets out from the billboard to catch it, accelerating at a constant rate of 3.00 m/s 2. How long does it take her to overtake the car? x car = t x trooper = V i t + ½ at 2 = 0 + ½(3t 2 ) = 1.5t 2 1.5t 2 = 45t t 2-45t 45 = 0 USE QUADRATIC FORMULA t = (-b ± (b 2 4ac)) / 2a = (-(-45) ± (45 2 4(1.5 * -45))) / 2(1.5) t = s, s

20 Freely Falling Objects All objects dropped near the earth s surface fall with the same constant acceleration in the absence of air resistance. Aristotle ( B.C.) believed that heavier objects fell faster than lighter ones. This fact was not fully accepted until the early 1600s. Galileo ( ) conducted experiments with balls on inclined planes and drew conclusions about freely falling objects since a freely falling ball is equivalent to a ball moving down a vertical incline.? What kinematic equation could be used to find the acceleration in this experimental procedure? x = v i t + 1 at 2 2 On August 2 nd, 1971 David Scott released a hammer and a feather on the surface of the moon. What was discovered?

21 By definition, a freely falling object is any object moving freely under the influence of gravity alone, regardless of its initial motion. Objects thrown upward or downward and those released from rest are all falling freely once they are released. Any freely falling object experiences an acceleration directed DOWNWARD, regardless of its initial motion. The value of g does change with increasing altitude, as we will see when we study Newton s Law of Universal Gravitation. At the earth s surface (and for our discussions in this course) the acceleration due to gravity is approximately m/s 2.

22 Our kinematic equations now become: v y f = v y i gt y v y f = 2 v y i t = v 2g y y i 2 1 gt 2 2

23 Name: Date: Period: Mr. Talboo - Physics SAMPLE PROBLEM A A stone is thrown downward from the top of a tall building with an initial speed of 5.00 m/s. What is the displacement of the stone after 3.00 s of free-fall? What is the final velocity of the stone when it strikes the ground? DATA TABLE y v yi v yf g t? 5? y = V i t ½ gt 2 = (-5)(3) - 1/2 (9.8)(3 2 ) = m V yf = V yi gt = 5 (9.8)(3) = m/s SAMPLE PROBLEM B You and a friend are going to toss a coin to determine who asks Mr. Talboo for help on a kinematics problem. You toss the coin up with an initial velocity of 6.00 m/s. How high does the coin go above the point of release? What is the total time the coin is in the air before returning to its release point? DATA TABLE y v yi v yf g t? ? y = (V f 2 V i 2 ) / -2g = ( ) / -2(9.8) = m t = (V f V i ) / -g t = (-6 6) / -9.8 t = s

24 SAMPLE PROBLEM C A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. The building is 50.0 m high, and the stone just misses the edge of the roof on its way down. Determine (a) the time at which the stone reaches its maximum height, (b) the maximum height, (c) the total time that it is in the air and (d) the velocity of the stone when it hits the ground. DATA TABLE y v yi v yf g t t = (V f V i ) / -g t = (0 20) / -9.8 t = s y = (V f 2 V i 2 ) / -2g = ( ) / -2(9.8) = m + 50 m = m y = V i t ½ gt 2 = (0) ½ gt 2 t = ((2 y)/-g) = ((2 * )/-9.8) = s s t = s V yf = V yi gt = 20 (9.8)(5.832) = m/s

Chapter 3. Motion in One Dimension

Chapter 3. Motion in One Dimension Chapter 3 Motion in One Dimension Outline 3.1 Position, Velocity and Speed 3.2 Instantaneous Velocity and Speed 3.3 Acceleration 3.4 Motion Diagrams 3.5 One-Dimensional Motion with Constant Acceleration