April 26, s1206 Lesson 7 sandy 2015.notebook. Acceleration. speeding up slowing down changing direction. Recall

Transcription

1 Acceleration speeding up slowing down changing direction Recall 1

2 The slope of a displacement time graph still represents velocity (tangent technique gives instantaneous velocity). The slope of a velocity time graph is acceleration. Area under a velocity time graph is displacement. 2

3 Velocity time graphs Recall that for these graphs, Slope represents acceleration (positive is speeding up, negative is slowing down) 0 means the object has stopped (0 m/s) positive portion of graph means motion to right, negative portion means motion to the left. Refer also to page in text. 3

4 4

5 Moving from one graph to another Moving from one graph to another Example 5

6 Acceleration 6

7 Instantaneous Speed: The speed at which an object is travelling at a particular instant in time. It is not affected by its previous speed or by how long it has been moving. Question: What are two devices that measure instantaneous speed? Answer: radar gun and speedometer. 7

8 Ex 3: What is the instantaneous velocity at 5 s on the graph? Ex 4: The following date was obtained for a skateboarder rolling down a slight grade of a parking lot. A) Plot a graph of distance vs. time on the grid provideḍ Table 1: Distance Time Data Graph 1: Graph of Distance vs.time Determine the instantaneous speed of the skateboarder at: (i) t = 20s (ii) t = 30s 8

9 Tabulate the results from question b in table 2 below. Graph the data using line of best fit on the grid provided below. Table 2: Speed Time Data Graph 2: Graph of Speed vs. Time d) Calculate the slope of the speed time graph. What does this represent? Ex 5: Acceleration is defined as the change in velocity divided by the change in time. I is a vector. An acceleration of +2.0 m/s/s means that each second, the velocity changes by +2.0 m/s. If you start from rest v = 0 m/s with acceleration of +2.0 m/s 2 then my velocity after 1.0 second is after 2.0 seconds is after 3.0 seconds is after 4.0 seconds is after 5.0 seconds is after 6.0 seconds is 1. Graph this motion 9

10 2. Is this motion uniform? How can you tell? 3. From the graph, What is the instantaneous velocity at 1.0 s = 3.0 s = 6.0 s = 4. What is the slope of the line? 5. The slope of the line from a v t graph tells me the acceleration. What is the acceleration of this object? (don t forget the unit) 6. Describe the motion of this object. Is it moving to the right or left? Is it speeding up or slowing down? 7.What does the area of a v t graph tell us? 8. The area under a v t graph tells us the displacement. Calculate the displacement under each v t graph. Don t forget your unit (meters) : : : 10

11 9. The slope of a v t graph means acceleration. What is the acceleration for each of the objects above? A. = B. = C. = 10. The initial velocity is the velocity at time zero (t = 0.0 seconds). You get this from the y intercept of the v t graph. What is the initial velocity for each of the objects above? A. = B. = C. = 11. How do you know if the object is speeding up or slowing down? Check to see if the velocity becomes bigger or smaller over time. For each of the graphs above describe its motion. (Speeding up or slowing down?) A. B. C. 12. How do you tell if the object is moving right or left? The velocity will tell you. Positive velocities are to the right, negative velocities are to theleft. Which way is each object moving above? (Be very careful with B!!) Moving left? Moving right? A. B. C. Note: The direction of velocity and acceleration will determine the size of the velocity (ie. If an object is speeding up or slowing down): Velocity Direction Acceleration Direction Size of velocity positive positive increases negative negative increases positive negative decreases negative positive decreases If they are in the same direction (both are positive or both are negative) then the object is speeding up If they are in opposite directions (one is positive and the other is negative), the object is slowing down 11

12 Practice: p. 450 #3 p. 456 # 2, 3, 6, 9, 10 Mathematical Relationship for Acceleration Recall: Recall: the slope = rise/run = m/s/s = m/s 2 Thus, a = v t 12

13 Definition: Acceleration is defined as the rate of change of speed or velocity. It is achieved through a change in an object s speed and/or direction. Note that in the lab completed, we dealt with a change in speed. Example: If an object has a constant acceleration of 2.0 m/s 2 it means that every second the object s speed increases by 2.0 m/s. Mathematically we write, a av = v = v f v i t t where v f = final speed vi = initial speed t = time Acceleration has units of m/s 2. Constant acceleration occurs when the same change in speed ( v) occurs in equal intervals of time ( t). Rearranging the acceleration formula a av = v f v i t 13

14 Example 1: What is meant if we say you have an average acceleration of 5.0 m/s 2? Solution: This means that for every second you increase your speed by 5.0 m/s. So at 1.0 s your speed would be 5.0 m/s, at 2.0 s it would be 10 m/s and so on. Example 2: A skier is moving at 1.8 m/s (down) near the top of a hill. 4.2 s later she is travelling at 8.3 m/s (down). What is her average acceleration? solution vi = 1.8 m/s vf = 8.3 m/s t = 4.2 s a =? a = vf vi t = 8.3m/s 1.8 m/s 4.2 s = 1.5 m/s 2 [down] 14

15 Example 3: Calculate the takeoff speed of a Boeing 747 if it accelerates at 4.0 m/s 2 and takes 40.0 s to reach this speed from rest. Ex 4: Canadian Myriam Bedard won two gold medals in the biathlon in the 1994 Winter Olympics. If she accelerates at an average of 2.5 m/s 2 (E) for 1.5 s, what is her change in velocity at the end of 1.5 s? a = v t Note that velocity and acceleration are not always in the same direction. 15

16 Ex5: A rabbit, eating in a field, scents a fox nearby and races off. It takes only 1.8 s to reach a top velocity of 7.5 m/s (N). What is the rabbit s acceleration during this time? Ex 6: An air puck on an air table is attached to a spring. The puck is fired across the table at an initial velocity of 0.45 m/s [right] and the spring accelerates the air puck at an average acceleration of 1.0 m/s 2 [left]. What is the velocity of the air puck after 0.60 s? 16

17 Ex 7: A person throws a ball straight up from the ground. The ball leaves the person s hand at an initial velocity of 10.0 m/s up. The acceleration of the ball is 9.81m/s2 down. Assume up is positive and down is negative. What is the velocity of the ball after 0.50s and after 1.5s? Ex 8: A skateboarder, travelling at a constant speed, approaches a ramp and accelerates at 0.75 m/s2 for 4.0s to a final speed of 6.0 m/s. What was the skateboarder s initial speed? 17

18 Practice Questions p. 388 # 1 16 p. 465 # 2, 3, 5, 6, 7, 8, 9 18