3. TopicSlidesMinutes 1 Displacement 927 2 Vectors 1339 3 Kinematics 1339 4 1030 5 Energy 1030 6 Power 515 7 Springs 412 8 Shadows 39 9 Field of Vision 721 10 Colors 39 11 Concave mirrors 721 12 Convex mirrors 412 13 Refraction 515 14 Lenses 1030 15 Optical Power 618 Graphs

5. Distance-Time Graphs

6. t(s) Example 0 1 2 3 4 5 6 7 8 1234567891011 121314 Distance-Time Graph Slope = velocity Area = nothing s (m) Click to continue a)At t = 2 s velocity = 1 m/s (slope = 1) b)At t = 5 s velocity = 0 (slope = 0) c)At t = 9 s velocity = -1 m/s (slope = -1) d)At t = 12 s velocity = 0 (slope = 0) Slope = m / s = velocity Area = m x s = ms = nothing ** NOTE **

7. 0 10 20 30 40 50 60 70 80 1234567891011 121314 s (m) t (s) In a distance versus time graph, the slope represents the velocity. And the area under the curve, does not represent anything. Area = nothing Slope = velocity This area T h u s : To find the velocity at any point, find the slope at that point. REMEMBER Click Distance –vs- Time graph

8. Velocity-Time Graphs

9. 0 10 20 30 40 50 60 70 80 1234567891011 121314 v (m/s) t (s) In a velocity versus time graph, the slope represents the acceleration. And the area under the curve, represents the distance traveled. Area = distance traveled Slope = acceleration This area T h u s : To find the acceleration at any point, find the slope at that point. To find the distance traveled between any two points, find the area under the curve between those two points. a n d REMEMBER Click

10. Area of sector D = 1 m/s x 6 s = 6 m t(s) A B C D Distance traveled = A + B + C + D = 4.5 m + 15 m + 2 m + 6 m = 27.5 m Area of sector A = (3 m/s x 3 s)/2 = 4.5 m Area of sector B = 3 m/s x 5 s = 15 m Area of sector C = (3 m/s x 2 s)/2 = 2 m Example-1 To find the total distance, we need to find the total area under the curve. There are a number of ways of finding the total area under a curve. In the example above, one way is to divide the total area into four sections (segment-A, segment-B, segment-C and segment-D). Answer First we find the area of each segment and then add up all the segments to obtain the total area and thus the answer. Since velocity is plotted versus time, we know that the slope represents the acceleration of the vehicle and the area under the curve represents the distance traveled by the vehicle. The graph on the left illustrates the velocity-time curve of a vehicle. Find the total distance traveled by the vehicle. TASK 0 1 2 3 4 5 6 7 8 1234567891011 121314 Velocity-Time Graph Slope = acceleration Area = distance v (m/s) Click to continue

11. 0 1 2 3 4 5 6 7 8 1234567891011 121314 v (m/s) t(s) Example-2 Velocity-Time Graph Slope = acceleration Area = distance Note that the first 5 seconds is from t = 0 to t = 5 s. Thus, we need to find the distance under the curve from t = 0 to t = 5 s.Since the distance for the first 5 s consists of two segments, we will find the area in two steps. Step-1 is from t = 0 to t = 3 s (segment A) and step-2 is from t = 3 s to t = 5 s (segment B). A B Step-1: Area of sector A = (3 m/s x 3 s)/2 = 4.5 m Step-2: Area of sector A = (3 m/s x 2 s) = 6 m Step 1Step 2 Distance traveled = A + B = 4.5 m + 8 m = 10.5 m The graph on the left illustrates the velocity-time curve of a vehicle. Find the distance traveled by the vehicle for the first 5 s. TASK Answer Click to continue

12. 0 1 2 3 4 5 6 7 8 1234567891011 121314 v (m/s) t(s) Example-3 Velocity-Time Graph Slope = acceleration Area = distance Note that the area under the curve for this problem is from t = 4 s to t = 7 s and consists of only one section. Thus, all we need to do is find the area of the rectangle. Area under the curve = 3 m/s x 3 s = 9 m Distance traveled = 9 m Answer The graph on the left illustrates the velocity-time curve of a vehicle. Find the distance traveled by the vehicle between t = 4 s and t = 7 s. TASK

13. 0 1 2 3 4 5 6 7 8 1234567891011 121314 v (m/s) t(s) Example-4 Velocity-Time Graph Slope = acceleration Area = distance Note that the area under the curve for this problem is from t = 9 s to t = 14 s. The area under the curve for this problem consists of two sections, a small triangle and a rectangle. Area of triangle = (1 m/s x 1 s)/2 = 0.5 m Distance traveled = 0.5 m + 5 m = 5.5 m The graph on the left illustrates the velocity-time curve of a vehicle. Find the distance traveled by the vehicle during the last 5 seconds. TASK 1 m/s by 1 s Area of rectangle = (1 m/s x 5 s) = 5 m 1 m/s by 5 s Answer

15. During the course of a laboratory experiment, a team of students obtained the graph on the right representing the force exerted as a function of the acceleration of a cart. Calculate the mass of the cart. A)3.0 kg B)2.0 kg C)1.0 kg D)0.50 kg E)0.25 kg Click Note that the slope represents mass. Click Graphs Slide: 4. 14. 1

16. Reminder The area under the curve is the distance travelled. 200 m 250 m + 100 m 350 m 200 m Answer: 350 m – 200 m = 150 m Click Graphs Slide: 4. 24. 2

17. Illustrated below is the velocity versus time graph of a particle. Click A)200 000 m B)250 000 m C)300 000 m D)350 000 m How far has the particle travelled in 5 minutes? Convert to seconds NOTE: The area under the curve represents distance. 5 min = 5 x 60 s = 300 s 50 000200 000 250 000 Graphs Slide: 4. 34. 3

18. Consider the position versus time graph below. Click A) Increasing velocity, constant velocity, increasing velocity B) Increasing velocity, zero velocity, increasing velocity C) Constant velocity, constant velocity, constant velocity D) Constant velocity, zero velocity, constant velocity Which one of the following statements best describes the motion illustrated by the above graph. Graphs Slide: 4. 44. 4

19. Click Which of these graphs represents the acceleration of the car? The velocity-time graph on the left represents the motion of a car during a 6 s interval of time. A) B) C) D) Positive acceleration Zero acceleration Negative acceleration Graphs Slide: 4. 54. 5

20. Fastest velocity here thus maximum KE at this point. This point shows that the car has zero KE which means it has zero velocity at point I which is incorrect. Click Graphs Slide: 4. 64. 6

21. Velocity = 25 m/s Velocity = 750 m/25 s = 30 m/s Area = velocity = 50 m/s Velocity = 600 m/25 s = 24 m/s Click Graphs Slide: 4. 74. 7

22. This means the first segment has to be a straight line (a=0). This means the second segment has a negative acceleration (slope). Click 4. 84. 8 Slide:

23. Since the distance between dots is increasing, the object is accelerating. Constant forward velocity Zero velocityConstant reverse velocity Click Graphs Slide: 4. 94. 9

24. Carlo Martini owns a Ferrari. Carlo performed a speed test on his car and plotted the graph below. Click Knowing that Carlo’s Ferrari has a mass of 1288 kg, calculate the net force of the car based on the above graph. Note that the slope represents acceleration. Graphs Slide: 4. 10

25. Click Both a car and a truck drive off at the same time and in the same direction. The graphs below illustrate their movement. Determine how far apart the two vehicles are after 30 seconds. Step-1 Distance of car = area under the curve = 275 m Step-2 Distance of car = V A t = (10 m/s)(30 s) = 300 m Area to x-axis = distance Slope = Average velocity = 10 m/s Step-3 Distance apart = 300 m – 275 m = 25 m Graphs Slide: 4. 13

26. … and good luck!