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Physics

Physics. Review. Graphs. Distance-Time Graphs. ** NOTE **. 8. 7. 6. 5. 4. 3. 2. 1. 0. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 11. 12. 13. 14. Slope = m / s = velocity. Area = m x s = ms = nothing. s (m). Distance-Time Graph Slope = velocity Area = nothing.

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Physics

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  1. Physics Review Graphs

  2. Distance-Time Graphs

  3. ** NOTE ** 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 Slope = m / s = velocity Area = m x s = ms = nothing s (m) Distance-Time Graph Slope = velocity Area = nothing t(s) • At t = 2 s velocity = 1 m/s (slope = 1) • At t = 5 s velocity = 0 (slope = 0) • At t = 9 s velocity = -1 m/s (slope = -1) • At t = 12 s velocity = 0 (slope = 0) Example Click to continue

  4. REMEMBER 80 70 60 50 40 30 20 10 0 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 Distance –vs- Time graph the slope represents the velocity. In a distance versus time graph, And the area under the curve, does not represent anything. s (m) Slope = velocity This area Area = nothing t (s) Thus: To find the velocity at any point, find the slope at that point. Click

  5. Velocity-Time Graphs

  6. REMEMBER 80 70 60 50 40 30 20 10 0 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 the slope represents the acceleration. In a velocity versus time graph, And the area under the curve, represents the distance traveled. v (m/s) Slope = acceleration This area Area = distance traveled t (s) Thus: To find the acceleration at any point, find the slope at that point. and To find the distance traveled between any two points, find the areaunder the curve between those two points. Click

  7. TASK The graph on the left illustrates the velocity-time curve of a vehicle. Find the total distance traveled by the vehicle. 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 Answer v (m/s) Velocity-Time Graph Slope = acceleration Area = distance C A B t(s) D 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 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). 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. First we find the area of each segment and then add up all thesegments to obtain the total area and thus the answer. Area of sector C = (3 m/s x 2 s)/2 = 2 m To find the total distance, we need to find the total area under the curve. Area of sector D = 1 m/s x 6 s = 6 m Distance traveled = A + B + C + D = 4.5 m + 15 m + 2 m + 6 m = 27.5 m Example-1 Click to continue

  8. TASK 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. 8 7 6 5 4 3 Answer 2 Step 1 Step 2 1 0 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 v (m/s) Velocity-Time Graph Slope = acceleration Area = distance A B t(s) Step-1: Area of sector A = (3 m/s x 3 s)/2 = 4.5 m Note that the first 5 seconds is 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). Thus, we need to find the distance under the curve from t = 0 to t = 5 s. Step-2: Area of sector A = (3 m/s x 2 s) = 6 m Distance traveled = A + B = 4.5 m + 8 m = 10.5 m Example-2 Click to continue

  9. TASK The graph on the left illustrates the velocity-time curve of a vehicle. Find the distance traveled by the vehicle between t = 4 sand t = 7 s. 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 Answer v (m/s) Velocity-Time Graph Slope = acceleration Area = distance t(s) 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. Area under the curve = 3 m/s x 3 s = 9 m Thus, all we need to do is find the area of the rectangle. Distance traveled = 9 m Example-3

  10. TASK 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. 8 7 6 5 4 3 1 m/s by 5 s 1 m/s by 1 s 2 1 0 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 Answer v (m/s) Velocity-Time Graph Slope = acceleration Area = distance t(s) The area under the curve for this problem consists of two sections, a small triangle and a rectangle. Note that the area under the curve for this problem is from t = 9 s to t = 14 s. Area of triangle = (1 m/s x 1 s)/2 = 0.5 m Area of rectangle = (1 m/s x 5 s) = 5 m Distance traveled = 0.5 m + 5 m = 5.5 m Example-4

  11. PRACTICE

  12. Graphs Slide: 4.1 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 Note that the slope represents mass. Click Click Click

  13. 4.2 Graphs Slide: Reminder The area under the curve is the distance travelled. 200 m 250 m 350 m 200 m + 100 m Answer: 350 m – 200 m = 150 m Click

  14. 4.3 Graphs Slide: Illustrated below is the velocity versus time graph of a particle. NOTE: The area under the curve represents distance. 250 000 50000 200000 How far has the particle travelled in 5 minutes? A) 200000 m B) 250000 m C) 300000 m D) 350000 m Convert to seconds 5 min = 5 x 60 s = 300 s Click

  15. 4.4 Graphs Slide: Consider the position versus time graph below. Which one of the following statements best describes themotion illustrated by the above graph. 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 Click

  16. Zero acceleration 4.5 Positive acceleration Negative acceleration A) B) C) D) Graphs Slide: The velocity-time graph on theleft represents the motion of a car during a 6 s interval of time. Which of thesegraphs represents the acceleration of the car? Click

  17. 4.6 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. Graphs Slide: Click

  18. Area = velocity = 50 m/s 4.7 Graphs Slide: Velocity = 750 m/25 s = 30 m/s Velocity = 600 m/25 s = 24 m/s Velocity = 25 m/s Click

  19. 4.8 Slide: This means the second segment has a negative acceleration (slope). This means the first segment has to be a straight line (a=0). The slope represents acceleration Click

  20. Zero velocity Constant reverse velocity Constant forward velocity 4.9 Graphs Slide: Since the distance between dots is increasing, the object is accelerating. Click

  21. 4.10 Graphs Slide: Carlo Martini owns a Ferrari. Carlo performed a speed test onhis car and plotted the graph below. Note that the slope representsacceleration. Answer Knowing that Carlo’s Ferrari has a mass of 1288 kg, calculatethe net force of the car based on the above graph. Click

  22. Area to x-axis = distance Slope = Average velocity = 10 m/s 4.13 Graphs Slide: Both a car and a truck drive off at the same time and in the samedirection. 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 = VAt = (10 m/s)(30 s) = 300 m Step-3 Distance apart = 300 m – 275 m = 25 m Answer Click

  23. The End … and good luck!

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