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Understanding Acceleration: Concepts and Equations

This chapter explains the concept of acceleration and provides examples and equations to calculate average and instantaneous acceleration, as well as constant acceleration. It also covers displacement and equations of motion related to acceleration.

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Understanding Acceleration: Concepts and Equations

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  1. Chapter 4 ACCELERATION

  2. What is Acceleration? Acceleration: change in velocity over the time interval over which it occurred

  3. Average Acceleration Average acceleration is the ratio of change in velocity over change in time. v2 – v1Δv t2 – t1 Δt a = =

  4. Average Acceleration Example: The velocity of a car increases from 2.0 m/s at 1.0 s to 16 m/s at 4.5 s. What is the car’s average acceleration? 16 m/s – 2.0 m/s14 m/s 4.5 s – 1. 0 s 3.5 s = = 4 m/s2 a =

  5. Average Acceleration Example: A car goes faster and faster backwards down a long driveway. We define forward velocity as positive, so backward velocity is negative. The car’s velocity changes from -2.0 m/s to -9.0 m/s in a 2.0-s time interval. Find the acceleration.

  6. Average Acceleration Solution: velocity change, Δv = -9.0 m/s – (-2.0 m/s) = -7.0 m/s average acceleration = Δv / Δt = -7.0 m/s / 2 s = -3.5 m / s^2

  7. Average acceleration vs. Instantaneous acceleration Average acceleration is calculated over a time interval. Instantaneous acceleration occurs at a specific time.

  8. Average Acceleration on Velocity-time graph Slope of the line = average acceleration Δv Δt

  9. Instantaneous Acceleration on Velocity-time graph The slope of the tangent to the curve is the instantaneous acceleration for the object at the specified time, t. v t

  10. Constant Acceleration Acceleration that does not change in time is uniform or constant acceleration. It is represented by a straight line on a velocity-time graph.

  11. Constant Acceleration vf = vi + at Slope = a vi t

  12. Constant Acceleration Example: If a car with a velocity of 2.0 m/s at t = 0 accelerates at a rate of +4.0 m/s^2 for 2.5 s, what is its velocity at time t = 2.5 s? Using the equation of constant acceleration: vf = vi + at = +2.0 m/s + (+4.0 m/s2)(2.5 s) = 12 m/s

  13. Displacement when velocity and time are known Average velocity: v = ½ (vf + vi) Average velocity: v = d/t Combining the two equations d/t = ½ (vf + vi) d = ½ (vf + vi) t

  14. Displacement when acceleration and time are known Starting with the displacement equation: d = ½ (vf + vi) t and using the final velocity equation: vf = vi + at Combining the two equations: d = ½ (vi + at+ vi) t = ½ (2vi + at) t d = vit + ½ at2

  15. Displacement when velocity and acceleration are known Starting with d = ½ (vf + vi) t and vf = vi + at Substituting t = (vf – vi)/a into d = ½ (vf + vi) t d = ½(vf + vi) (vf – vi) /a = (vf2 – vi2) / 2a Solve for vf2: vf2= vi2 + 2ad

  16. Equations of Motion for Uniform Acceleration Equations Variables vf = vi + at vi vf a t d = ½ (vf + vi) t vi vf d t d = vit + ½ at2vi a d t vf2=vi 2 + 2ad vi vf a d

  17. Example: What is the displacement of a train as it is accelerated uniformly from +11 m/s to +33 m/s in a 20.0-s interval?

  18. Example: A car starting from rest accelerates uniformly at +6.1 m/s2 for 7.0 s. How far does the car move?

  19. Example: An airplane must reach a velocity of 71 m/s for takeoff. If the runway is 1.0 km long, what must the constant acceleration be?

  20. Example: The time the Demon Drop ride at Cedar Point, Ohio is freely falling for 1.5 s. a. What is the velocity at the end of this time? b. How far does it fall?

  21. Problems 1. Find the uniform acceleration that causes a car’s velocity to change from 32 m/s to 96 m/s in an 8.0-s period. 96 m/s – 32 m/s = 8.0 m/s^2 8.0 s

  22. Problems 2. Rocket-powered sleds are used to test the responses of humans to acceleration. Starting from rest, one sled can reach a speed of 444 m/s in 1.80 s and can be brought to a stop again in 2.15 s.

  23. Problems • Calculate the acceleration of the sled when starting and compare it to the acceleration due to gravity, 9.80 m/s2. a = 444 m/s – 0 = 247 m/s2 1.80 s 247 / 9.8 = 25.2

  24. Problems • Find the acceleration of the sled when braking and compare it to the magnitude of the acceleration due to gravity. a = 0 – 444 m/s = - 207 m/s2 2.15 s 207 m/s2 = 21.1 9.8 m/s2

  25. Problems • Find the acceleration of the moving object. 30 Velocity m/s 20 5 10 20 25 Time s

  26. Problems • During the first five seconds of travel. a = 30 m/s – 0 m/s = 6 m/s2 5 s b. Between the fifth and the tenth second of travel. a. 30 m/s – 30 m/s = 0 5 s

  27. Problems c. Between the tenth and the fifteenth second of travel. a = 20 m/s – 30 m/s = -2 m/s2 5 s d. Between the twentieth and the twenty-fifth second of travel. a = 0 – 20 m/s =-4 m/s2 5 s

  28. Problems 4. Position-time graph d d d t t t c b a

  29. Problems a. Velocity-time graphs v v v 0 0 0 t t t b c a

  30. Problems b. Acceleration-time graphs 0 a a a 0 t t t b c a

  31. Problems 5. A car with a velocity of 22 m/s is accelerated uniformly at the rate of 1.6 m/s^2 for 6.8 s. What is its final velocity? vf = vi + at = 22 m/s + (1.6 m/s2)(6.8 s) = 33 m/s

  32. Problems • The velocity of an automobile changes over an 8.0-s time period as shown. time velocity time velocity 0.0 0.0 5.0 20.0 1.0 4.0 6.0 20.0 2.0 8.0 7.0 20.0 3.0 12.0 8.0 20.0 4.0 16.0

  33. Problems a. Plot the velocity-time graph of the motion. 20 16 12 8 4 2 4 6 8

  34. Problems b. Determine the displacement of the car the first 2.0 s. (displacement is the area under the curve) d = ½ bh = ½ (2.0 s)(8.0 m/s – 0) = 8 m

  35. Problems c. What displacement does the car have during the first 4.0 s? d = ½ bh = ½ (4.0 s)(16.0 m/s – 0) = 32 m

  36. Problems d. What displacement does the car have during the entire 8.0 s? d = ½ bh + bh = ½ (5.0 s)(20.0 m/s – 0) + (8.0 s – 5.0 s)(20.0 m/s) = 110 m

  37. Problems e. Find the slope of the line between t = 0 s and t = 4.0 s. What does this slope represent? a = Δv/ Δt = (16 m/s – 0 m/s)/(4.0s – 0s) = 4 m/s2

  38. Problems f. Find the slope of the line between t = 5.0 s and t = 7.0 s. What does this slope indicates. a = Δv/ Δt = (20 m/s – 20 m/s)/(7.0s – 5.0s) = 0 constant velocity

  39. Problems 7. Figure 4-20 shows the position-time and velocity-time graphs of a karate expert using a fist to break wooden boards.

  40. Problems 7a. Use the velocity-time graph to describe the motion of the expert’s fist during the first 10 ms. The fist moves downward at about -13 m/s for about 5 ms. It then suddenly comes to a halt (accelerates).

  41. Problems 7b. Estimate the slope of the velocity-time graph to determine the acceleration of the fist when it suddenly stops. a = Δv/ Δt = (0 – (-13 m/s))/(7.5 ms – 5.0 ms)) = 5200 m/s2

  42. Problems 7c. Express the acceleration as a multiple of the gravitational acceleration, g = 9.8 m/s^2. (5200 m/s^2)/(9.8 m/s^2) = 530

  43. Problems 7d. Determine the area under the velocity-time curve to find the displacement of the fist in the first 6 ms. Compare with the position-time graph. The area is almost rectangular: (-13 m/s)(0.006s) = -8 cm. This is in agreement with the position-time graph where the hand moves from +8 cm to 0 cm, for a net displacement of – 8cm.

  44. Problems 8. A supersonic jet flying at 145 m/s is accelerated uniformly at the rate of 23.1 m/s^2 for 20 s.

  45. Problems 8a. What is its final velocity? vf = vi + at = 145 m/s + (23.1 m/s^2)(20.0 s) = 607 m/s

  46. Problems b. The speed of sound in air is 331 m/s. How many times the speed of sound is the plane’s final speed? x = (607 m/s) / (331 m/s) = 1.83 times the speed of sound

  47. Problems • Determine the final velocity of a proton that has an initial velocity of 2.35 x 10^5 m/s and then is accelerated uniformly in an electric field at the rate of -1.10 x 10^12 m/s^2 for 1.50 x 10^-7 s. vf = vi + at = 235 m/s + (-1.10 x 10^12 m/s^2 )(1.50 x 10^-7 s) = 7.0 x 10^4 m/s

  48. Problems 10. Determine the displacement of a plane that is uniformly accelerated from 66 m/s to 88 m/s in 12 s. d = (vf + vi)t / 2 = (88 m/s + 66 m/s)(12 s) / 2 = 924 m

  49. Problems 11. How far does a plane fly in 15 s while its velocity is changing from +145 m/s to +75 m/s at a uniform rate of acceleration? d = (vf + vi)t / 2 = (75 m/s + 145 m/s)(15 s) / 2 = 1650 m

  50. Problems 12. A car moves at 12 m/s and coasts up a hill with a uniform acceleration of -1.6 m/s^2.

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