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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48 Problem 49 Problem 50 Problem 51 Problem 52 Problem 53 Problem 54 Problem 55 Problem 56 Problem 57 Problem 58 Problem 59 Problem 60 Problem 61 Problem 62 Problem 63 Problem 64 Problem 65 Problem 66 Problem 67 Problem 68 Problem 69 Problem 70 Problem 71 Problem 72 Problem 73 Problem 74 Problem 75 Problem 76 Problem 77 Problem 78 Problem 79 Problem 80 Problem 81 Problem 82 Problem 83 Problem 84 Problem 85 Problem 86 Problem 87 Problem 88

Problem 15 Easy Difficulty

(a) Suppose that a NASCAR race car is moving to the right with a constant velocity of $+82 \mathrm{m} / \mathrm{s}$ . What is the average acceleration of the car? (b) Twelve seconds later, the car is halfway around the track and traveling in the opposite direction with the same speed. What is the average acceleration of the car?

Answer

(a) $a=0$
(b) $a=13.67 \mathrm{m} \cdot \mathrm{s}^{-2}$

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Physics 101 Mechanics

Physics

Chapter 2

Kinematics in One Dimension

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Video Transcript

in this problem, a NASCAR race car has a velocity of 82 meters per second at a track. The question asked us to find the average acceleration off the car at this point when it has this constant velocity of 82 meters per second, will by definition accelerate acceleration is the change in velocity over time. So if we have a car going at 82 meters per second and it's not changing its speed by fall to the velocity, change is syrup. So regardless of how long time you never will looking at as long as that velocity is constant, the acceleration is zero. The velocity is not changing, so this is the answer to part a. Part B says that at a later point, the cars traveling on the opposite side of the track with the same velocity, so moving in the opposite direction, so a label. This philosophy here also is 82 meters per second. But because it's going in the opposite direction, it's a negative value. Velocity is a vector. So if we use the equation from Kinnah, Matics B is equal to re not plus 80 which is just the definition of acceleration rearranged. I have negative 82 meters per second equals positive 82 meters per second plus acceleration times 12 seconds. So we're looking for the average acceleration in that time period. So, rearranging the equation, we can subtract the 82 from both sides. We get negative 164 meters per second equals a times 12 seconds. Man dividing both sides By 12 we get negative 13.67 meters per second squared second square is equal to the acceleration and that's the answer. Rounded to two significant figures which were given in the problem. It would be negative 14 meters second squared.

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