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(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?

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

Physics 101 Mechanics

Chapter 2

Kinematics in One Dimension

Motion Along a Straight Line

University of Michigan - Ann Arbor

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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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