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bebehoong

  • 4 years ago

Suppose that a NASCAR race car is moving to the right with a constant velocity of +82 m/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? I understand (a) but I don't understand how to solve for (b) can someone please help me?

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  1. abhinav_cherukuri
    • 4 years ago
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    average acceleration = average velocity/time = 82 / 12 = 41 / 6 = 6. 83

  2. abhinav_cherukuri
    • 4 years ago
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    is my answer correct

  3. dspieg
    • 4 years ago
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    Acceleration = dv/dt thus it is the change in velocity over the change in time or the derivative of the velocity time graph. Thus if you have a constant velocity (including both speed AND direction) then you have 0 acceleration.

  4. dspieg
    • 4 years ago
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    in part b you have to realize that velocity includes direction. Velocity is a vector. Because of this, when the car is going in the opposite direction, its velocity is now -82 m/s

  5. dspieg
    • 4 years ago
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    so, in those 12 seconds the car went from moving at +82 m/s to -82 m/s. The difference which is equivalent to the change is velocity is obviously 82*2=164 m/s

  6. dspieg
    • 4 years ago
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    so to find average acceleration we look change in velocity over change in time = 164m/s/12s = 13.67 m/s^2

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