His average acceleration during the last four quarter-miles of the race was +0.0

Question

His average acceleration during the last four quarter-miles of the race was +0.0105 m/s2. His velocity at the start of the final mile (x = +1609 m) was about 16.58 m/s. The acceleration, although small, was very important to his victory. To assess its effect, determine the difference between the time he would have taken to run the final mile at a constant velocity of +16.58 m/s and the time he actually took. Although the track is oval in shape, assume it is straight for the purpose of this problem.

Explanation / Answer

at constant velocity t= 1609/16.58 =97 s actually 1609=16.58t+0.5x0.0105t^2 t=94.233 s

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