A bicycle racer sprints near the end of a race to clinch a victory. The racer ha

Question

A bicycle racer sprints near the end of a race to clinch a victory. The racer has an initial velocity of 10.5 m/s and accelerates at the rate of 0.510 m/s2 for 6.09 s. What is his final velocity?

The racer continues at this velocity to the finish line. If he was 239 m from the finish line when he started to accelerate, how much time did he save by accelerating?

One other racer was 4.10 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at a constant speed of 12.1 m/s until the finish line. How far behind (in meters) is he from the winner when the winner crosses the finish line?

How many more seconds did the other racer takes to cross the finish line?

Explanation / Answer

final velocity vi = vo + a*t = 10.5+(0.5*6.09) = 13.545 m/s <<<------answer

++++++++++

distance x = 239 m

if the racer moves with constant velocity

time taken t1 = x/v1 = 239/13.5 = 17.7 s

if the racer accelerates

x = v1*t2 + 0.5*a*t2^2

239 = 13.55*t2+(0.5*0.51*t2^2)

t2 = 14.0 s

time saves = 17.7-14 = 3.7 s <<<------answer

+++++++++++

other racer

position of the other racer from the finish line = 239-4.1 = 234.9 m

distance by the other racer in 14 s = 12.1*14 = 169.4 m

racer is 234.9-169.4 = 65.5 m far behind the winner <<<------answer

extra time = 65.5/12.1 = 5.41 s <<<------answer

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