College Physics by Openstax| Problem 2.38

A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of 0.500 m/s^2 for 7.00 s.

(a) What is his final velocity?

(b) The racer continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save?

(c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?

SOLUTION:

Part a

v=v_o+at

v=11.5\:m/s+\left(0.500\:m/s^2\right)\left(7.00\:s\right)

v=15.0\:m/s

 

Part b

Let t_{const} be the time it takes to reach the finish line without accelerating:

t_{const}=\frac{x}{v_o}

t_{const}=\frac{300\:m}{11.5\:m/s}

t_{const}=26.09\:s

Now let be the distance traveled during the 7 seconds of acceleration. We know t=7.00 s so

d=v_ot+\frac{1}{2}at^2

d=\left(11.5\:m/s\right)\left(7.00\:s\right)+\frac{1}{2}\left(0.500\:m/s^2\right)\left(7.00\:s\right)^2

d=92.75\:m

Let t’ be the time it will take the rider at the constant final velocity to complete the race:

t'=\frac{x-d}{v}

t'=\frac{300\:m-92.75\:m}{15.0\:m/s}

t'=13.82\:s

So the total time it will take the accelerating rider to reach the finish line is 

T=t+t'

T=\:7\:s+13.82\:s

T=20.82\:s

Finally, let T* be the time saved. So 

T*=26.09\:s-20.82\:s

T*=5.27\:s

 

Part c

Let t_2 be the time it takes for rider 2 to reach the finish line.

t_2=\frac{x'}{v_o'}

t_2=\frac{295\:m}{11.8\:m/s}

t_2=25.0\:s

The time difference is

=t_2-T

=25.0\:s-20.817\:s

=4.2\:s

Therefore, he finishes 4.2 s after the winner.

When the other racer reaches the finish line, the winner has been traveling at 15 m/s for 4.2 seconds, so the other racer finishes

x=\left(4.2\:s\right)\left(15\:m/s\right)

x=63\:m

behind the other racer.

 

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