# College Physics 2.38 – A bicycle racer sprints at the end of a race to clinch a victory

## 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 tconst be the time it takes to reach the finish line without accelerating:

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

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

$t_{const}=26.09\:s$

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:

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

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

$t'=13.82\:s$

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 t2 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.