# College Physics 4.2 – The time for the race of a sprinter subjected to a force

## Solution:

For the first segment of the sprint, we are given $\text{m}=63.0\:\text{kg}$, $\text{a}=4.20\:\text{m/s}^2$, and $\text{v}_o=0\:\text{m/s}$, The final velocity for this segment is

$\text{v}^2=\left(\text{v}_o\right)^2+2\text{ax}$

$\text{v}^2=0+2\left(4.20\:\text{m/s}^2\right)\left(20\:\text{m}\right)$

$\text{v}=\sqrt{168}$

$\text{v}=12.961\:\text{m/s}$

Given the final velocity after running for 20 meters, we can find the time it took him to do so.

$\text{t}=\frac{\text{v}-\text{v}_o}{\text{a}}$

$\text{t}=\frac{12.961\:\text{m/s}-0\:\text{m/s}}{4.20\:\text{m/s}^2}$

$\text{t}=3.086\:\text{s}$

For the second segment of the sprint, we know that $\text{a}=0\:\text{m/s}^2;\:\text{v}=12.961\:\text{m/s},\:\text{x}=80\:\text{m}$. We are going to solve for the time the sprinter took to run this segment.

$\text{t}=\frac{\text{d}}{\text{v}}$

$\text{t}=\frac{80\:\text{m}}{12.961\:\text{m/s}}$

$\text{t}=6.172\:\text{s}$

Therefore, the total time it took the sprinter to run the whole 100-m dash is

$\text{Total time}=3.086\:\text{s}+6.172\:\text{s}$

$\text{Total time}=9.258\:\text{s}$