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


If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder of the 100-m dash, what will be his time for the race?


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}