# Net External Force on a Sprinter| Newton’s Second Law of Motion: Concept of a System| Dynamics| College Physics| Problem 4.2

### 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 $m=63.0\:kg$, $a=4.20\:m/s^2$, and $v_o=0\:m/s$, The final velocity for this segment is $v^2=\left(v_o\right)^2+2ax$ $v^2=0+2\left(4.20\:m/s^2\right)\left(20\:m\right)$ $v=\sqrt{168}$ $v=12.961\:m/s$

Given the final velocity after running for 20 meters, we can find the time it took him to do so. $t=\frac{v-v_o}{a}$ $t=\frac{12.961\:m/s-0\:m/s}{4.20\:m/s^2}$ $t=3.086\:s$

For the second segment of the sprint, we know that $a=0\:m/s^2;\:v=12.961\:m/s,\:x=80\:m$. We are going to solve for the time the sprinter took to run this segment. $t=\frac{d}{v}$ $t=\frac{80\:m}{12.961\:m/s}$ $t=6.172\:s$

Therefore, the total time it took the sprinter to run the whole 100-m dash is $Total\:time=3.086\:s+6.172\:s$ $Total\:time=9.258\:s$