Answer to Question #135205 in Mechanics | Relativity for Miles J

Question

Hello. Hope all is well. For Question #91240, I understand most of the solution except the final time calculation. Would you please explain why (t1) is multiplied by 2, when (t2) is not? I feel like the answer should be 9.3 s instead of 10.8 s.

Answer 1

Question: A 63.0kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him?

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

Solution:

Total time:

$t=t_1+t_2$

$t_1$ is a time of acceleration.

$S_1=\frac{at_1^2}{2}$ where $t_1=\sqrt{\frac{2S_1}{a}}$

$t_2$ is at time of constant motion with velocity $v=\sqrt{2aS_1}$

$t_2=\frac{S_2}{v}=\frac{S_2}{\sqrt2aS_1}$

$t=\sqrt{\frac{2S_1}{a}}+\frac{S_2}{\sqrt2aS_1}$

$t=\sqrt{\frac{2*20}{4.2}}+\frac{80}{\sqrt{2*20*4.2}}=3.09+6.17=9.26$ s

So, you are alright. Your answer is true. It's very good, that you are want to deal with physics. Good luck!