Question

An 84.0 kg sprinter starts a race with an acceleration of 1.76 m/s2. If the sprinter accelerates at that rate for 11 m, and then maintains that velocity for the remainder of the 100 m dash, what will be his time (in s) for the race?

Answer 1

Answer:

t=17.838s

Explanation:

The displacement is divided in two sections, the first is a section with constant acceleration, and the second one with constant velocity. Let's consider the first:

The acceleration is, by definition:

$a=\frac{dv}{dt}=1.76$

So, the velocity can be obtained by integrating this expression:

$v=1.76t$

The velocity is, by definition: $v=\frac{dx}{dt}$, so

$dx=1.76tdt\\x=1.76\frac{t^{2}}{2}$.

Do x=11 in order to find the time spent.

$11=1.76\frac{t^2}{2}\\ t^2=\frac{2*11}{76} \\t=\sqrt{12.5}=3.5355s$

At this time the velocity is: $v=1.76t=1.76*3.5355s=6.2225\frac{m}{s}$

This velocity remains constant in the section 2, so for that section the movement equation is:

$x=v*t\\t=\frac{x}{v}$

The left distance is 89 meters, and the velocity is $6.2225\frac{m}{s}$, so:

$t=\frac{89}{6.2225}=14.303s$

So, the total time is 14.303+3.5355s=17.838s