Question

A world-class sprinter can reach a top speed (of about 11.8 m/s ) in the first 19.0 m of a race. what is the average acceleration of this sprinter

Answer 1

We use the kinematic equation of motion, to calculate the average acceleration

$v^{2} =u^{2} +2a (h-h_{0} )$

or          $a=\frac{v^{2} -v_{0}^2}{2(h-h_{0}) }$

Here, $v_{0}$ is initial speed , v is final speed  h is covered distance $h_{0}$ is initial position.

Given  $v= 11.8 m/s$ ,$v_{0} =0$ ,$h = 19.0 m$ and $h_{0} =0$.

Substituting these values in above equation we get,

$a=\frac{(11.8)^2-0}{2(19.0-0)} \\\\\ a =3.66 m/s^2$

Thus, the average acceleration of the sprinter is $3.66 m/s^2$.