Answer to Question #88097 in Mechanics | Relativity for angela pang

Question

Starting from rest,a sprinter reaches his top velocity in 3 seconds.He runs a distance of 24m in the 3 secinds.What is his acceleration?

Answer 1

We can use the formula for distance assuming constant acceleration:

$s=0.5at^2$

Thus,

$a=\frac{2s}{t^2}=\frac{2(24)}{3^2}=5.3 \frac{m}{s^2}$