Question:
An athlete takes $2.0 \mathrm{~s}$ to reach his maximum speed of $18.0 \mathrm{~km} / \mathrm{h}$. What is the magnitude of his average acceleration?
Solution:
$u=0 ; v=18^{\times \frac{5}{18}}=5 \mathrm{~m} / \mathrm{s} ; \mathrm{t}=2 \mathrm{sec}$
$v=u+a t$
$5=0+a(2)$
$a=2.5 \mathrm{~m} / \mathrm{s}^{2}$