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Solve the following :

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:

$u=0 ; v=18^{\times \frac{5}{18}}=5 \mathrm{~m} / \mathrm{s} ; \mathrm{t}=2 \mathrm{sec}$

$v=u+a t$

$5=0+a(2)$

$a=2.5 \mathrm{~m} / \mathrm{s}^{2}$

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