Question:
A person travelling at $43.2 \mathrm{~km} / \mathrm{h}$ applies the brake giving a deceleration of $6.0 \mathrm{~m} / \mathrm{s}^{2}$ to his scooter. How far will it travel before stopping?
Solution:
$\mathrm{u}=43.2^{\times \frac{5}{18}}=12 \mathrm{~m} / \mathrm{s} ; \mathrm{v}=0 ; \mathrm{a}=-6 \mathrm{~m} / \mathrm{s}^{2}$
Using,
$\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}$
$\mathrm{O}^{2}=(12)^{2}-2^{X^{X_{s}}} 6^{\mathrm{s}}$
$\mathrm{s}=12 \mathrm{~m}$
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