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Q. On turning a corner, a motorist rushing at $40\, m / s$, finds a child on the road $108 \,m$ ahead. He instantly stops the engine and applies the brakes so as to stop it within $1 \,m$ of the child, what time is required to stop it?

Motion in a Straight Line

Solution:

2 as $=v^{2}-u^{2}$
$a=\frac{v^{2}-u^{2}}{2 s}=\frac{0-(40)^{2}}{2 \times 108}=\frac{-1600}{216}$
$a=-7.4 \,m / s ^{2}$
And.
$v = u + a$
$v-u=a t$
$\therefore t=\frac{v-u}{a}=\frac{0-(40)}{-7.4}=\frac{-40}{-7.4}=5.4 s$