Question

A truck is moving at a speed of $72\, kmh ^{-1}$ on a straight road. The driver can produce deceleration of $2\, ms ^{-2}$ by applying brakes. The stopping distance of truck is $13 x m$, if the reaction time of the driver is $0.2\, s$. The value of $x$ is_____.

Answer 1

$\because v^{2}=u^{2}+2 a s$
$\Rightarrow 0^{2}=\left(72 \times \frac{5}{18}\right)^{2}-2 \times 2 s_{0}$
$\therefore s_{0}=\frac{400}{4}=100\, m$
Thus, stopping distance
$=\left(72 \times \frac{5}{18}\right) \times 0.2+s_{0}$
$=4+100=104=13 x$
$\therefore x=\frac{104}{13}=8$