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Q. A point initially at rest moves along $x$-axis. Its acceleration varies with time as $ a=(61+5) $ in $m/s$. If it starts from origin, the distance covered in $2\,s$ is

Uttarkhand PMTUttarkhand PMT 2011

Solution:

Given, $a=\frac{d v}{d t}=6 t+5$
$\int\limits_{0}^{v} d v=\int\limits_{0}^{t}(6 t+5) d t$
$v=\frac{6 t^{2}}{2}+5 t $
$\frac{d s}{d t}=\left(\frac{6 t^{2}}{2}+5 t\right) d t$
$s=\frac{3 t^{2}}{3}+\frac{5 t^{2}}{2}$
where, $t=2 \,s$,
$S=3 \times \frac{2^{3}}{3}+\frac{5 \times 2^{2}}{2}=18 \,m$