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Q.
A point initially at rest moves along x-axis. Its acceleration varies with time as $a = (6t+ 5) m/s^{2}$. If it starts from origin, the distance covered in $2\, s$ is
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Solution:
Given acceleration $a = 6t + 5$
$\therefore a=\frac{dv}{dt}=6t+5$,
$dv=(6t+5)dt$
Integrating it, we have $\int_{0}^{v}dv=\int_{0}^{t}\left(6t+5\right)dt $
$v=3t^{2}+5t+C$, where C is constant of integration
When $ t=0, v=0$ so $C=0$
$\therefore v=\frac{ds}{dt}=3t^{2}+5t$
or, $ ds=\left(3t^{2}+5t\right)dt$
Integrating it within the conditions of motion, i.e.as t changes from 0 to 2 s, s changes from 0 to s,we have
$\int\limits_{0}^{s}ds=\int\limits_{0}^{2} \left(3t^{2}+5t\right)dt$
$\therefore s=t^{3}+\frac{5}{2}t^{2} |_{0}^{2}
=8+10=18\,m$