Question

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 $2s$ is

• A. $20$ $m$
• B. $18$ $m$
• C. $16$ $m$
• D. $25$ $m$

Answer 1

The correct option is B $18$ $m$

Given acceleration $a=6t+5$
$\therefore a=\frac{dv}{dt}$
$=6t+5,$
$dv=(6t+5)dt$
Integrating it, we have
${\int }_0^{v}dv={\int }_0^t(6t+5)dt$
$v=3t^2+5t+C$, where $C$ is constant of integration.
When $t=0,v=0$ so $C=0$
$\therefore v=\frac{ds}{dt}$
$=3t^2+5t$
or $ds=(3t^2+5t)dt$
Integrating it within the conditions of motion, i.e., as $t$ changes from $0$ to $2s$, $s$ changes from $0$ to $s$, we have
${\int }_0^{s}ds={\int }_0^2(3t^2+5t)dt$
$\therefore s=t^3+\frac{5}{2}{t}^2$ ${|}_0^2$
$=8+10$
$=18m$