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

Answer 1

Answer: (B)

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
$\underset{0}{\overset{s}{\int }}ds=\underset{0}{\overset{2}{\int }}\left(3t^2+5t\right)dt$
$\therefore s=t^3+\frac{5}{2}{t}^2{|}_0^2<br>=8+10=18m$