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)

$18m$

Given,

$a=(6t+5)m/s^2$

Velocity, $v=\int adt$

$v=\int (6t+5)dt=\frac{6}{2}{t}^2+5t$

$v=3t^2+5t$

Distance covered in $2sec$ is

$S={\int }_0^{2}vdt$

$S={\int }_0^2(3t^2+5t)dt$

$S=[t^3+\frac{5t^2}{2}{]}_0^2$

$S=(2\times 2\times 2\times +\frac{5\times 2\times 2}{2})-(0^3+\frac{5(0{)}^2}{2})$

$S=8+10$

$S=18m$

The correct option is B.