Question

A point initially at run moves along x-axis. Its acceleration varies with time $a=(6t+5)$ in m/s. If its starts from origin, the distance covered in $2s$ is :

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

Answer 1

Answer: (B)

$18m$

Acceleration of point

$a=\frac{dv}{dt}=6t+5$

${\int }_0^{v}dv={\int }_0^t(6t+5)dt$

$v=\frac{6t^2}{2}+5t$

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

$\Rightarrow ds=(3t^2+5t)dt$

$s=\frac{3t^2}{3}+\frac{5t^2}{2}$

t = 2s

$s=3\times \frac{2^2}{3}+\frac{5\times 2^2}{2}=18m$