Question

The acceleration 'a' in $m/s^2$ of a particle is given by a $a=3t^2+2t+2$ where t is the time. If the particle starts out with a velocity u = $2m/s$ at $t=0$, then the velocity at the end of 2 second is

• A. $12m/s$
• B. $18m/s$
• C. $27m/s$
• D. $36m/s$

Answer 1

The correct option is B $18m/s$

We have, $a=\frac{dV}{dt}$

Given $a=3t^2+2t+2$

$dV=a\times dt$

$dV=(3t^2+2t+2)dt$

Integrating bothe sides, we get

$V=\frac{3\times t^3}{3}+\frac{2t^2}{2}+t=t^3+t^2+2t+C$ .....(1)

where $C$ is constant

Putting $t=0$ in equation (1), we get

$2=0+C$

Then,

$C=2$

Substituting the value in equation (2), we get

$V=t^3+t^2+2t+2$

At $t=2$

$V=8+4+4+2=18m/s$