Question

If $a=(3t^2+2t+1)m/s^2$ is the expression according to which the acceleration of a particle moving along a straight line varies, then the change in velocity after 3 seconds of its start is

• A. $30m/s$
• B. $39m/s$
• C. $3m/s$
• D. $20m/s$

Answer 1

The correct option is B $39m/s$

$a=(3t^2+2t+1)$
${\int }_0^{v}dv={\int }_0^t(3t^2+2t+1dt)(\because a=\frac{dv}{dt})$
$v=t^3+t^2+t$
$v(t=0)=0$
$v(t=3s)=(3{)}^3+(3{)}^2+3$
$=27+9+3$
$=39m/s$
$\Delta v=39-0=39m/s$