The acceleration a in $m/s^2$ of a particle is given by $a = 3t^2 + 2t + 2$ where t is the time. If the particle starts out with a velocity $u = 2 m/s$ at $t = 0$, then the velocity at the end of 2 second is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 18 m/s
$\frac{dv}{dt}=3t^2 + 2t + 2$
or $\int_2^vdv=\int_0^2(3t^2 + 2t + 2)dt$
$v-2=\left|3\frac{t^3}{3}+2\frac{t^2}{2}+2t\right|_0^2$
$∴ v = 18 m/s$