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 $u = 2m/s$ at $t = 0,$ then the velocity at the end of 2 $seconds$ is

Question

Dr. Abubacker · Accepted Answer

(b) $v = u + \int_{}^{}{adt = u + \int_{}^{}{(3t^{2} + 2t + 2)dt}}$
$= u + \frac{3t^{3}}{3} + \frac{2t^{2}}{2} + 2t = u + t^{3} + t^{2} + 2t$
$= 2 + 8 + 4 + 4 = 18\ m/s$ (As $t = 2\ sec$)

Q. 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 $u = 2m/s$ at $t = 0,$ then the velocity at the end of 2 $seconds$ is

Explanation

Reviewed by Dr. Abubacker

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