Question: The acceleration $'a'$ in $m/s^{2}$ of a particle is given by $a = 3t^{2} + 2t + 2$ where \(t\...

Question

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 second is

Answer 1

Solution

$v = u + \int_{}^{}{a ⥂ dt = u + \int_{}^{}{(3t^{2} + 2t + 2)dt}}$

$= u + \frac{3t^{3}}{3} + \frac{2t^{2}}{2} + 2t = u + t^{3} + t^{2} + 2t$

$= 2 + 8 + 4 + 4 = 18\mspace{6mu} m/s$ (As t = 2 sec)

Question: The acceleration \('a'\) in \(m/s^{2}\) of a particle is given by \(a = 3t^{2} + 2t + 2\) where \(t\...

Solution