Q. The acceleration $a$ of a particle starting from rest varies with time according to relation $a = \alpha t + \beta$. The velocity of the particle after a time $t$ will be
A) $\frac{\alpha t^{2}}{2} + \beta$
B) $\frac{\alpha t^{2}}{2} + \beta t$
C) $\alpha t^{2} + \frac{1}{2}\beta t$
D) $\frac{(\alpha t^{2} + \beta t)}{2}$
Explanation
(b) According to given relation acceleration $a = \alpha t + \beta$
As $a = \frac{dv}{dt} \Rightarrow \alpha t + \beta = \frac{dv}{dt}$
Since particle starts from rest, its initial velocity is zero
$i.e.,$ At time $t = 0$, velocity $= 0$
$\Rightarrow \int_{0}^{v}{dv} = \int_{0}^{t}{(\alpha t + \beta)dt} \Rightarrow v = \frac{\alpha t^{2}}{2} + \beta t$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.