Q. A body starts from rest, with uniform acceleration $a$. The acceleration of a body as function of time $t$ is given by the equation $a = pt$ where $p$ is constant, then the displacement of the particle in the time interval $t = 0$ to $t = t_{1}$ will be
A) $\frac{1}{2}pt_{1}^{3}$
B) $\frac{1}{3}pt_{1}^{2}$
C) $\frac{1}{4}pt_{1}^{2}$
D) $\frac{1}{6}pt_{1}^{3}$
Explanation
(a) Given, $\frac{dv}{dt} = pt$
$dv = pt\ dt$
$\int_{0}^{v}{dv = P\int_{0}^{t_{1}}{t\ dt}}$
$v = \frac{1}{2}pt_{1}^{2}$
Again, $x = \frac{1}{2}p\int_{0}^{t_{1}}{t_{1}^{2}dt = \frac{1}{2}pt_{1}^{3}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.