Q. The acceleration of a particle increasing linearly with time $t$ is $bt$. The particle starts from the origin with an initial velocity $v_{0}$. The distance travelled by the particle in time $t$ will be
A) $v_{0}t + \frac{1}{6}bt^{3}$
B) $v_{0}t + \frac{1}{3}bt^{2}$
C) $v_{0}t + \frac{1}{3}bt^{3}$
D) $v_{0}t + \frac{1}{3}bt^{2}$
Explanation
(a) $\frac{dv}{dt} = bt$ or $dv = bt\ dt$
$\int_{v_{0}}^{v}{dv = \int_{0}^{t}{bt\ dt}\ \ }$or $v - v_{0} = \frac{bt^{2}}{2}$
or $v = v_{0} = + \frac{bt^{2}}{2}$
or $dx = v_{0}dt + \frac{bt^{2}}{2}dt$
$\int_{0}^{x}{dx} = \int_{0}^{t}{v_{0}dt} + \frac{b}{2}\int_{0}^{t}{t^{2}dt}$
or $x = v_{0}t + \frac{1}{2}\frac{bt^{3}}{3} = v_{0}t + \frac{bt^{3}}{6}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.