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