Question

The acceleration of a particle is increasing linearly with time $t$ as $bt$. The particle starts from origin with an initial velocity $v_0$. The distance travelled by the particle in time t will be

• A. $v_0t+\frac{1}{3}bt^2$
• B. $v_0 t+\frac{1}{2}bt^2$
• C. $v_0 t+\frac{1}{6}bt^3$
• D. $v_0 t+\frac{1}{3}bt^3$

Answer 1

Answer: (C)

$v_0 t+\frac{1}{6}bt^3$

Answer 2

Acceleration $\propto bt.$ i.e., $\frac{d^2 x}{dt^2} = a\, \propto bt$
Integrating, $\frac{dx}{dt} =\frac{bt^2}{2}+C$
Initially, $t = 0, dx/dt=v_0$
Therefore, $\frac{dx}{dt} =\frac{bt^2}{2}+v_0$

Integrating again, $x = \frac{bt^3}{6}+v_0 t + C$
When $t = 0, x = 0 \Rightarrow C = 0.$
i.e., distance travelled by the particle in time t
$= v_0 t + \frac{bt^3}{6}$.