Question

The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity ${v_0}$. What will be the distance travelled by the particle in time t?
${\text{A}}{\text{. }}{v_0}t + \dfrac{1}{3}b{t^2} \\\ {\text{B}}{\text{. }}{v_0}t + \dfrac{1}{3}b{t^3} \\\ {\text{C}}{\text{. }}{v_0}t + \dfrac{1}{6}b{t^3} \\\ {\text{D}}{\text{. }}{v_0}t + \dfrac{1}{2}b{t^2} \\\$

Answer 1

- Hint- Here, we will proceed by using the formula i.e., Acceleration a = $\dfrac{{{\text{Change in velocity}}}}{{{\text{Change in time }}}} = \dfrac{{dv}}{{dt}}$ and then integrating this. Then, we will apply the formula i.e., Velocity v = $\dfrac{{{\text{Change in displacement(or distance)}}}}{{{\text{Change in time }}}} = \dfrac{{dx}}{{dt}}$ and integrate it.

Complete step-by-step solution -

Let us suppose the acceleration of the particle be denoted by variable a
Given that the acceleration of a particle is increasing linearly with time as bt
i.e., a = bt
Initial velocity = ${v_0}$
Let us assume the velocity of the particle at any time t be $v$
As we know that the acceleration is equal to the rate of change of velocity with time
i.e., Acceleration a = $\dfrac{{{\text{Change in velocity}}}}{{{\text{Change in time }}}} = \dfrac{{dv}}{{dt}}$
$\Rightarrow \dfrac{{dv}}{{dt}} = bt \\\ \Rightarrow dv = btdt \\\$
Integrating the above equation on both sides, we get

$$\Rightarrow \int_{{v_0}}^v {dv} = \int_0^t {btdt} \\\ \Rightarrow \left[ v \right]_{{v_0}}^v = b\left[ {\dfrac{{{t^2}}}{2}} \right]_0^t \\\ \Rightarrow \left[ {v - {v_0}} \right] = \dfrac{b}{2}\left( {{t^2} - {0^2}} \right) \\\ \Rightarrow v - {v_0} = \dfrac{{b{t^2}}}{2} \\\ \Rightarrow v = \dfrac{{b{t^2}}}{2} + {v_0}{\text{ }} \to {\text{(1)}} \\\$$

Let us assume the total distance travelled by the particle in time t be x
It is given that the particle starts from the origin i.e., Initial distance = 0
Also, we know that velocity is equal to the rate of change of displacement
i.e., Velocity v = $\dfrac{{{\text{Change in displacement(or distance)}}}}{{{\text{Change in time }}}} = \dfrac{{dx}}{{dt}}$
Using the above formula in equation (1), we have

$$\Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{{b{t^2}}}{2} + {v_0} \\\ \Rightarrow dx = \left( {\dfrac{{b{t^2}}}{2} + {v_0}} \right)dt \\\$$

By integrating the above equation on both sides, we get

$$\Rightarrow \int_0^x {dx} = \int_0^t {\left( {\dfrac{{b{t^2}}}{2} + {v_0}} \right)dt} \\\ \Rightarrow \left[ x \right]_0^x = \int_0^t {\left( {\dfrac{{b{t^2}}}{2}} \right)dt} + \int_0^t {{v_0}dt} \\\ \Rightarrow \left[ {x - 0} \right] = \dfrac{b}{2}\int_0^t {{t^2}dt} + {v_0}\int_0^t {dt} \\\ \Rightarrow x = \dfrac{b}{2}\left[ {\dfrac{{{t^3}}}{3}} \right]_0^t + {v_0}\left[ t \right]_0^t \\\ \Rightarrow x = \dfrac{b}{{2 \times 3}}\left[ {{t^3}} \right]_0^t + {v_0}\left( {t - 0} \right) \\\ \Rightarrow x = \dfrac{b}{6}\left( {{t^3} - 0} \right) + {v_0}t \\\ \Rightarrow x = \dfrac{{b{t^3}}}{6} + {v_0}t \\\$$

Therefore, the total distance travelled by the particle in time t is ${v_0}t + \dfrac{1}{6}b{t^3}$.
Hence, option C is correct.

Note- In this particular problem, definite integration (integration with limits) is used in order to avoid the occurrence of integration constant. The upper and lower limits for the velocity of the particle are v (any general velocity at time t) and ${v_0}$ respectively and those for the distance are x and 0 respectively.