Question

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
\[\begin{align}
 & A){{v}_{0}}t+\dfrac{b{{t}^{3}}}{6} \\
 & B){{v}_{0}}t+\dfrac{b{{t}^{2}}}{3} \\
 & C){{v}_{0}}t+\dfrac{b{{t}^{3}}}{3} \\
 & D){{v}_{0}}t+\dfrac{b{{t}^{2}}}{2} \\
\end{align}\]

Answer 1

Hint: We will use the formula for acceleration which is given as rate of change of velocity and relate it with the given acceleration to find the change in velocity. Then we will integrate this change in velocity which will give us the velocity at that particular time. Then we will find the distance travelled by using the relation rate of change displacement gives velocity.

Formula used:
\[\begin{align}
  & a=\dfrac{\delta v}{\delta t} \\
 & v=\dfrac{\delta x}{\delta t} \\
\end{align}\]

Complete step-by-step answer:
Let us take the acceleration of the particle as \[a\]. Given acceleration of the particle as \[bt\]. That is \[a=bt\].
Let us assume the velocity of a particle at any time \[t\] be \[v\]. As we know the relation between acceleration and velocity,
      \[a=\dfrac{\delta v}{\delta t}\]
       \[\begin{align}
  & \Rightarrow \dfrac{\delta v}{\delta t}=bt \\
 & \Rightarrow \delta v=bt\delta t \\
\end{align}\]
Integrating this equation on both sides with velocity limit \[{{v}_{0}}\] to \[v\] and time zero to \[t\], we will get,
     \[\int\limits_{{{v}_{0}}}^{v}{\delta v}=\int\limits_{0}^{t}{bt\delta t}\]
       \[\begin{align}
  & \Rightarrow \left[ \delta v \right]_{{{v}_{0}}}^{v}=b\left[ \dfrac{{{t}^{2}}}{2} \right]_{0}^{t} \\
 & \Rightarrow \left[ v-{{v}_{0}} \right]=\dfrac{b}{2}\left( {{t}^{2}}-{{0}^{2}} \right) \\
 & \Rightarrow v-{{v}_{0}}=\dfrac{b{{t}^{2}}}{2} \\
\end{align}\]
       \[\Rightarrow v=\dfrac{b{{t}^{2}}}{2}+{{v}_{0}}\] --- (1)
Now, let us assume that total distance travelled by the particle in time \[t\] be \[x\]. The initial distance is given as zero and also we know the velocity is related to displacement as \[v=\dfrac{\delta x}{\delta t}\].
Using this formula in equation (1), we have
    \[\begin{align}
  & \Rightarrow \dfrac{\delta x}{\delta t}=\dfrac{b{{t}^{2}}}{2}+{{v}_{0}} \\
 & \Rightarrow \delta x=\left( \dfrac{b{{t}^{2}}}{2}+{{v}_{0}} \right)\delta t \\
\end{align}\]
By integrating this equation on both sides with displacement limit \[0\] to \[x\] and time zero to \[t\],
   \[\begin{align}
  & \Rightarrow \int\limits_{0}^{x}{\delta x}=\int\limits_{0}^{t}{\left( \dfrac{b{{t}^{2}}}{2}+{{v}_{0}} \right)\delta t} \\
 & \Rightarrow \left[ x \right]_{0}^{x}=\int\limits_{0}^{t}{\left( \dfrac{b{{t}^{2}}}{2} \right)\delta t}+\int\limits_{0}^{t}{{{v}_{0}}\delta t} \\
 & \Rightarrow \left[ x-0 \right]=\dfrac{b}{2}\int\limits_{0}^{t}{{{t}^{2}}}\delta t+{{v}_{0}}\int\limits_{0}^{t}{\delta t} \\
 & \Rightarrow x=\dfrac{b}{2}\left[ \dfrac{{{t}^{3}}}{3} \right]_{0}^{t}+{{v}_{0}}\left[ t \right]_{0}^{t} \\
 & \Rightarrow x=\dfrac{b}{2\times 3}\left[ {{t}^{3}} \right]_{0}^{t}+{{v}_{0}}\left[ t-0 \right] \\
 & \Rightarrow x=\dfrac{b}{6}\left( {{t}^{3}}-0 \right)+{{v}_{0}}t \\
 & \Rightarrow x=\dfrac{b{{t}^{3}}}{6}+{{v}_{0}}t \\
\end{align}\]
Therefore, the total distance travelled by the particle in time \[t\] is \[{{v}_{0}}t+\dfrac{b{{t}^{3}}}{6}\].

So, the correct answer is “Option A”.

Note: In this problem, we used definite integration in order to avoid the occurrence of integration constant. The upper and lower limits for the velocity of the particle are taken as \[v\] and \[{{v}_{0}}\]respectively and those fore distances are \[x\] and \[0\]respectively. So we must be very careful to use the limits as per given in the question.