Question

A particle located at \[x=0\]at time \[t=0\] starts moving along x – direction with velocity u that varies as \[v=a\sqrt{x}\]. Find the displacement of the particle varies with time.
(a) \[\dfrac{{{a}^{2}}t}{4}\]
(b) \[\dfrac{{{a}^{2}}{{t}^{2}}}{4}\]
(c) \[\dfrac{{{a}^{2}}{{t}^{2}}}{2}\]
(d) \[\dfrac{a{{t}^{2}}}{4}\]

Answer 1

Hint:We use the relation, velocity is equal to rate of change of displacement, that is,
\[v=\dfrac{dx}{dt}\]
Where v implies the velocity of the particle, dx denotes the change in displacement and dt denotes the change on time. Equating with the relation of velocity with displacement given in the question, we can find displacement with respect to time.

Complete answer:
As we know that a particle is located at \[x=0\]at time \[t=0\]. Also, the variation of velocity u with displacement is given by \[v=a\sqrt{x}\]….(1)
Now, velocity,\[v=\dfrac{dx}{dt}\] …… (2)
On comparing the equation (1) to (2), we get,
\[\begin{align}
  & \dfrac{dx}{dt}=a\sqrt{x} \\
 & \Rightarrow \dfrac{dx}{\sqrt{x}}=adt \\
 & \Rightarrow {{x}^{-\dfrac{1}{2}}}dx=adt \\
\end{align}\]
Integrating on both sides, we get
\[\begin{align}
  & \Rightarrow \dfrac{{{x}^{-\dfrac{1}{2}+1}}}{-\dfrac{1}{2}+1}=at+c \\
 & \Rightarrow 2\sqrt{x}=at+c....(3) \\
\end{align}\]
Now, at \[t=0\] ,\[x=0\]; so equation (3) becomes
\[\begin{align}
  & \Rightarrow 2\times 0=a\times 0+c \\
 & \Rightarrow c=0 \\
\end{align}\]
Therefore, \[2\sqrt{x}=at\]
Squaring on both sides, we get
\[\begin{align}
  & \Rightarrow {{\left( 2\sqrt{x} \right)}^{2}}={{\left( at \right)}^{2}} \\
 & \Rightarrow x=\dfrac{{{a}^{2}}{{t}^{2}}}{4} \\
\end{align}\]
Therefore, the displacement of the particle varies with time as \[\dfrac{{{a}^{2}}{{t}^{2}}}{4}\]

Hence, option B is the correct answer.

Note:
From the relation obtained, we can conclude that displacement is proportional to the square of time.
In addition, if we plot a time-displacement graph, its slope gives the velocity of the particle.