Question

A particle located at \[x = 0\] at time \[t = 0\] , starts moving along the positive $x - $ direction with a velocity $v$ that varies as $v = a\sqrt x $ .Where $a$ is a constant. Then the displacement of the particle varies with time as
\[
  A.{\text{ }}{t^3} \\
  B.{\text{ }}{t^2} \\
  C.{\text{ }}t \\
  D.{\text{ }}{t^{\dfrac{1}{2}}} \\
 \]

Answer 1

- Hint: In order to find the relation between time and displacement, first we will use the definition of velocity, the obtained equation will be in time and displacement, now we will proceed further by integrating that equation according to the given limits.

Complete step-by-step solution -

Given velocity equation is $v = a\sqrt x $
As we know that the velocity is the time rate of change of displacement.
\[v = \dfrac{{dx}}{{dt}}\]
Substitute value of $v$ in given equation, so we have
\[ \Rightarrow \dfrac{{dx}}{{dt}} = a\sqrt x \]
By cross multiplying the terms, we obtain
\[ \Rightarrow \dfrac{{dx}}{{\sqrt x }} = \left( a \right)dt\]
Now, we will integrate both the sides, we will get
\[ \Rightarrow \int\limits_{x = {x_1}}^x {\dfrac{{dx}}{{\sqrt x }}} = \int\limits_{t = {t_1}}^t {\left( a \right)dt} \]
Now, let us integrate both sides by using the formula for integration.
\[
   \Rightarrow \int\limits_{x = {x_1}}^x {{x^{\dfrac{{ - 1}}{2}}}dx} = a\int\limits_{t = {t_1}}^t {1.dt} \\
   \Rightarrow \left[ {\dfrac{{{x^{\left( {\dfrac{{ - 1}}{2}} \right) + 1}}}}{{\left( {\dfrac{{ - 1}}{2}} \right) + 1}}} \right]_{x = {x_1}}^x = \left[ {a \cdot t} \right]_{t = {t_1}}^t{\text{ }}\left[ {\because \int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}} \right] \\
   \Rightarrow \left[ {\dfrac{{{x^{\dfrac{1}{2}}}}}{{\left( {\dfrac{1}{2}} \right)}}} \right]_{x = {x_1}}^x = \left[ {at} \right]_{t = {t_1}}^t \\
   \Rightarrow \left[ {2{x^{\dfrac{1}{2}}}} \right]_{x = {x_1}}^x = \left[ {at} \right]_{t = {t_1}}^t \\
 \]
Given that at particle located at \[x = 0\] at time \[t = 0\]
Let us now substitute the initial value given in the equation.
\[
   \Rightarrow \left[ {2{x^{\dfrac{1}{2}}}} \right]_0^x = \left[ {at} \right]_0^t \\
   \Rightarrow \left[ {2{x^{\dfrac{1}{2}}} - 2 \times {0^{\dfrac{1}{2}}}} \right] = \left[ {at - a \times 0} \right] \\
   \Rightarrow 2{x^{\dfrac{1}{2}}} = at \\
   \Rightarrow 2\sqrt x = at \\
 \]
Now let us square both sides in order to find the relation between x and t.
\[
   \Rightarrow {\left( {2\sqrt x } \right)^2} = {\left( {at} \right)^2} \\
   \Rightarrow 4x = {a^2}{t^2} \\
   \Rightarrow x = \dfrac{{{a^2}{t^2}}}{4} \\
  \therefore x\left( t \right) = {\left( {\dfrac{a}{2}} \right)^2}{t^2} \\
 \]
So, we get the relation between the displacement and time as:
\[x\left( t \right) \propto \left( {{t^2}} \right)\]

Therefore the correct answer is option B.

Note: In order to solve such problems students must remember that by single differentiation of displacement equation with respect to time is velocity and double differentiation of displacement equation with respect to time is acceleration. A displacement is a vector whose length is the shortest distance from the initial to the final position of a point P undergoing motion.