Question

A particle is moving rectilinearly with acceleration a, whose value is given as a function of distance by the equation $a = \alpha \sqrt x $ find displacement as a function of time. [At $t = 0$ particle is at rest at $x = 0$]
(A) $4\alpha {t^{32}}$
(B) $\alpha \sqrt 6 t$
(C) $\alpha \dfrac{1}{{236}}{t^4}$
(D) ${\alpha ^2}\dfrac{1}{{144}}{t^4}$

Answer 1

Hint:In order to solve this question, we will write acceleration in terms of derivative form of velocity and displacement and then we will find the velocity in term of distance using integration and then we will again write velocity in its distance time derivative form and solve for displacement relation with time.


Formula used:
The acceleration is written in derivative form as $a = \dfrac{{dv}}{{dt}}$ and $v = \dfrac{{dx}}{{dt}}$ where v is velocity, dx is change in displacement of the body and ‘dt’ is change in time.


Complete answer:
According to the question, we have given that acceleration is related to displacement as $a = \alpha \sqrt x $ now, we have $a = \dfrac{{dv}}{{dt}}$ or we can write it as
$
  a = \dfrac{{dv}}{{dt}}.\dfrac{{dx}}{{dx}} \\
  a = v\dfrac{{dv}}{{dx}} \\
 $
$ \Rightarrow v\dfrac{{dv}}{{dx}} = \alpha \sqrt x $
Now, integrating above equation we get,
$\int {vdv = \alpha \int {\sqrt x dx} } $
using integration formula as $\int {{r^n}dr = \dfrac{{{r^{n + 1}}}}{{n + 1}}} $ we get,
$
  \dfrac{{{v^2}}}{2} = \alpha \dfrac{{2{x^{\dfrac{3}{2}}}}}{3} \\
   \Rightarrow {v^2} = \dfrac{4}{3}\alpha {x^{\dfrac{3}{2}}} \\
   \Rightarrow v = {x^{\dfrac{3}{4}}}\sqrt {\dfrac{4}{3}\alpha } \\
 $
Now, we have the relation $v = {x^{\dfrac{3}{4}}}\sqrt {\dfrac{4}{3}\alpha } $ and we can write
$
  v = \dfrac{{dx}}{{dt}} = {x^{\dfrac{3}{4}}}\sqrt {\dfrac{4}{3}\alpha } \\
   \Rightarrow \dfrac{{dx}}{{dt}} = {x^{\dfrac{3}{4}}}\sqrt {\dfrac{4}{3}\alpha } \\
 $
Again, integrating the above equation, we get
$\int {{x^{\dfrac{{ - 3}}{4}}}} dx = \sqrt {\dfrac{4}{3}\alpha } \int {dt} $ and using the formula $\int {{r^n}dr = \dfrac{{{r^{n + 1}}}}{{n + 1}}} $ we get,
$
  \dfrac{{{x^{\dfrac{{ - 3}}{4} + 1}}}}{{\dfrac{{ - 3}}{4} + 1}} = t\sqrt {\dfrac{4}{3}\alpha } \\
  {x^{\dfrac{1}{4}}} = \dfrac{1}{4}t\sqrt {\dfrac{4}{3}\alpha } \\
 $
Taking fourth power on both sides, we get
$
  x = \dfrac{1}{{256}}{t^4}\dfrac{{16}}{9}{\alpha ^2} \\
  x = \dfrac{1}{{144}}{\alpha ^2}{t^4} \\
 $
So, the relation between the displacement and time of the particle is given by ${\alpha ^2}\dfrac{1}{{144}}{t^4}$
Hence, the correct answer is option (D) ${\alpha ^2}\dfrac{1}{{144}}{t^4}$

Hence, the correct option is Option (D).


Note:While solving such numerical questions, always remember all the basic integration formulas used and solve integration equations carefully in order to avoid any minor errors, also remember the basic derivative forms of velocity, acceleration and force which are commonly used in kinematics questions.