Question

A particle initially (i.e., at $t = 0$) moving with a velocity u is subjected to a retarding force, as a result of which it decelerates $a = - k\sqrt v $ at a rate where v is the instantaneous velocity and k is a positive constant. The time T taken by the particle to come to rest is given by :
(A) $T = \dfrac{{2\sqrt u }}{k}$
(B) $T = \dfrac{{2u}}{k}$
(C) $T = \dfrac{{2{u^{\dfrac{3}{2}}}}}{k}$
(D) $T = \dfrac{{2{u^2}}}{k}$

Answer 1

Hint:In order to solve this question, we will use the derivative form of acceleration in terms of velocity and time and then using integration formulas we will find a relation between velocity and time and finally we will solve for time at which final velocity of the body becomes zero.


Formula used:
Acceleration ‘a’ can be written as $a = \dfrac{{dv}}{{dt}}$ where, ‘dv’ is the change in velocity an ‘dt’ is the change in time

Complete answer:
According to the question, we have given that at $t = 0$ initial velocity of the particle is ‘u’ and acceleration is given to us as $a = - k\sqrt v $ so, we can write acceleration in derivative form as $a = \dfrac{{dv}}{{dt}}$ so, we get
$\dfrac{{dv}}{{dt}} = - k\sqrt v $
Now, integrating both sides we get with the limits of velocity as $
  {v_1} = u,{t_1} = 0 \\
  {v_2} = v,{t_2} = t \\
 $
$\int\limits_u^v {{v^{ - \dfrac{1}{2}}}} dv = - k\int\limits_0^t {dt} $
Using the integration formula $\int {{r^n}dr} = \dfrac{{{r^{n + 1}}}}{{n + 1}}$ we get,
$
  [2\sqrt v ]_u^v = - k[t]_0^t \\
  2[\sqrt v - \sqrt u ] = - kt \\
 $
Now, we need to find at what time ‘t’ the final velocity of the particle becomes zero which means it became at rest, so putting the value of $v = 0$ and at this time denoting time t by ‘T’ we get,
$2[\sqrt 0 - \sqrt u ] = - kT$
solving for T we get,
$T = \dfrac{{2\sqrt u }}{k}$
So, after the time $T = \dfrac{{2\sqrt u }}{k}$ the particle will finally come to rest.

Hence, the correct option is Option (A).


Note:While solving such questions, involving differentiation and integration, always remember the basic integration formulas and derivative forms of physical quantities and avoid the integration constant using given limits of the variable, also remember constant terms do not come inside integrals as they are not variables.