Answer

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**Hint:**As a first step, you could recall that the time rate of change of velocity will give us the acceleration of the particle. As the given particle is known to be decelerating, do assign negative signs before solving. Now you could integrate the relation after necessary rearrangements by assigning the limits accordingly. After integration, apply the limits and then rearrange to get the answer.

**Formula used:**

Acceleration,

$a=\dfrac{dv}{dt}$

**Complete Step by step solution:**

In the question, we are given the deceleration of the particle as,

$a=-k\sqrt{v}$

Negative sign is an indication of deceleration. We know that acceleration is the time rate of change of velocity. A body is said to be decelerating when its velocity is being decreased.

Now we could rewrite the expression as,

$\dfrac{dv}{dt}=-k\sqrt{v}$

$\Rightarrow \dfrac{dv}{\sqrt{v}}=-kdt$

$\Rightarrow {{v}^{-\dfrac{1}{2}}}dv=-kdt$

Now let $t=0$ be the time at which the body started moving and t be the time at which it came to rest. During this period, the body’s velocity is being reduced from ${{v}_{0}}$ to 0.

Let us integrate the above relation on both sides by assigning limits accordingly.

$\int\limits_{{{v}_{0}}}^{0}{{{v}^{-\dfrac{1}{2}}}}dv=-k\int\limits_{0}^{t}{dt}$

$\Rightarrow \left[ 2{{v}^{\dfrac{1}{2}}} \right]_{{{v}_{0}}}^{0}=-k\left[ t \right]_{0}^{t}$

$\Rightarrow 0-2{{v}_{0}}^{\dfrac{1}{2}}==-kt$

$\therefore t=\dfrac{2\sqrt{{{v}_{0}}}}{k}$

Therefore, we found that the time after which the given particle will stop its motion is given by,

$t=\dfrac{2\sqrt{{{v}_{0}}}}{k}$

**Hence, option C is found to be the correct answer.**

**Note:**

For the case of one dimensional motion, deceleration is known to occur when the signs of velocity and acceleration are opposite to each other. That is, it is deceleration when acceleration is negative and velocity is positive and also when velocity is negative and acceleration is positive. To summarize we could say that deceleration occurs when velocity is directed opposite to acceleration.

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