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=-ksqrt 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 :

Options

A. ` T=(2sqrt u)/(k) `
B. ` T=(2u)/(k)`
C. ` T=(2u^(3//2) )/(k)`
D. ` T=(2u^(2))/(k) `

Explanation

To solve the problem, we need to find the time \( T \) taken by a particle to come to rest when it is subjected to a retarding force that causes it to decelerate at a rate given by \( a = -k \sqrt{v} \).

### Step-by-step Solution:

1. **Understand the relationship between acceleration, velocity, and time**:
The acceleration \( a \) can be expressed as:
\[ a = \frac{dv}{dt} \]
Given that \( a = -k \sqrt{v} \), we can write:
\[ \frac{dv}{dt} = -k \sqrt{v} \]

2. **Rearranging the equation**:
We can rearrange the equation to separate variables:
\[ \frac{dv}{\sqrt{v}} = -k \, dt \]

3. **Integrate both sides**:
We will integrate both sides. The left side will be integrated with respect to \( v \) and the right side with respect to \( t \):
\[ \int \frac{dv}{\sqrt{v}} = -k \int dt \]
The left-hand side integrates to \( 2\sqrt{v} \) and the right-hand side integrates to \( -kt + C \), where \( C \) is the constant of integration.

4. **Set up the limits of integration**:
Initially, at \( t = 0 \), the velocity \( v = u \) and at \( t = T \), the velocity \( v = 0 \):
\[ 2\sqrt{0} - 2\sqrt{u} = -k(T - 0) \]
This simplifies to:
\[ -2\sqrt{u} = -kT \]

5. **Solve for \( T \)**:
Rearranging gives:
\[ kT = 2\sqrt{u} \]
Therefore, we find:
\[ T = \frac{2\sqrt{u}}{k} \]

### Final Answer:
The time \( T \) taken by the particle to come to rest is:
\[ T = \frac{2\sqrt{u}}{k} \]