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 of finding the time \( T \) taken by a particle to come to rest under the given conditions, we will follow these steps:

### Step 1: Understand the given information
The particle starts with an initial velocity \( u \) and experiences a retarding force that causes it to decelerate according to the equation:
\[ a = -k \sqrt{v} \]
where \( k \) is a positive constant and \( v \) is the instantaneous velocity.

### Step 2: Relate acceleration to velocity
We know that acceleration \( a \) can be expressed in terms of velocity \( v \) and time \( t \) as:
\[ a = \frac{dv}{dt} \]
Thus, we can rewrite the equation as:
\[ \frac{dv}{dt} = -k \sqrt{v} \]

### Step 3: Separate variables for integration
We can separate the variables \( v \) and \( t \) to facilitate integration:
\[ \frac{dv}{\sqrt{v}} = -k \, dt \]

### Step 4: Integrate both sides
Now we will integrate both sides. The left side integrates from \( v = u \) to \( v = 0 \), and the right side integrates from \( t = 0 \) to \( t = T \):
\[ \int_{u}^{0} v^{-1/2} \, dv = -k \int_{0}^{T} dt \]

The left side integrates to:
\[ \left[ 2\sqrt{v} \right]_{u}^{0} = 2\sqrt{0} - 2\sqrt{u} = -2\sqrt{u} \]
The right side integrates to:
\[ -k[T - 0] = -kT \]

### Step 5: Set the integrals equal to each other
Equating the results from both integrals gives us:
\[ -2\sqrt{u} = -kT \]

### Step 6: Solve for time \( T \)
Rearranging the equation to solve for \( T \):
\[ T = \frac{2\sqrt{u}}{k} \]

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