An object is subjected to retardation, `(dv)/(dt)=-5sqrtv` , which has initial velocity of `4ms^(-1)`. The time taken by the object to come to rest would be

Options

A. 2 s
B. `5/4 s`
C. `4/5 s`
D. `1/2 s`

Answer

To solve the problem, we need to find the time taken by an object to come to rest when it is subjected to a retardation given by the equation:

\[ \frac{dv}{dt} = -5\sqrt{v} \]

with an initial velocity \( v_0 = 4 \, \text{m/s} \).

### Step-by-Step Solution:

1. **Rearranging the equation**:
We start with the equation of retardation:
\[ \frac{dv}{dt} = -5\sqrt{v} \]
We can rearrange this to separate variables:
\[ \frac{dv}{\sqrt{v}} = -5 \, dt \]

2. **Integrating both sides**:
Now we will integrate both sides. The left side will be integrated with respect to \( v \) and the right side with respect to \( t \):
\[ \int \frac{dv}{\sqrt{v}} = \int -5 \, dt \]
The integral of \( \frac{1}{\sqrt{v}} \) is \( 2\sqrt{v} \), so we have:
\[ 2\sqrt{v} = -5t + C \]
where \( C \) is the constant of integration.

3. **Finding the constant of integration**:
To find \( C \), we use the initial condition. When \( t = 0 \), \( v = 4 \):
\[ 2\sqrt{4} = -5(0) + C \implies 2 \times 2 = C \implies C = 4 \]
Thus, our equation becomes:
\[ 2\sqrt{v} = -5t + 4 \]

4. **Finding the time to come to rest**:
The object comes to rest when \( v = 0 \). Substituting \( v = 0 \) into the equation:
\[ 2\sqrt{0} = -5t + 4 \implies 0 = -5t + 4 \]
Solving for \( t \):
\[ 5t = 4 \implies t = \frac{4}{5} \, \text{s} \]

### Final Answer:
The time taken by the object to come to rest is:
\[ t = \frac{4}{5} \, \text{s} \quad \text{or} \quad 0.8 \, \text{s} \]