An object moving with a speed of `6.25 m//s`, is deceleration at a rate given by :
`(dv)/(dt) = -25 sqrt(v)` ,
where `v` is instantaneous speed. The time taken by the obeject, to come to rest, would be :

Options

A. 1 s
B. 2 s
C. 4 s
D. 8 s

Explanation

To solve the problem, we need to determine the time taken by the object to come to rest given its initial speed and the rate of deceleration. The deceleration is defined by the equation:

\[ \frac{dv}{dt} = -25 \sqrt{v} \]

### Step-by-Step Solution:

1. **Rearranging the Equation**:
We start with the equation of deceleration:
\[ \frac{dv}{dt} = -25 \sqrt{v} \]
To separate variables, we can rearrange this to:
\[ \frac{dv}{\sqrt{v}} = -25 dt \]

2. **Integrating Both Sides**:
Now, we integrate both sides. The left side will be integrated with respect to \(v\) and the right side with respect to \(t\):
\[ \int \frac{dv}{\sqrt{v}} = \int -25 dt \]

3. **Calculating the Integrals**:
The integral of \(\frac{1}{\sqrt{v}}\) is \(2\sqrt{v}\), and the integral of \(-25\) is \(-25t\):
\[ 2\sqrt{v} = -25t + C \]
where \(C\) is the constant of integration.

4. **Finding the Constant of Integration**:
To find \(C\), we use the initial condition. At \(t = 0\), the speed \(v = 6.25 \, \text{m/s}\):
\[ 2\sqrt{6.25} = C \]
Since \(\sqrt{6.25} = 2.5\), we have:
\[ C = 2 \times 2.5 = 5 \]

5. **Substituting Back**:
Now we substitute \(C\) back into the equation:
\[ 2\sqrt{v} = -25t + 5 \]

6. **Finding the Time to Come to Rest**:
The object comes to rest when \(v = 0\):
\[ 2\sqrt{0} = -25t + 5 \]
This simplifies to:
\[ 0 = -25t + 5 \]
Rearranging gives:
\[ 25t = 5 \implies t = \frac{5}{25} = \frac{1}{5} = 0.2 \, \text{seconds} \]

7. **Final Calculation**:
We need to check the calculations again. The correct integration should yield:
\[ 2\sqrt{v} = -25t + 5 \]
Solving for \(t\):
\[ 0 = -25t + 5 \implies 25t = 5 \implies t = \frac{5}{25} = 0.2 \, \text{seconds} \]

### Conclusion:
The time taken by the object to come to rest is \(0.2 \, \text{seconds}\).