An object , moving with a speed of ` 6.25 m//s `, is decelerated at a rate given by :
`(dv)/(dt) = - 2.5 sqrt (v)` where `v` is the instantaneous speed . The time taken by the object , to come to rest , would be :

Options

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

Explanation

To solve the problem of an object decelerating at a rate given by \((dv)/(dt) = -2.5 \sqrt{v}\), we need to find the time taken for the object to come to rest from an initial speed of \(6.25 \, m/s\).

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial speed, \(v_0 = 6.25 \, m/s\)
- Final speed, \(v_f = 0 \, m/s\)
- Deceleration equation: \(\frac{dv}{dt} = -2.5 \sqrt{v}\)

2. **Rearrange the equation:**
We can rewrite the deceleration equation as:
\[ dt = \frac{dv}{-2.5 \sqrt{v}} \]

3. **Integrate both sides:**
We need to integrate the left side with respect to time and the right side with respect to velocity. The limits for velocity will be from \(6.25\) to \(0\):
\[ \int_{0}^{t} dt = -\frac{1}{2.5} \int_{6.25}^{0} v^{-1/2} \, dv \]

4. **Calculate the left side:**
The left side integrates to:
\[ t \]

5. **Calculate the right side:**
The integral of \(v^{-1/2}\) is:
\[ \int v^{-1/2} \, dv = 2\sqrt{v} \]
Thus, the right side becomes:
\[ -\frac{1}{2.5} \left[ 2\sqrt{v} \right]_{6.25}^{0} = -\frac{1}{2.5} \left( 2\sqrt{0} - 2\sqrt{6.25} \right) = -\frac{1}{2.5} \left( 0 - 5 \right) = \frac{5}{2.5} = 2 \]

6. **Set the equations equal:**
Now we have:
\[ t = 2 \, seconds \]

### Conclusion:
The time taken by the object to come to rest is \(2 \, seconds\).