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

Explanation

To solve the problem step by step, we need to analyze the given information and apply the appropriate mathematical techniques.

### Step 1: Understand the given information
We have an object moving with an initial speed \( v_0 = 6.25 \, \text{m/s} \) and it is decelerating at a rate described by the equation:
\[ \frac{dv}{dt} = -2.5 \sqrt{v} \]
We need to find the time taken for the object to come to rest (i.e., when \( v = 0 \)).

### Step 2: Rearranging the equation
We can rearrange the equation to separate the variables:
\[ \frac{dv}{\sqrt{v}} = -2.5 \, dt \]

### Step 3: Integrate 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 \( t \):
\[ \int \frac{dv}{\sqrt{v}} = \int -2.5 \, dt \]

### Step 4: Perform the integration
The integral of \( \frac{1}{\sqrt{v}} \) is \( 2\sqrt{v} \). Thus, we have:
\[ 2\sqrt{v} = -2.5t + C \]
where \( C \) is the constant of integration.

### Step 5: Apply initial conditions
At \( t = 0 \), the speed \( v = 6.25 \, \text{m/s} \). We can use this to find \( C \):
\[ 2\sqrt{6.25} = -2.5(0) + C \]
Calculating \( \sqrt{6.25} = 2.5 \):
\[ 2 \times 2.5 = C \implies C = 5 \]

### Step 6: Substitute back to find \( t \)
Now we substitute \( C \) back into our equation:
\[ 2\sqrt{v} = -2.5t + 5 \]
We need to find the time when \( v = 0 \):
\[ 2\sqrt{0} = -2.5t + 5 \]
This simplifies to:
\[ 0 = -2.5t + 5 \]
Rearranging gives:
\[ 2.5t = 5 \implies t = \frac{5}{2.5} = 2 \, \text{s} \]

### Final Answer
The time taken by the object to come to rest is \( t = 2 \, \text{seconds} \).

---