A body of mass m starts moving from rest along x-axis so that it velocity varies as v=`asqrt(s)` where a is a constant and s is the distance covered by the body .The total work done by all the forces acting on the body in the first second after the start of the motion is :

Options

A. `(1)/(8)ma^(4)t^(2)`
B. `4ma^(4)t^(2)`
C. `8ma^(4)t^(2)`
D. `(1)/(4)ma^(4)t^(2)`

Explanation

To solve the problem, we need to find the total work done by all the forces acting on a body of mass \( m \) in the first second after it starts moving from rest, given that its velocity varies as \( v = a \sqrt{s} \), where \( a \) is a constant and \( s \) is the distance covered by the body.

### Step-by-Step Solution:

1. **Understanding the Velocity Equation**:
The velocity of the body is given by:
\[ v = a \sqrt{s} \]
where \( s \) is the distance covered.

2. **Relating Velocity to Displacement**:
Since velocity is the rate of change of displacement, we can write:
\[ v = \frac{ds}{dt} \]
Thus, we have:
\[ \frac{ds}{dt} = a \sqrt{s} \]

3. **Separating Variables**:
Rearranging gives:
\[ \frac{ds}{\sqrt{s}} = a \, dt \]

4. **Integrating Both Sides**:
We integrate both sides. The left side integrates from \( s = 0 \) to \( s \) and the right side from \( t = 0 \) to \( t \):
\[ \int_0^s \frac{ds}{\sqrt{s}} = \int_0^t a \, dt \]
The integral of \( \frac{1}{\sqrt{s}} \) is \( 2\sqrt{s} \), so we have:
\[ 2\sqrt{s} \bigg|_0^s = at \bigg|_0^t \]
This simplifies to:
\[ 2\sqrt{s} = at \]
Therefore,
\[ \sqrt{s} = \frac{at}{2} \]
Squaring both sides gives:
\[ s = \frac{a^2 t^2}{4} \]

5. **Finding Acceleration**:
To find the acceleration \( a \), we differentiate \( s \) with respect to \( t \):
\[ v = \frac{ds}{dt} = \frac{d}{dt} \left( \frac{a^2 t^2}{4} \right) = \frac{a^2 t}{2} \]
Differentiating again gives:
\[ a = \frac{dv}{dt} = \frac{d}{dt} \left( \frac{a^2 t}{2} \right) = \frac{a^2}{2} \]

6. **Calculating the Force**:
The force \( F \) acting on the body is given by:
\[ F = m \cdot a = m \cdot \frac{a^2}{2} \]

7. **Calculating Work Done**:
The work done \( W \) is given by:
\[ W = F \cdot s \]
Substituting the expressions for \( F \) and \( s \):
\[ W = \left( m \cdot \frac{a^2}{2} \right) \cdot \left( \frac{a^2 t^2}{4} \right) = \frac{m a^4 t^2}{8} \]

8. **Substituting \( t = 1 \) Second**:
To find the work done in the first second, we substitute \( t = 1 \):
\[ W = \frac{m a^4 (1)^2}{8} = \frac{m a^4}{8} \]

### Final Answer:
The total work done by all the forces acting on the body in the first second after the start of the motion is:
\[ W = \frac{m a^4}{8} \]