A particle of mass m moves on a straight line with its velocity varying with the distance travelled according to the equation `v=asqrtx`, wher ea is a constant. Find the total work done by all the forces during a displacement from `x=0 to x=d`.

Explanation

To solve the problem, we need to find the total work done by all forces during the displacement of a particle from \( x = 0 \) to \( x = d \) when its velocity varies according to the equation \( v = a\sqrt{x} \).

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Mass of the particle: \( m \)
- Velocity as a function of distance: \( v = a\sqrt{x} \), where \( a \) is a constant.

2. **Calculate Initial Velocity**:
- At \( x = 0 \):
\[ v_i = a\sqrt{0} = 0 \]
- Therefore, the initial velocity \( v_i = 0 \).

3. **Calculate Final Velocity**:
- At \( x = d \):
\[ v_f = a\sqrt{d} \]
- Therefore, the final velocity \( v_f = a\sqrt{d} \).

4. **Use the Work-Energy Theorem**:
- According to the work-energy theorem, the total work done \( W \) is equal to the change in kinetic energy (\( \Delta KE \)):
\[ W = KE_f - KE_i \]

5. **Calculate Initial and Final Kinetic Energy**:
- Initial Kinetic Energy (\( KE_i \)):
\[ KE_i = \frac{1}{2} m v_i^2 = \frac{1}{2} m (0)^2 = 0 \]
- Final Kinetic Energy (\( KE_f \)):
\[ KE_f = \frac{1}{2} m v_f^2 = \frac{1}{2} m (a\sqrt{d})^2 = \frac{1}{2} m a^2 d \]

6. **Calculate Total Work Done**:
- Substitute the values of \( KE_f \) and \( KE_i \) into the work-energy theorem:
\[ W = KE_f - KE_i = \frac{1}{2} m a^2 d - 0 = \frac{1}{2} m a^2 d \]

### Final Answer:
The total work done by all the forces during the displacement from \( x = 0 \) to \( x = d \) is:
\[ W = \frac{1}{2} m a^2 d \]