A particle travels along the arc of a circle of radius `r`. Its speed depends on the distance travelled `l` as `v=asqrtl` where 'a' is a constant. The angle `alpha` between the vectors of net acceleration and the velocity of the particle is

Options

A. `alpha = tan^(-1) ((2l)/r)`
B. `alpha = cos^(-1) ((2l)/r)`
C. `alpha = sin^(-1) ((2l)/r)`
D. `alpha = cot^(-1) ((2l)/r)`

Answer

To solve the problem, we need to find the angle \( \alpha \) between the net acceleration vector and the velocity vector of a particle moving along the arc of a circle with a radius \( r \). The speed of the particle depends on the distance traveled \( l \) as given by the equation \( v = a \sqrt{l} \).

### Step-by-Step Solution:

1. **Identify the Components of Acceleration**:
The net acceleration \( \vec{A} \) of the particle has two components:
- Tangential acceleration \( A_T \) (along the direction of velocity)
- Centripetal acceleration \( A_C \) (perpendicular to the direction of velocity)

2. **Calculate Tangential Acceleration**:
The tangential acceleration \( A_T \) is given by the rate of change of speed with respect to time:
\[ A_T = \frac{dv}{dt} \]

3. **Relate Speed to Distance**:
Given \( v = a \sqrt{l} \), we can square both sides to get:
\[ v^2 = a^2 l \]
Differentiating both sides with respect to \( l \):
\[ 2v \frac{dv}{dl} = a^2 \]
Rearranging gives:
\[ \frac{dv}{dl} = \frac{a^2}{2v} \]

4. **Express \( \frac{dv}{dt} \)**:
Using the chain rule, we can express \( \frac{dv}{dt} \) as:
\[ \frac{dv}{dt} = \frac{dv}{dl} \cdot \frac{dl}{dt} = \frac{a^2}{2v} \cdot v = \frac{a^2}{2} \]
Thus, the tangential acceleration \( A_T \) is:
\[ A_T = \frac{a^2}{2} \]

5. **Calculate Centripetal Acceleration**:
The centripetal acceleration \( A_C \) is given by:
\[ A_C = \frac{v^2}{r} \]
Substituting \( v^2 = a^2 l \):
\[ A_C = \frac{a^2 l}{r} \]

6. **Find the Ratio of Accelerations**:
The angle \( \alpha \) between the net acceleration and the velocity can be found using the tangent of the angle:
\[ \tan \alpha = \frac{A_C}{A_T} = \frac{\frac{a^2 l}{r}}{\frac{a^2}{2}} = \frac{2l}{r} \]

7. **Express the Angle**:
Therefore, the angle \( \alpha \) is:
\[ \alpha = \tan^{-1} \left( \frac{2l}{r} \right) \]

### Final Answer:
The angle \( \alpha \) between the vectors of net acceleration and the velocity of the particle is:
\[ \alpha = \tan^{-1} \left( \frac{2l}{r} \right) \]