A particle of charge `q` and mass `m` moves rectilinearly under the action of an electric field `E=A-B x`, where `B` is a positive constant and `x` is a distance from the point where the particle was initially at rest. The distance travelled by the particle till it comes to rest is `(alpha A)/(beta B)`. Find `(alpha+beta)`.

Answer

To solve the problem, we need to analyze the motion of a charged particle in a non-uniform electric field given by \( E = A - Bx \). Here’s a step-by-step breakdown of the solution:

### Step 1: Understanding the Electric Field
The electric field \( E \) is given as a function of position \( x \):
\[ E = A - Bx \]
This indicates that the electric field decreases linearly with increasing \( x \).

### Step 2: Finding the Force on the Particle
The force \( F \) acting on the particle of charge \( q \) is given by:
\[ F = qE = q(A - Bx) \]

### Step 3: Setting Up the Equation of Motion
Using Newton's second law, we can express the motion of the particle:
\[ m \frac{d^2x}{dt^2} = q(A - Bx) \]
This can be rearranged to:
\[ \frac{d^2x}{dt^2} = \frac{q}{m}(A - Bx) \]

### Step 4: Finding the Velocity
To find the distance traveled until the particle comes to rest, we first need to find the velocity. We can use the work-energy principle. The work done on the particle as it moves from an initial position \( x_0 \) to a position \( x \) is equal to the change in kinetic energy.

### Step 5: Work Done by the Electric Field
The work done \( W \) by the electric field when the particle moves from \( x_0 \) to \( x \) is:
\[ W = \int_{x_0}^{x} qE \, dx = \int_{x_0}^{x} q(A - Bx) \, dx \]
Calculating this integral gives:
\[ W = q \left[ Ax - \frac{Bx^2}{2} \right]_{x_0}^{x} \]

### Step 6: Setting Initial Conditions
Assuming the particle starts from rest, the initial kinetic energy is zero. Therefore, the work done is equal to the final kinetic energy when the particle comes to rest:
\[ W = \frac{1}{2} mv^2 \]

### Step 7: Finding the Distance Traveled
The distance traveled by the particle until it comes to rest can be found by setting the work done equal to the change in kinetic energy. When the particle comes to rest, the velocity \( v = 0 \), and we can solve for the distance \( x \).

### Step 8: Analyzing the Electric Field
The electric field becomes zero when:
\[ A - Bx = 0 \Rightarrow x = \frac{A}{B} \]
This indicates that the particle will experience a change in direction of the force at this point.

### Step 9: Total Distance Traveled
The particle will travel to the point \( x = \frac{A}{B} \) and then come back to the point where the electric field is zero. Therefore, the total distance traveled until it comes to rest is:
\[ \text{Total distance} = 2 \cdot \frac{A}{B} = \frac{2A}{B} \]

### Step 10: Identifying \( \alpha \) and \( \beta \)
From the problem statement, we have:
\[ \text{Distance} = \frac{\alpha A}{\beta B} \]
Comparing this with our result:
\[ \frac{2A}{B} = \frac{\alpha A}{\beta B} \]
We can see that \( \alpha = 2 \) and \( \beta = 1 \).

### Final Calculation
Now, we need to find \( \alpha + \beta \):
\[ \alpha + \beta = 2 + 1 = 3 \]

### Conclusion
The final answer is:
\[ \boxed{3} \]