A body moving in a straight line with constant acceleration covers distances a and b in successive equal time interval of t. The acceleration of the body is

Options

A. `(a+b)/(t^(2))`
B. `(b-a)/(t^(2))`
C. `(2b-a)/(2t^(2))`
D. `(b-2a)/(t^(2))`

Answer

To solve the problem, we need to find the acceleration of a body that covers distances \( a \) and \( b \) in successive equal time intervals \( t \) under constant acceleration.

### Step-by-step Solution:

1. **Understanding the Motion**:
- The body covers distance \( a \) in the first time interval \( t \).
- It covers distance \( b \) in the second time interval \( t \).

2. **Using the Equations of Motion**:
- The equation for distance covered under constant acceleration is given by:
\[ s = ut + \frac{1}{2} a t^2 \]
- For the first interval (distance \( a \)):
\[ a = ut + \frac{1}{2} A t^2 \quad \text{(1)} \]
- For the second interval (distance \( b \)), the final velocity after the first interval becomes the initial velocity for the second interval. The final velocity after the first interval is:
\[ v = u + At \]
- The distance covered in the second interval can be expressed as:
\[ b = vt + \frac{1}{2} A t^2 \]
- Substituting \( v \) in the equation:
\[ b = (u + At)t + \frac{1}{2} A t^2 \]
- Simplifying this gives:
\[ b = ut + At^2 + \frac{1}{2} A t^2 \]
\[ b = ut + \frac{3}{2} A t^2 \quad \text{(2)} \]

3. **Setting Up the Equations**:
- We now have two equations:
- From (1): \( a = ut + \frac{1}{2} A t^2 \)
- From (2): \( b = ut + \frac{3}{2} A t^2 \)

4. **Subtracting the Equations**:
- Subtract equation (1) from equation (2):
\[ b - a = \left(ut + \frac{3}{2} A t^2\right) - \left(ut + \frac{1}{2} A t^2\right) \]
- This simplifies to:
\[ b - a = \left(\frac{3}{2} A t^2 - \frac{1}{2} A t^2\right) \]
\[ b - a = A t^2 \]

5. **Finding Acceleration**:
- Rearranging gives us:
\[ A = \frac{b - a}{t^2} \]

### Final Answer:
The acceleration of the body is:
\[ A = \frac{b - a}{t^2} \]