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 find the acceleration of a body moving in a straight line with constant acceleration that covers distances \( a \) and \( b \) in successive equal time intervals of \( t \), we can follow these steps:

### 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 \( t \).
- The total time for both intervals is \( 2t \).

2. **Using the Equation of Motion**:
- For the first interval (distance \( a \)):
\[ a = ut + \frac{1}{2} \alpha t^2 \quad \text{(Equation 1)} \]
- For the second interval (distance \( b \)), the initial velocity for this interval is \( u + \alpha t \) (the final velocity from the first interval):
\[ b = (u + \alpha t)t + \frac{1}{2} \alpha t^2 \quad \text{(Equation 2)} \]

3. **Expanding Equation 2**:
- Expanding the second equation:
\[ b = ut + \alpha t^2 + \frac{1}{2} \alpha t^2 \]
\[ b = ut + \frac{3}{2} \alpha t^2 \]

4. **Setting Up the Equations**:
- From Equation 1, we have:
\[ a = ut + \frac{1}{2} \alpha t^2 \]
- From the expanded Equation 2:
\[ b = ut + \frac{3}{2} \alpha t^2 \]

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

6. **Solving for Acceleration**:
- Rearranging gives us:
\[ \alpha = \frac{b - a}{t^2} \]

### Final Answer:
Thus, the acceleration \( \alpha \) of the body is:
\[ \alpha = \frac{b - a}{t^2} \]