If a body travles 30 m in an interval of 2 s and 50 m in the next interval of 2s. Then the acceleration of the body is

Options

A. `10 m//s^(2)`
B. `5 m//s^(2)`
C. `20 m//s^(2)`
D. `25 m//s^(2)`

Explanation

To solve the problem step by step, we will use the equations of motion under constant acceleration.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The body travels 30 m in the first 2 seconds.
- It travels 50 m in the next 2 seconds (total time = 4 seconds).

2. **Using the Equation of Motion**:
- The equation of motion we will use is:
\[ S = ut + \frac{1}{2} a t^2 \]
- Where:
- \( S \) = distance traveled
- \( u \) = initial velocity
- \( a \) = acceleration
- \( t \) = time

3. **Setting Up the First Equation**:
- For the first interval (0 to 2 seconds):
\[ S_1 = 30 \, \text{m}, \quad t_1 = 2 \, \text{s} \]
\[ 30 = u \cdot 2 + \frac{1}{2} a \cdot (2^2) \]
\[ 30 = 2u + 2a \]
Dividing the whole equation by 2:
\[ 15 = u + a \quad \text{(Equation 1)} \]

4. **Setting Up the Second Equation**:
- For the second interval (2 to 4 seconds):
- Total distance traveled in 4 seconds is:
\[ S_2 = 30 + 50 = 80 \, \text{m}, \quad t_2 = 4 \, \text{s} \]
\[ 80 = u \cdot 4 + \frac{1}{2} a \cdot (4^2) \]
\[ 80 = 4u + 8a \]
Dividing the whole equation by 4:
\[ 20 = u + 2a \quad \text{(Equation 2)} \]

5. **Solving the Equations**:
- Now we have two equations:
- Equation 1: \( 15 = u + a \)
- Equation 2: \( 20 = u + 2a \)
- We can subtract Equation 1 from Equation 2:
\[ (20 - 15) = (u + 2a) - (u + a) \]
\[ 5 = a \]

6. **Conclusion**:
- The acceleration \( a \) of the body is \( 5 \, \text{m/s}^2 \).