A body travelling with uniform acceleration crosses two point `A` and `B` with velocities `20 m s^-1` and `30 m s^-1` respectively. The speed of the body at the mid-point of `A` and `B` is.

Options

A. `24 m s^-1`
B. `25 m s^-1`
C. `25.5 m s^-1`
D. `10 sqrt(6) m s^-1`

Explanation

To find the speed of the body at the midpoint between points A and B, we can use the concept of uniformly accelerated motion. The average speed can be calculated using the velocities at points A and B.

### Step-by-Step Solution:

1. **Identify the given data:**
- Velocity at point A, \( v_A = 20 \, \text{m/s} \)
- Velocity at point B, \( v_B = 30 \, \text{m/s} \)

2. **Calculate the average speed at the midpoint:**
- The average speed \( v_{avg} \) between two points in uniformly accelerated motion can be calculated using the formula:
\[ v_{avg} = \frac{v_A + v_B}{2} \]
- Substitute the values of \( v_A \) and \( v_B \):
\[ v_{avg} = \frac{20 \, \text{m/s} + 30 \, \text{m/s}}{2} \]

3. **Perform the calculation:**
- Calculate the sum:
\[ v_{avg} = \frac{50 \, \text{m/s}}{2} = 25 \, \text{m/s} \]

4. **Conclusion:**
- The speed of the body at the midpoint between points A and B is \( 25 \, \text{m/s} \).