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

Explanation

To solve the problem step by step, we can follow these steps:

### Step 1: Understand the Problem
We have a body moving with uniform acceleration that passes two points A and B with velocities \( v_A = 20 \, \text{m/s} \) and \( v_B = 30 \, \text{m/s} \) respectively. We need to find the speed of the body at the midpoint C between A and B.

### Step 2: Define the Midpoint
Let the distance between points A and B be \( 2s \). Therefore, the distance from A to C (midpoint) is \( s \) and from C to B is also \( s \).

### Step 3: Apply the Equation of Motion
Using the equation of motion:
\[ v_B^2 = v_A^2 + 2a \cdot d \]
where:
- \( v_B = 30 \, \text{m/s} \)
- \( v_A = 20 \, \text{m/s} \)
- \( d = 2s \) (the total distance from A to B)

Substituting the values:
\[ 30^2 = 20^2 + 2a \cdot (2s) \]
\[ 900 = 400 + 4as \]
\[ 4as = 900 - 400 = 500 \]
\[ as = \frac{500}{4} = 125 \]

### Step 4: Find the Velocity at Point C
Now, we will find the velocity at point C using the same equation of motion from A to C:
\[ v_C^2 = v_A^2 + 2a \cdot s \]
Substituting the known values:
\[ v_C^2 = 20^2 + 2a \cdot s \]
We know from the previous calculation that \( as = 125 \), so:
\[ v_C^2 = 400 + 2 \cdot 125 \]
\[ v_C^2 = 400 + 250 = 650 \]
\[ v_C = \sqrt{650} \]
Calculating \( \sqrt{650} \):
\[ v_C \approx 25.5 \, \text{m/s} \]

### Final Answer
The speed of the body at the midpoint C is approximately \( 25.5 \, \text{m/s} \).

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