A body is moving with constant acceleration, `a= 5 m/s^2` from A to B in straight line AB. The velocity of particle at A and B are 5 m/s and 15 m/s respectively. The velocity at mid-point of AB is

Options

A. `sqrt125 m/s`
B. `sqrt250 m/s`
C. `sqrt550 m/s`
D. `sqrt600 m/s`

Answer

To find the velocity at the midpoint of the straight line AB, we can follow these steps:

### Step 1: Identify the given values
- Initial velocity at point A, \( V_A = 5 \, \text{m/s} \)
- Final velocity at point B, \( V_B = 15 \, \text{m/s} \)
- Constant acceleration, \( a = 5 \, \text{m/s}^2 \)

### Step 2: Use the third equation of motion to find the distance \( S \) from A to B
The third equation of motion is given by:
\[ V^2 = U^2 + 2aS \]
Where:
- \( V \) is the final velocity (at B)
- \( U \) is the initial velocity (at A)
- \( a \) is the acceleration
- \( S \) is the distance

Substituting the known values:
\[ 15^2 = 5^2 + 2 \cdot 5 \cdot S \]
Calculating the squares:
\[ 225 = 25 + 10S \]
Rearranging the equation:
\[ 225 - 25 = 10S \]
\[ 200 = 10S \]
Dividing both sides by 10:
\[ S = 20 \, \text{m} \]

### Step 3: Find the midpoint distance
Since the total distance \( S \) from A to B is 20 m, the distance to the midpoint O is:
\[ \text{Distance to midpoint} = \frac{S}{2} = \frac{20}{2} = 10 \, \text{m} \]

### Step 4: Use the third equation of motion again to find the velocity at the midpoint \( V_O \)
We will use the same equation, but now we will calculate the velocity at the midpoint:
\[ V_O^2 = V_A^2 + 2a \cdot \text{Distance to midpoint} \]
Substituting the known values:
\[ V_O^2 = 5^2 + 2 \cdot 5 \cdot 10 \]
Calculating the squares:
\[ V_O^2 = 25 + 100 \]
\[ V_O^2 = 125 \]
Taking the square root to find \( V_O \):
\[ V_O = \sqrt{125} \, \text{m/s} \]
\[ V_O = 11.18 \, \text{m/s} \, (\text{approximately}) \]

### Final Answer
The velocity at the midpoint of AB is approximately \( 11.18 \, \text{m/s} \).

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