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 where a body is moving with constant acceleration, we can use the equations of motion. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Initial velocity at point A (u) = 5 m/s
- Final velocity at point B (v) = 15 m/s
- Constant acceleration (a) = 5 m/s²

### Step 2: Use the equation of motion to find the distance (S)
We can use the equation of motion:
\[ v^2 = u^2 + 2aS \]

Substituting the known values:
\[ (15)^2 = (5)^2 + 2 \cdot (5) \cdot S \]

### Step 3: Calculate the distance (S)
Calculating the left side:
\[ 225 = 25 + 10S \]

Now, rearranging the equation to solve for S:
\[ 225 - 25 = 10S \]
\[ 200 = 10S \]
\[ S = \frac{200}{10} = 20 \text{ m} \]

### Step 4: Find the distance to the midpoint
The midpoint of AB is at half the distance:
\[ \text{Midpoint distance} = \frac{S}{2} = \frac{20}{2} = 10 \text{ m} \]

### Step 5: Use the equation of motion to find the velocity at the midpoint
Using the same equation of motion:
\[ v^2 = u^2 + 2aS \]
Where \( S = 10 \text{ m} \) (the distance to the midpoint).

Substituting the values:
\[ v^2 = (5)^2 + 2 \cdot (5) \cdot (10) \]

### Step 6: Calculate the velocity at the midpoint
Calculating:
\[ v^2 = 25 + 100 \]
\[ v^2 = 125 \]

Taking the square root to find v:
\[ v = \sqrt{125} = 11.18 \text{ m/s} \]

### Final Answer
The velocity at the midpoint of AB is approximately **11.18 m/s**.

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