A car moving with constant acceleration crosses two points A and B with velocities 10 m/s and 20 m/s in its path. Find the velocity of the car midway between A and B. (in m/s).

Options

A. `sqrt500`
B. `5sqrt5`
C. `5sqrt10`
D. `sqrt300` m/s

Answer

To find the velocity of the car midway between points A and B, we can follow these steps:

### Step 1: Identify the given values
- Velocity at point A (U) = 10 m/s
- Velocity at point B (V) = 20 m/s

### Step 2: Use the kinematic equation to find acceleration (A)
We can use the kinematic equation:
\[ V^2 = U^2 + 2AD \]
Where:
- \( V \) = final velocity at point B = 20 m/s
- \( U \) = initial velocity at point A = 10 m/s
- \( A \) = acceleration (unknown)
- \( D \) = distance between points A and B (unknown)

Rearranging the equation gives:
\[ A = \frac{V^2 - U^2}{2D} \]

Substituting the known values:
\[ A = \frac{20^2 - 10^2}{2D} \]
\[ A = \frac{400 - 100}{2D} \]
\[ A = \frac{300}{2D} \]
\[ A = \frac{150}{D} \]

### Step 3: Find the velocity at the midpoint (C)
Let the velocity at point C (midpoint) be \( V' \). The distance from A to C is \( \frac{D}{2} \).

Using the kinematic equation again for the distance from A to C:
\[ V'^2 = U^2 + 2A \left(\frac{D}{2}\right) \]

Substituting the known values:
\[ V'^2 = 10^2 + 2A \left(\frac{D}{2}\right) \]
\[ V'^2 = 100 + A \cdot D \]

Now, substituting \( A \) from Step 2:
\[ V'^2 = 100 + \frac{150}{D} \cdot D \]
\[ V'^2 = 100 + 150 \]
\[ V'^2 = 250 \]

### Step 4: Calculate \( V' \)
Taking the square root of both sides:
\[ V' = \sqrt{250} \]
\[ V' = 5\sqrt{10} \, \text{m/s} \]

### Final Answer
The velocity of the car midway between points A and B is \( 5\sqrt{10} \, \text{m/s} \).

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