A car is moving along a straight road with a uniform acceleration. It passes through two points P and Q separated by a distance with velocity `30km//h` and `40 km//h` respectively. The velocity of the car midway between P and Q is

Options

A. `33.3 km//h`
B. `20sqrt(2)km//h`
C. `25 sqrt(2) km//h`
D. `0.35 km//h`

Answer

To find the velocity of the car midway between points P and Q, we can use the equations of motion. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Initial velocity at point P (U) = 30 km/h
- Final velocity at point Q (V) = 40 km/h

### Step 2: Convert velocities to m/s
To use the equations of motion, we need to convert the velocities from km/h to m/s:
- \( U = 30 \, \text{km/h} = \frac{30 \times 1000}{3600} = \frac{30000}{3600} \approx 8.33 \, \text{m/s} \)
- \( V = 40 \, \text{km/h} = \frac{40 \times 1000}{3600} = \frac{40000}{3600} \approx 11.11 \, \text{m/s} \)

### Step 3: Use the equation of motion
We can use the equation of motion:
\[ V^2 - U^2 = 2aX \]
Where:
- \( V \) = final velocity
- \( U \) = initial velocity
- \( a \) = acceleration
- \( X \) = distance between points P and Q

### Step 4: Calculate \( V^2 - U^2 \)
Calculating \( V^2 \) and \( U^2 \):
- \( V^2 = (11.11)^2 \approx 123.46 \)
- \( U^2 = (8.33)^2 \approx 69.39 \)

Now, calculate \( V^2 - U^2 \):
\[ V^2 - U^2 = 123.46 - 69.39 = 54.07 \]

### Step 5: Express \( 2aX \)
From the equation:
\[ 54.07 = 2aX \]
This means:
\[ aX = \frac{54.07}{2} = 27.035 \]

### Step 6: Find the velocity at the midpoint
Now, we need to find the velocity at the midpoint (Vm). At the midpoint, we can use the same equation of motion:
\[ V_m^2 - U^2 = 2a\left(\frac{X}{2}\right) \]
This simplifies to:
\[ V_m^2 - U^2 = aX \]
Substituting \( aX = 27.035 \):
\[ V_m^2 - (8.33)^2 = 27.035 \]

### Step 7: Calculate \( V_m^2 \)
Now substituting \( U^2 \):
\[ V_m^2 - 69.39 = 27.035 \]
\[ V_m^2 = 27.035 + 69.39 = 96.425 \]

### Step 8: Find \( V_m \)
Taking the square root:
\[ V_m = \sqrt{96.425} \approx 9.82 \, \text{m/s} \]

### Step 9: Convert back to km/h
Convert \( V_m \) back to km/h:
\[ V_m = 9.82 \times \frac{3600}{1000} \approx 35.39 \, \text{km/h} \]

### Final Answer
The velocity of the car midway between points P and Q is approximately **35.39 km/h**.