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"^(-1)`
B. `20sqrt2 " km h"^(-1)`
C. `25sqrt2 " km h"^(-1)`
D. `35 " km h"^(-1)`

Answer

To solve the problem step by step, we will use the equations of motion.

### Step 1: Identify the given values
- Initial velocity at point P (U) = 30 km/h
- Final velocity at point Q (V) = 40 km/h
- Distance between points P and Q = S (unknown for now)

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

### Step 3: Use the third equation of motion
We will use the third equation of motion, which states:
\[ V^2 = U^2 + 2aS \]
Here, we need to find the acceleration (a) and the distance (S) between points P and Q.

### Step 4: Substitute known values into the equation
Substituting the known values into the equation:
\[ (11.11)^2 = (8.33)^2 + 2aS \]
Calculating the squares:
\[ 123.4561 = 69.3889 + 2aS \]
Rearranging gives:
\[ 2aS = 123.4561 - 69.3889 = 54.0672 \]
Thus,
\[ aS = \frac{54.0672}{2} = 27.0336 \quad \text{(Equation 1)} \]

### Step 5: Find the velocity at the midpoint
Now, we need to find the velocity at the midpoint (O) between points P and Q. The distance from P to O is \( \frac{S}{2} \).

Using the same equation of motion for the distance \( \frac{S}{2} \):
\[ V_O^2 = U^2 + 2a\left(\frac{S}{2}\right) \]
Substituting the values:
\[ V_O^2 = (8.33)^2 + 2a\left(\frac{S}{2}\right) \]
This simplifies to:
\[ V_O^2 = (8.33)^2 + aS \]
From Equation 1, we know that \( aS = 27.0336 \):
\[ V_O^2 = 69.3889 + 27.0336 \]
Calculating this gives:
\[ V_O^2 = 96.4225 \]
Taking the square root to find \( V_O \):
\[ V_O = \sqrt{96.4225} \approx 9.82 \, \text{m/s} \]

### Step 6: Convert back to km/h
Finally, converting \( V_O \) back to km/h:
\[ V_O = 9.82 \times \frac{3600}{1000} \approx 35.39 \, \text{km/h} \]

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

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