A car weight W is on an inclined road that rises by 100 m over a distance of 1 km and applies a constant frictional force `W/20 ` on the car . While moving uphill on the road at a speed of `10 " m "s^(-1)` , the car needs power P . If it needs power `P/2 ` while moving downwill at speed v then value of v is :

Options

A. `20 " m"s^(-1)`
B. `15 " m "s^(-1)`
C. `10 " m "s^(-1)`
D. `5 " m "s^(-1)`

Explanation

To solve the problem step by step, we will analyze the forces acting on the car while it is moving uphill and downhill, and then calculate the required power in each case.

### Step 1: Understand the Forces Acting on the Car

1. **Weight of the Car (W)**: The weight acts vertically downwards.
2. **Frictional Force**: The frictional force opposing the motion is given as \( \frac{W}{20} \).
3. **Inclined Plane**: The car is moving on an incline that rises by 100 m over a distance of 1 km (1000 m).

### Step 2: Calculate the Angle of Incline

Using the height and distance, we can find the angle \( \theta \):
\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{100}{1000} = 0.1 \]
For small angles, \( \sin \theta \approx \tan \theta \).

### Step 3: Calculate the Component of Weight Parallel to the Incline

The component of the weight acting parallel to the incline is:
\[ W_{\parallel} = W \sin \theta = W \cdot 0.1 \]

### Step 4: Calculate the Power Required While Moving Uphill

When the car is moving uphill at a speed of \( 10 \, \text{m/s} \), the total power \( P \) required is the sum of the power to overcome friction and the power to overcome the component of weight parallel to the incline:
\[ P = \text{Frictional Force} \times \text{Velocity} + W_{\parallel} \times \text{Velocity} \]
Substituting the values:
\[ P = \left(\frac{W}{20}\right) \times 10 + (W \cdot 0.1) \times 10 \]
\[ P = \frac{W \cdot 10}{20} + W \cdot 1 = \frac{W}{2} + W = \frac{3W}{2} \]

### Step 5: Calculate the Power Required While Moving Downhill

When the car is moving downhill at speed \( v \), the power required is \( \frac{P}{2} \):
\[ \frac{P}{2} = \frac{3W}{4} \]
The power required to overcome only the frictional force while moving downhill is:
\[ \frac{W}{20} \times v = \frac{3W}{4} \]

### Step 6: Solve for \( v \)

Now, we can solve for \( v \):
\[ \frac{W}{20} \times v = \frac{3W}{4} \]
Dividing both sides by \( W \) (assuming \( W \neq 0 \)):
\[ \frac{v}{20} = \frac{3}{4} \]
Multiplying both sides by 20:
\[ v = 20 \times \frac{3}{4} = 15 \, \text{m/s} \]

### Final Answer

The speed \( v \) while moving downhill is:
\[ \boxed{15 \, \text{m/s}} \]