A car moves up a hill of incline 1 in 50 with a velocity of 15KMPH. The resistance due to friction is `1/25` th of weight of the car. The speed of the car while moving down hill using same power is

Options

A. 15 KMPH
B. 30 KMPH
C. 45 KMPH
D. 60 KMPH

Answer

To solve the problem step by step, we need to analyze the forces acting on the car when it is moving uphill and then use that information to find the speed of the car while moving downhill using the same power.

### Step 1: Understand the incline and forces acting on the car
The incline of the hill is given as 1 in 50, which means for every 50 units of horizontal distance, the car rises 1 unit vertically.

Using this information, we can find the angle of incline (θ) using the formula:
\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{1^2 + 50^2}} = \frac{1}{\sqrt{2501}} \approx 0.02 \]
This approximation indicates a small angle.

### Step 2: Calculate the power used by the car while moving uphill
The power (P) used by the car can be calculated using the formula:
\[ P = F \cdot v \]
where \( F \) is the total force acting against the motion of the car and \( v \) is the velocity.

The forces acting on the car while moving uphill include:
1. The component of gravitational force acting down the incline: \( mg \sin \theta \)
2. The frictional force, which is \( \frac{1}{25} \) of the weight of the car: \( \frac{mg}{25} \)

Thus, the total force \( F \) can be expressed as:
\[ F = mg \sin \theta + \frac{mg}{25} \]

Now, substituting \( v = 15 \text{ km/h} = \frac{15 \times 1000}{3600} \text{ m/s} = 4.17 \text{ m/s} \):
\[ P = \left( mg \sin \theta + \frac{mg}{25} \right) v \]

### Step 3: Substitute values to find power
Assuming \( m = 25 \text{ kg} \) (for simplicity, we can assume a mass), and \( g = 10 \text{ m/s}^2 \):
\[ \sin \theta \approx 0.02 \rightarrow mg \sin \theta \approx 25 \times 10 \times 0.02 = 5 \text{ N} \]
\[ \frac{mg}{25} = \frac{25 \times 10}{25} = 10 \text{ N} \]
Thus,
\[ F = 5 + 10 = 15 \text{ N} \]
Now, calculating power:
\[ P = 15 \times 4.17 = 62.55 \text{ W} \]

### Step 4: Calculate the speed while moving downhill
When the car is moving downhill, the forces acting on it are:
1. The component of gravitational force acting down the incline: \( mg \sin \theta \)
2. The frictional force acting against the motion: \( \frac{mg}{25} \)

The power while moving downhill can be expressed as:
\[ P = \left( mg \sin \theta - \frac{mg}{25} \right) v' \]
where \( v' \) is the speed while moving downhill.

Setting the powers equal:
\[ 62.55 = \left( 5 - 10 \right) v' \]
\[ 62.55 = -5 v' \]
\[ v' = \frac{62.55}{5} = 12.51 \text{ m/s} \approx 45 \text{ km/h} \]

### Final Answer
The speed of the car while moving downhill using the same power is approximately **45 km/h**.

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