A car of mass `1000kg` negotiates a banked curve of radius `90m` on a fictionless road. If the banking angle is `45^(@)` the speed of the car is:

Options

A. `20 ms^(-1)`
B. `30 ms^(-1)`
C. `5 ms^(-1)`
D. `10 ms^(-1)`

Explanation

To find the speed of a car negotiating a banked curve on a frictionless road, we can use the relationship between the banking angle, the radius of the curve, and the speed of the car. Here’s a step-by-step solution:

### Step 1: Understand the relationship
For a banked curve without friction, the relationship between the banking angle (θ), the radius of the curve (R), and the speed of the vehicle (V) can be given by the formula:
\[ \tan(\theta) = \frac{V^2}{Rg} \]
where:
- \( \theta \) = banking angle
- \( R \) = radius of the curve
- \( g \) = acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \))

### Step 2: Identify the given values
From the problem, we have:
- Mass of the car (m) = 1000 kg (not needed for this calculation)
- Radius of the curve (R) = 90 m
- Banking angle (θ) = 45 degrees
- Acceleration due to gravity (g) = 10 m/s²

### Step 3: Calculate the tangent of the banking angle
Since \( \theta = 45^\circ \):
\[ \tan(45^\circ) = 1 \]

### Step 4: Substitute the values into the formula
Now substituting the known values into the equation:
\[ 1 = \frac{V^2}{90 \times 10} \]

### Step 5: Simplify the equation
This simplifies to:
\[ 1 = \frac{V^2}{900} \]

### Step 6: Solve for V²
Multiplying both sides by 900 gives:
\[ V^2 = 900 \]

### Step 7: Solve for V
Taking the square root of both sides, we find:
\[ V = \sqrt{900} = 30 \, \text{m/s} \]

### Conclusion
The speed of the car is \( 30 \, \text{m/s} \).

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