A road is 8 m wide. Its radius of curvature is 40 m. The outer edge is above the lower edge by a distance of 1.2 m. This road is most suited for a velocity of

Options

A. `5.7m//s`
B. `7.4m//s`
C. `36.1m//s`
D. `9.7m//s`

Answer

To solve the problem, we need to determine the velocity at which a vehicle can safely navigate a curved road with a given radius of curvature and banking angle. Here are the steps to find the solution:

### Step 1: Understand the Geometry of the Road
The road is banked, meaning that the outer edge is higher than the inner edge. The width of the road is 8 m, and the outer edge is elevated by 1.2 m compared to the inner edge.

### Step 2: Calculate the Banking Angle
We can find the banking angle (θ) using the height difference and the width of the road. The relationship can be expressed as:

\[ \tan(\theta) = \frac{\text{height difference}}{\text{width}} = \frac{1.2 \, \text{m}}{8 \, \text{m}} \]

Calculating this gives:

\[ \tan(\theta) = 0.15 \]

Now, we can find the angle θ:

\[ \theta = \tan^{-1}(0.15) \]

### Step 3: Use the Formula for Banking of Roads
The formula for the maximum velocity (v) on a banked curve without friction is given by:

\[ v = \sqrt{g \cdot r \cdot \tan(\theta)} \]

Where:
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( r \) is the radius of curvature (40 m),
- \( \tan(\theta) \) is the value we calculated earlier (0.15).

### Step 4: Substitute the Values into the Formula
Now we can substitute the values into the formula:

\[ v = \sqrt{9.81 \, \text{m/s}^2 \cdot 40 \, \text{m} \cdot 0.15} \]

Calculating this step-by-step:

1. Calculate \( g \cdot r \cdot \tan(\theta) \):
\[ 9.81 \cdot 40 \cdot 0.15 = 58.86 \, \text{m}^2/\text{s}^2 \]

2. Now take the square root:
\[ v = \sqrt{58.86} \approx 7.67 \, \text{m/s} \]

### Final Answer
The road is most suited for a velocity of approximately **7.67 m/s**.

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