The circumference of a circular track is 1.256 km . What is the tangent of the angle of banking of the track if the maximum speed at which a car can be safely driven along it is 20 m/s and `g= 10 m//s^(2)`?

Options

A. `(1)/(2) `
B. `1/3`
C. `1/4`
D. `1/5`

Explanation

To solve the problem of finding the tangent of the angle of banking of a circular track, we can follow these steps:

### Step 1: Find the Radius of the Circular Track
Given the circumference \( C \) of the circular track is 1.256 km, we first convert this to meters:
\[ C = 1.256 \, \text{km} = 1.256 \times 1000 \, \text{m} = 1256 \, \text{m} \]
The formula for the circumference of a circle is:
\[ C = 2 \pi r \]
We can rearrange this to find the radius \( r \):
\[ r = \frac{C}{2 \pi} = \frac{1256}{2 \pi} \]
Using \( \pi \approx 3.14 \):
\[ r \approx \frac{1256}{2 \times 3.14} \approx \frac{1256}{6.28} \approx 200 \, \text{m} \]

### Step 2: Use the Formula for Tangent of the Angle of Banking
The formula for the tangent of the angle of banking \( \theta \) is given by:
\[ \tan \theta = \frac{v^2}{g r} \]
where:
- \( v \) is the maximum speed (20 m/s),
- \( g \) is the acceleration due to gravity (10 m/s²),
- \( r \) is the radius of the circular track (200 m).

### Step 3: Substitute the Values into the Formula
Now we can substitute the known values into the formula:
\[ \tan \theta = \frac{(20)^2}{10 \times 200} \]
Calculating the numerator:
\[ (20)^2 = 400 \]
Calculating the denominator:
\[ 10 \times 200 = 2000 \]
Thus:
\[ \tan \theta = \frac{400}{2000} = \frac{1}{5} \]

### Step 4: Conclusion
The tangent of the angle of banking of the track is:
\[ \tan \theta = \frac{1}{5} \]

### Final Answer
The correct option is \( \frac{1}{5} \).

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