An unbanked circular highway curve on the level ground makes a turn of `90^(@)`. The highway carries traffic at `"108 km h"^(-1)`, and the centripetal force on a vehicle is not to exceed `(1)/(10)` of its weight. What is the approximate minimum length of he curve, in km?

Options

A. 1.41 km
B. 1 km
C. 0.6 km
D. None of these

Answer

To solve the problem, we need to determine the minimum length of a 90-degree circular highway curve given the speed of traffic and the maximum allowed centripetal force.

### Step-by-Step Solution:

1. **Identify Given Data**:
- Speed of the vehicle, \( v = 108 \, \text{km/h} \)
- Maximum centripetal force, \( F_c \leq \frac{1}{10} W \) (where \( W \) is the weight of the vehicle)
- Weight of the vehicle, \( W = mg \)

2. **Convert Speed to Meters per Second**:
\[ v = 108 \, \text{km/h} = 108 \times \frac{1000 \, \text{m}}{3600 \, \text{s}} = 30 \, \text{m/s} \]

3. **Centripetal Force Equation**:
The centripetal force required to keep the vehicle moving in a circular path is given by:
\[ F_c = \frac{mv^2}{r} \]
According to the problem, this force should not exceed \( \frac{1}{10} W \):
\[ \frac{mv^2}{r} \leq \frac{1}{10} mg \]

4. **Simplify the Equation**:
Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{v^2}{r} \leq \frac{1}{10} g \]
Rearranging gives:
\[ r \geq \frac{10v^2}{g} \]

5. **Substitute Values**:
Use \( g \approx 10 \, \text{m/s}^2 \):
\[ r \geq \frac{10 \times (30)^2}{10} = 900 \, \text{m} \]

6. **Calculate the Length of the Curve**:
The length of a quarter circle (90 degrees) is given by:
\[ L = \frac{1}{4} \times 2\pi r = \frac{\pi r}{2} \]
Substituting \( r = 900 \, \text{m} \):
\[ L = \frac{\pi \times 900}{2} \approx \frac{3.14 \times 900}{2} \approx 1413 \, \text{m} \]

7. **Convert Length to Kilometers**:
\[ L \approx 1.413 \, \text{km} \]

8. **Final Answer**:
The approximate minimum length of the curve is:
\[ \text{Length} \approx 1.41 \, \text{km} \]