A motorcyclist goes round a circular race curve at `162 kmh^(-1)`. To keep his balance he leans inward 40° from the vertical. What is the radius of the circular path ?

Explanation

To solve the problem, we need to find the radius of the circular path that the motorcyclist is taking while leaning inward at an angle of 40° from the vertical. We can use the relationship between the speed of the motorcycle, the angle of lean, and the radius of the circular path.

### Step-by-Step Solution:

1. **Convert Speed from km/h to m/s**:
The speed of the motorcyclist is given as 162 km/h. We need to convert this speed into meters per second (m/s) using the conversion factor:
\[ \text{Speed in m/s} = \text{Speed in km/h} \times \frac{5}{18} \]
\[ v = 162 \times \frac{5}{18} = 45 \, \text{m/s} \]

2. **Identify the Angle of Lean**:
The motorcyclist leans inward at an angle of 40° from the vertical. This angle will be used in our calculations.

3. **Use the Formula Relating Speed, Radius, and Angle**:
The relationship between the speed \( v \), the radius \( r \), and the angle \( \theta \) is given by:
\[ \tan(\theta) = \frac{v^2}{rg} \]
Rearranging this formula to solve for the radius \( r \):
\[ r = \frac{v^2}{g \tan(\theta)} \]

4. **Calculate the Tangent of the Angle**:
We need to find \( \tan(40°) \). Using a calculator or trigonometric tables, we find:
\[ \tan(40°) \approx 0.8391 \]

5. **Substitute Values into the Formula**:
Now we can substitute the values into the formula for \( r \):
\[ g = 9.8 \, \text{m/s}^2 \quad (\text{acceleration due to gravity}) \]
\[ r = \frac{(45)^2}{9.8 \times 0.8391} \]
\[ r = \frac{2025}{8.2278} \approx 246.2 \, \text{m} \]

6. **Final Answer**:
The radius of the circular path is approximately:
\[ r \approx 246.2 \, \text{meters} \]