A vehicle whose wheel track is 1.7m wide and whose centre of gravity 1 m above the road and central between the wheels, taken a curve whose radius 50m, on a level road. The speed at which the inner wheel would leave the road is close to `n xx 5m//s`.

Answer

To solve the problem, we need to determine the speed at which the inner wheel of the vehicle would leave the road while taking a curve. We will use the formula for the critical velocity to prevent overturning of the vehicle.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Wheel track (width between wheels), \( a = 1.7 \, \text{m} \)
- Height of the center of gravity, \( h = 1 \, \text{m} \)
- Radius of the curve, \( r = 50 \, \text{m} \)

2. **Calculate Half of the Wheel Track**:
- Since the center of gravity is central between the wheels, we need to find half of the wheel track:
\[ \frac{a}{2} = \frac{1.7}{2} = 0.85 \, \text{m} \]

3. **Use the Formula for Critical Velocity**:
- The formula for the critical velocity \( v \) to prevent overturning is given by:
\[ v = \sqrt{\frac{g \cdot \frac{a}{2} \cdot r}{h}} \]
where \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

4. **Substitute the Values into the Formula**:
- Plugging in the values:
\[ v = \sqrt{\frac{9.8 \cdot 0.85 \cdot 50}{1}} \]

5. **Calculate the Value Inside the Square Root**:
- First, calculate \( 9.8 \cdot 0.85 \cdot 50 \):
\[ 9.8 \cdot 0.85 = 8.33 \]
\[ 8.33 \cdot 50 = 416.5 \]

6. **Take the Square Root**:
- Now, calculate the square root:
\[ v = \sqrt{416.5} \approx 20.4 \, \text{m/s} \]

7. **Express the Speed in Terms of \( n \)**:
- According to the problem, the speed can be expressed as \( v = n \cdot 5 \, \text{m/s} \).
- Setting \( 20.4 = n \cdot 5 \):
\[ n = \frac{20.4}{5} \approx 4.08 \]

### Final Answer:
The value of \( n \) is approximately \( 4.08 \).

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