For a circular path of radius 300 m, the super elevation is restricted to 0.1 m for a width of 1.6 m. The maximum speed, in m/s, of vehicle to avoid overturn is ____ (round off up... For a circular path of radius 300 m, the super elevation is restricted to 0.1 m for a width of 1.6 m. The maximum speed, in m/s, of vehicle to avoid overturn is ____ (round off up to 2 decimals) Read more

Steps to Solve

1. Identify Given Values

The problem provides the following values:

• Radius of the circular path, $R = 300 \text{ m}$
• Super elevation, $e = 0.1 \text{ m}$
• Width of the super elevation, $w = 1.6 \text{ m}$

2. Calculate Effective Height of Super Elevation

To find the effective height of the road, we can use the following formula: $$ h = e + \frac{w}{2} $$ Calculating: $$ h = 0.1 + \frac{1.6}{2} = 0.1 + 0.8 = 0.9 \text{ m} $$

3. Use the Formula for Maximum Safe Speed

The formula for the maximum speed ( V ) that a vehicle can travel without overturning, considering centrifugal force and gravitational force, is given by: $$ V = \sqrt{g \cdot R \cdot \left( \frac{e + \frac{w}{2}}{R} \right)} $$ Where ( g \approx 9.81 \text{ m/s}^2 ).

4. Substitute the Values into the Formula

Substituting the calculated effective height and given radius into the formula: $$ V = \sqrt{9.81 \cdot 300 \cdot \left( \frac{0.9}{300} \right)} $$ Simplifying: $$ V = \sqrt{9.81 \cdot 300 \cdot 0.003} $$ $$ V = \sqrt{9.81 \cdot 0.9} $$

5. Calculate the Value of V

Now calculating the value: $$ V = \sqrt{8.829} \approx 2.973 \text{ m/s} $$

6. Round the Answer

Finally, round the answer to two decimal places: $$ V \approx 2.97 \text{ m/s} $$