Question

A car of 800 kg is taking a turn on a banked road of radius 300 m and angle of banking 30°. If the coefficient of static friction is 0.2, then the maximum speed with which the car can negotiate the turn safely: (Given $g = 10 \, \text{m/s}^2$, $\sqrt{3} = 1.73$).

• A. 70.4 m/s
• B. 51.4 m/s
• C. 264 m/s
• D. 102.8 m/s

Answer 1

The Correct Option is B

Given:
\[m = 800 \, \text{kg}, \quad r = 300 \, \text{m}, \quad \theta = 30^\circ, \quad \mu = 0.2\]
The maximum speed \( V_{\text{max}} \) for safe negotiation of the turn is given by:
\[V_{\text{max}} = \sqrt{r g \frac{\tan \theta + \mu}{1 - \mu \tan \theta}}\]
Substitute the values:
\[V_{\text{max}} = \sqrt{300 \times 10 \times \frac{\tan 30^\circ + 0.2}{1 - 0.2 \times \tan 30^\circ}}\]
\[= \sqrt{300 \times 10 \times \frac{0.57 + 0.2}{1 - 0.2 \times 0.57}}\]
\[= \sqrt{3000 \times \frac{0.77}{0.886}}\]
\[36\]
\[V_{\text{max}} = 51.4 \, \text{m/s}\]