Question

A motor cyclist moving with a velocity of $72km/hr$ on a flat road takes a turn on the road at a point where the radius of curvature of the road is $20$ meters. The acceleration due to gravity is $10m/s^2$. In order to avoid skidding, he must not bend with respect to the vertical plane by an angle greater than

• A. $q=ta{n}^{-1}6$
• B. $q=ta{n}^{-1}2$
• C. $q=ta{n}^{-1}25.92$
• D. $q=ta{n}^{-1}4$

Answer 1

Answer: (A)

Given : $R=20$ m

Velocity of the motor cyclist $v=72km/hr=72\times \frac{5}{18}=20m/s$

Let the motorcyclist bends w.r.t vertical plane by angle $\theta$

To avoid skidding, $\frac{v^2}{gR}\ge tan\theta$

$\therefore$ $tan\theta \le \frac{{20}^2}{10\times 20}$

$⟹\theta \le ta{n}^{-1}\left(2\right)$