Question

A car turns a corner on a slippery road at a constant speed of $10m/s$. If the coefficient of friction is $0.5$, the minimum radius of the arc in meter in which the car turns is

• A. $20m$
• B. $10m$
• C. $5m$
• D. $4m$

Answer 1

The correct option is A

$20m$

Explanation for the correct option:

Step 1: Given data:

Velocity of the car,$v_{max}=10m/s$

coefficient of friction, $\mu =0.5$

Let gravitational acceleration be $g=10m/s^2$

Step 1: Finding radius of the arc:

Using $v_{max}=\sqrt{\mu gR}$

where $R$ is the radius of the arc

Squaring on both sides, we get

$$\begin{gathered} R=\frac{{v_{max}}^2}{\mu g} \\ =\frac{100}{(0.5x10)} \\ =20m \end{gathered}$$

Hence, option A is correct.