Question

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

• A. 20
• B. 10
• C. 5
• D. 4

Answer 1

Answer: (A)

20

Answer 2

$v = \sqrt { \mu g r } \Rightarrow r = \frac { v ^ { 2 } } { \mu g } = \frac { 100 } { 0.5 \times 10 } = 20$