Question

A car is moving at high speed along a highway when the driver makes an emergency braking. The wheels become locked (stop rolling), and the resulting skid marks are 32.0 meters long. If the coefficient of kinetic friction between tires and road is 0.550, and the acceleration was constant during braking, how fast was the car going when the wheels became locked?

(a) 20 m/s
(b) 25 m/s
(c) 30 m/s
(d) 35 m/s

Answer 1

The car was going at approximately 30 m/s when the wheels became locked. Therefore, the correct option is c.

Step-by-step explanation:

During emergency braking, when the wheels become locked, the car experiences kinetic friction between the tires and the road. The equation that relates the distance traveled during braking (skid marks), initial velocity, final velocity, acceleration, and coefficient of kinetic friction is:

`\[ d = (v_i^2)/(2 \mu g) \]`

`\[ v_i = √(2 \mu g d) \]`

Substituting the given values:

`\[ v_i = √(2 * 0.550 * 9.8 * 32.0) \]`

Calculating the result:

`\[ v_i \approx 29.84 \, \text{m/s} \]`

Therefore, the car was going at approximately 30 m/s when the wheels became locked, corresponding to option (c).

Therefore, the correct option is c.