Question

The driver of a 1560-kg Toyota Avalon traveling at 24 m/s (53 mph) on a level, paved road hits the brakes to stop for a red light, locking up the brakes and skidding. Determine the minimum distance needed to stop the car if the coefficient of kinetic friction between the car tires and road is 0.80.

Answer 1

Answer:

The minimum distance needed to stop the car S = 36.68 meter

Explanation:

Mass of the car m = 1560 kg

Initial velocity u = 24 $\frac{m}{sec}$

Final velocity v = 0

Coefficient of kinetic friction between the car tires and road is =  0.80

We know that acceleration of the car is given by

$a = \mu g$ ----- (1)

Where $\mu$  =  Coefficient of kinetic friction between the car tires and road.

& g = 9.81 $\frac{m}{s^{2} }$

⇒ Acceleration a = 0.8 × 9.81 = 7.848 $\frac{m}{s^{2} }$

From the third law of motion

$V^{2} = u ^{2} - 2 a S$ ------ (1)

Negative sign is due to speed is decreasing of the car.

Put all the values in equation (1),

⇒ $0 = 24^{2}$ - 2 × 7.848 × S

⇒ 15.7 S = 576

⇒ S = 36.68 meter

This is the minimum distance needed to stop the car.