Question

Your older brother approaches a stop sign at a velocity of 14 m/s [E] as the light turns amber, He applies the brakes to get the maximum stopping force while avoiding skidding. The car has a mass of 1500 kg, and the coefficient of friction between the tires and the road is 1.07. Ignoring your brother's reaction time, calculate:
a) The maximum deceleration of the car
b) The stopping distance​

Answer 1

The maximum deceleration of the car is 10.5 m/$s^{2}$ approximately. While the stopping distance is 9.3 m

What is Stopping Force ?

Stopping force is the force required to stop the object in motion. Stopping force is tantamount to frictional force.

Given that a brother approaches a stop sign at a velocity of 14 m/s [E] as the light turns amber, He applies the brakes to get the maximum stopping force while avoiding skidding. The car has a mass of 1500 kg, and the coefficient of friction between the tires and the road is 1.07. Ignoring your brother's reaction time.

The given parameters are;

• Velocity V = 14 m/s

• Mass m = 1500 kg

• Coefficient of friction μ = 1.07

• Deceleration a = ?

• Stopping distance S = ?

The normal reaction N = mg

N = 1500 x 9.8

N = 14700 N

The stopping force = μN = ma

1.07 x 14700 = 1500a

15729 = 1500a

a = 15729/1500

a = 10.486 m/$s^{2}$

From 3rd equation of motion

$V^{2}$ = $U^{2}$ + 2aS

where U = 0

$14^{2}$ = 2 x 10.486 x S

196 = 20.972S

S = 196/20.972

S = 9.3 m

Therefore, the maximum deceleration of the car is 10.5 m/$s^{2}$ approximately. While the stopping distance is 9.3 m

Learn more about Force here: brainly.com/question/8119756

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