The maximum deceleration of the car is 10.5 m/[tex]s^{2}[/tex] 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;
- Coefficient of friction μ = 1.07
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/[tex]s^{2}[/tex]
From 3rd equation of motion
[tex]V^{2}[/tex] = [tex]U^{2}[/tex] + 2aS
where U = 0
[tex]14^{2}[/tex] = 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/[tex]s^{2}[/tex] approximately. While the stopping distance is 9.3 m
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