Question

A car has a mass of 1500 kg. If the driver applies the brakes while on a gravel road, the maximum friction force that the tires can provide without skidding is about 7000 N.

If the car is moving at 22 m/s, what is the shortest distance in which the car can stop safely?

Answer 1

he horizontal forces acting on the car are: the braking (friction) force opposing car's movement and the inertial force which tends to keep the car moving. The two forces have equal intensity and opposite directions. That means that the friction force equals the inertial force:
Ff = Fin
Ff = m*a
Solving for acceleration:
a = Ff/m = 7000/1500 = 4.67 m/s^2
Velocity (car speed), uniform acceleration and space are linked by the following formula:
v^2 = 2as
Solving for s:
s = V^2/(2a) = 22^2/(2*4.67) = 51.82 m
which is the shortest safe stopping distance for the car.

Answer 2

The horizontal forces acting on the car are: the braking (friction) force opposing car's movement and the inertial force which tends to keep the car moving. The two forces have equal intensity and opposite directions. That means that the friction force equals the inertial force:
Ff = Fin
Ff = m*a
Solving for acceleration:
a = Ff/m = 7000/1500 = 4.67 m/s^2
Velocity (car speed), uniform acceleration and space are linked by the following formula:
v^2 = 2as
Solving for s:
s = V^2/(2a) = 22^2/(2*4.67) = 51.83 m
which is the shortest safe stopping distance for the car.

Answer 3

maximum possible accelaration=7000/1500=4.6667m/s²

v²=u²+2as

0=22²-2*4.6667*s

s=51.8534m