The shortest distance over which a car moving at 33.6 m/s may be stopped by locking the brakes is 61.93 m
The kinetic friction coefficient is 0.93.
using N, where N is the normal reaction = mg, frictional force can be evaluated.
Fr =- 0.93 x 9.8 = -9.114 m/s²
Friction, which takes the form of a decelerating force, helps the car come to a stop.
This frictional force equals the ma force,
where ma = 7.84 m and a = 7.84 m/s2.
However, this acceleration is negative at 7.84 m/s2 since it is decelerating.
So, using the equations of motion,
v² = u² + 2ax
where;
v = final velocity
u = initial velocity
a = acceleration
0 = 33.6² + 2(-9.114)x
0 = 1128.96 - 18.228x
18.228x = 1128.96
x = 1128.96/18.228
x = 61.93 m
Therefore, the shortest distance is 61.93 m
To know more about shortest distance refer to: brainly.com/question/16751593
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