Question

If the coefficient of kinetic friction between tires and dry pavement is 0.93. What is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 33.6 m/s ?

Answer 1

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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