If the coefficient of kinetic friction between tires and dry pavement is 0.84, what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 29.0 m/s

The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

Explanation:

Given;

coefficient of kinetic friction, μ = 0.84

speed of the automobile, u = 29.0 m/s

To determine the  the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;

v² = u² + 2ax

where;

v is the final velocity

u is the initial velocity

a is the acceleration

x is the shortest distance

First we determine a;

From Newton's second law of motion

∑F = ma

F is the kinetic friction that opposes the motion of the car

-Fk = ma

but, -Fk = -μN

-μN = ma

-μmg = ma

-μg = a

- 0.8 x 9.8 = a

-7.84 m/s² = a

Now, substitute in the value of a in the equation above

v² = u² + 2ax

when the automobile stops, the final velocity, v = 0

0 = 29² + 2(-7.84)x

0 = 841 - 15.68x

15.68x = 841

x = 841 / 15.68

x = 53.64 m

Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

f=ma newton

f=100x2.5=250n

a.   mass number from atomic number