Question

If the coefficient of kinetic friction between tires and dry pavement is 0.800, what is the shortest distance in which an automobile can be stopped by locking the brakes when traveling at 29.4 m/s ?

Answer 1

54.02m

Step-by-step explanation:

According to Newton's second law of motion;

∑F = m x a ------------------(i)

Where;

∑F = effective force acting on a body

m = mass of the body

a = acceleration of the body

(i) According to the question, the only force acting on the horizontal (direction of motion) is the frictional force (
`F_(R)`) between the tires of the automobile and the dry pavement and it is directed opposite the direction of motion. Therefore, it is negative.

i.e ∑F = -
`F_(R)`

Where;

`F_(R)` = μ
`_(K)` x N

μ
`_(K)` = coefficient of kinetic friction.

N = normal reaction = m x g

m = mass of the automobile as depicted by the tires and

g = acceleration due to gravity = 10m/s²

=> ∑F = - μ
`_(K)` x m x g

Substitute the value of ∑F into equation (i) as follows;

- μ
`_(K)` x m x g = m x a

Divide through by m;

- μ
`_(K)` x g = a ------------------(ii)

Given;

μ
`_(K)` = Coefficient of kinetic friction between tires and dry = 0.800

Substitute this value into equation (ii) as follows;

- 0.800 x 10 = a

a = - 8m/s²

Therefore, the acceleration of the automobile is -8m/s² [The negative sign shows deceleration]

(ii) Now, the distance (s) is gotten from one of the equations of motion as follows;

v² = u² + 2as ---------------------(iii)

Where

v = final velocity of the automobile = 0 [since the automobile is stopped]

u = initial velocity of the automobile = 29.4m/s

a = acceleration of the automobile = - 8m/s²

Substitute these values into equation (iii) as follows;

0² = 29.4² + [2(-8) x s]

0 = 864.36 - 16s

16s = 864.36

Solve for s;

s =
`(864.36)/(16)`

s = 54.02m

Therefore, the shortest distance in which the automobile can be stopped by locking the brakes when traveling at 29.4m/s is 54.02m

Answer 2

The shortest distance in which an automobile can be stopped by locking the brakes when traveling at 29.4 m/s = 55.125 m = 55.13 m

Step-by-step explanation:

Coefficient of kinetic friction = 0.8

Frictional force can be evaluated using μN

where N is the normal reaction = mg

Fr = 0.8 m × 9.8 = 7.84 m

The car stops with th help of friction, this friction manifests in a form of decelerating force.

This frictional force will be equal to the ma force,

ma = 7.84 m

a = 7.84 m/s²

But Since it's deceleration, this acceleration is - 7.84 m/s²

So, using the equations of motion,

u = initial velocity = 29.4 m/s

v = final velocity = 0 m/s (since it comes to a stop finally)

x = distance travelled before stopping

a = - 7.84 m/s²

v² = u² + 2ax

0² = 29.4² + 2(-7.84)(x)

15.68 x = 864.36

x = 55.125 m = 55.13 m