A car initially traveling at 23.5 m/s comes to a sliding stop on a wet street. The coefficient of kinetic friction between the street and the tires is 0.1. How long will it take for the car to come to a stop? How far will it travel during this time?

t = 23.96 s

s = 281.5 m

Explanation:

We can calculate the deceleration caused by the kinetic friction knowing its coefficient is 0.1 and applying Newton's 2nd law:

where m is the car mass. Friction force is the product of coefficient and normal force

and normal force has the same magnitude as gravity force, which is product of mass m and gravitational acceleration g = 9.81 m/s2

The time it would take to deceleration from 23.5 m/s to 0 m/s at the rate of 0.981 m/s2 is

We can use this to calculate the distance it travels:

It will take 23.5s for the car to come to a stop.

The car will travel a distance of 276.125m during this time.

According to Newton's second law of motion, the net force (F) acting on a body is the product of the mass(m) of the body and the acceleration(a) of the body. i.e

F = m x a

In other words, the net horizontal force () on the body is given by;

= m x a

And the net vertical force () on the body is given by;

= m x a

Now, from the question, the motion of the car is such that there is no motion in the vertical direction but the motion in the horizontal direction is opposed by a frictional force().

But;

= - μ x N

Where;

The negative sign shows that the force opposes motion

N = normal reaction = m x g [since there is no vertical motion]

m = mass of the body (car in this case)

g = acceleration due to gravity.

μ = coefficient of kinetic friction between the street and the tires = 0.1

Since the only force acting in the horizontal direction is the frictional force, then it implies that the net horizontal force is the frictional force. i.e

= =  - μ x N = m x a        [replace N with m x g]

- μ x m x g = m x a           [divide through by m]

- μ x g = a                ------------------------(ii)

Let's calculate the acceleration of the car by substituting the values of μ = 0.1 and take g = 10m/s² into equation (ii) as follows;

- 0.1 x 10 = a

=> a = -1m/s²       [The negative sign confirms that the body is decelerating]

(a) Now, let's calculate the time taken for the car to come to a stop by using one of the equations of motion as follows;

v = u + at           -------------------(iii)

Where;

v = final velocity of the car = 0   [since it comes to a stop]

u = initial velocity of the car  = 23.5m/s    [given in the question]

a = -1m/s                            [calculated above]

Substitute these values into equation (iii) as follows;

0 = 23.5 + (-1)t

=> 0 = 23.5 - t

=> t = 23.5s

Therefore, the car will come to a stop in 23.5s

(b)  Also, let's calculate how far (distance covered) the car will travel during this time by using one of the equations of motion as follows;

s = ut + at²                ------------(iv)

Where;

s = distance covered by the car

u = initial velocity of the car = 23.5m/s                [given in the question]

t = time taken for the motion = 23.5s                    [as calculated above]

a = acceleration of the car = -1m/s²                     [as calculated above]

Now substitute these values into equation (iv) as follows;

s = 23.5(23.5) + (-1)(23.5)²

s = 23.5(23.5) - (1)(23.5)²

s = 552.25 - (552.25)

s = 552.25 - 276.125

s = 276.125m

Therefore, the distance traveled by the car during this time is 276.125m

the cellphone, they would then know if the killer would have contacted them or always hated them. or if it was a breakup. but if that is not the case, then it would be the clothing. to put in a bullet, it would have to be put in by hands. hands have finger prints which, would have giving you the answer of who done it.

explanation:

a longitudinal wave moves energy that is to the direction of the wave’s movement. - true