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 total distance traveled by the car is

30km + 49km = 79 km .

massxgravity x heght=weightxheight= pe

4900=490xh

h=10