Answer:
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:
[tex]a_f = F_f/m[/tex]
where m is the car mass. Friction force is the product of coefficient and normal force
[tex]a_f = \frac{\mu N}{m}[/tex]
and normal force has the same magnitude as gravity force, which is product of mass m and gravitational acceleration g = 9.81 m/s2
[tex]a_f = \frac{\mu mg}{m} = \mu g = 0.1*9.81 = 0.981 m/s^2[/tex]
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
[tex]t = \Delta v / a_f = (23.5 - 0)/0.981 = 23.96 s[/tex]
We can use this to calculate the distance it travels:
[tex]s = vt + at^2/2 = 23.5 * 23.96 - 0.981*23.96^2/2 = 281.5 m[/tex]
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 ([tex]F_{X}[/tex]) on the body is given by;
[tex]F_{X}[/tex] = m x a
And the net vertical force ([tex]F_{Y}[/tex]) on the body is given by;
[tex]F_{Y}[/tex] = 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([tex]F_{R}[/tex]).
But;
[tex]F_{R}[/tex] = - 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
[tex]F_{X}[/tex] = [tex]F_{R}[/tex] = - 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 + [tex]\frac{1}{2}[/tex]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) + [tex]\frac{1}{2}[/tex](-1)(23.5)²
s = 23.5(23.5) - [tex]\frac{1}{2}[/tex](1)(23.5)²
s = 552.25 - [tex]\frac{1}{2}[/tex](552.25)
s = 552.25 - 276.125
s = 276.125m
Therefore, the distance traveled by the car during this time is 276.125m
