Answer:
Part a) Force on car = 2833.84 Newtons
Part b) Time to stop the car = 3.8 seconds
Part c) Factor for stopping distance is 4.
Part d) Factor for stopping time is 1.
Explanation:
The deceleration produced when the car is brought to rest in 20 meters can be found by third equation of kinematics as
where
v = final speed of the car ( = 0 in our case since the car stops)
u = initial speed of the car = 38 km/hr =
a = deceleration produces
s = distance in which the car stops
Applying the given values we get
Now the force can be obtained using newton's second law as
Applying values we get
The negative direction indicates that the force is opposite to the motion of the object.
Part b)
The time required to stop the car can be found using the first equation of kinematics as
with symbols having the same meanings
Applying values we get
Part c)
From the developed relation of stopping distance we can see that the for same force( Same acceleration) the stopping distance is proportional to the square of the initial speed thus doubling the initial speed increases the stopping distance 4 times.
Part d)
From the relation of stopping time and the initial speed we can see that the stopping distance is proportional initial speed thus if we double the initial speed the stopping time also doubles.