1. A car of mass 4 * 10 3 kg traveling at 72 km/hr on a horizontal road is brought to rest in a distance of 100m by the action of brakes and frictional forces. Calculate: (a) average stopping force (b) Time taken to stop the car
Answers
Answer:
Motion of vehicles
Stopping vehicles as quickly as possible in an emergency is important but many factors affect this. The driver’s reactions and the road and vehicle conditions play a part, as well as mass and speed.
Part of
Physics (Single Science)Motion and forces
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Page 2 of 2
Calculating stopping distances
It is important to be able to:
estimate how the stopping distance for a vehicle varies with different speeds
calculate the work done in bringing a moving vehicle to rest
The diagram shows some typical stopping distances for an average car in normal conditions.
Bar chart showing the thinking and braking distances of a car at different speeds. The greater the speed, the longer the thinking and braking takes.
It is important to note that the thinking distance is proportional to the starting speed. This is because the reaction time is taken as a constant, and distance = speed × time.
Braking forces
However, the braking distance increases four times each time the starting speed doubles. This is because the work done in bringing a car to rest means removing all of its kinetic energy.
Work done = kinetic energy
Work done = braking force × distance
W=F×d
kineticenergy=12×mass×(velocity)2
KE=12×m×v2
This means that:
F×d=12×m×v2
So for a fixed maximum braking force, the braking distance is proportional to the square of the velocity.
Example thinking distance calculation
A car travels at 12 m/s. The driver has a reaction time of 0.5 s and sees a cat run into the road ahead. What is the thinking distance as the driver reacts?
distance = speed × time
d=v×t
d=12m/s×0.5s
thinkingdistance=6m
Example braking distance calculation
The car in the previous example has a total mass of 900 kg. With a braking force of 2,000 N, what will the braking distance be?
F×d=12×m×v2
d=12×m×v2F
d=12×900×1222,000
brakingdistance=32m
Example stopping distance calculation
What is the stopping distance for the car above?
stopping distance = thinking distance + braking distance
stopping distance = 6 + 32
stopping distance = 38 m
Question
Explanation:
Given,
Mass of car, M=2000kg
Velocity, V=72×
18
5
=20ms
−1
Apply kinematic equation of motion
v
2
−u
2
=2as
0−20
2
=2a×20
a=−10ms
−2
Breaking force, F=ma=2000×10=20 kN
Apply first kinematic equation
v=u+at
t=
a
v−u
=
−10
0−20
=2 sec
Breaking force is 20 kNand time is 2sec.