5. Reaction times and stopping distances e.g. road vehicles
Doc Brown's Physics Revision
Notes
Suitable for GCSE/IGCSE Physics/Science courses or
their equivalent
This page will help answer questions such as ...
What is the formula for stopping
distance?
What factors affect thinking distance?
What factors affect braking distance?
What is the relationship between
braking distance and kinetic energy?
Can you devise a simple experiment to
measure somebody's reaction time?
Introduction  stopping distances
and the speed of
road vehicles
When driving a car you obviously need to be
alert to any sudden change in your situation, particularly if you have to do an
emergency brake to come to a halt.
In this situation you want to stop the
car (or any other road vehicle) in the shortest time as possible to
effect an appropriate emergency stop!
This means applying the maximum force
on the brake pedal.
The longer it takes for you to react
and longer it takes to come to a halt, the greater the risk of crashing
into an object in your path. Everybody's 'thinking' reaction time to a situation
requiring a quick physical response is different, though typically in the range
0.2 to 0.8 seconds. In biology you may have studied
the nervous system including the reflex arc.
The distance it takes to stop a road vehicle
in an emergency situation is given by the following formula:
STOPPING DISTANCE = THINKING DISTANCE
+ BRAKING DISTANCE
The thinking distance is how far
you travel during your reaction time, which is the time interval from you
perceiving a hazard and actually starting to apply the brakes.
The braking distance is the actual
distance you travel from when you first apply the brakes until you come to a
halt.
The chart above gives typical or average values
for thinking distances, braking distances and stopping distance and are
quoted from the UK Highway Code
guidance booklet.
You can see the thinking distance is
quite an appreciable portion of the total stopping distance, particularly at
lower speeds, BUT, look how dramatically the total stopping distance
increases with increase in speed.
These values should be doubled for
wet roads and multiplied by 10 for ice coated roads. Snow would be
somewhere in between, but where?, so just take great care in driving in
any of these adverse driving conditions.
Later in this page I've used this
data to do some graphs and calculations relating stopping distance to
speed and the kinetic energy of a car.
How to calculate thinking distances and braking distances
from velocity  time graphs.
Graphs
1a
You have probably encountered velocity time graphs by now,
so you should know that the area under a section of a velocity time graph
is equal to the distance travelled in that section (in terms of units m/s x
s = m).
The graphs assume the same car and driver
so that the deceleration on maximum braking is the same, which is why the
negative gradient is the same value on both graphs.
The graph on the left of 1a shows an initial situation of a
driver's quicker response time travelling at a lower speed.
Rectangular area A1 = initial
velocity v1 x reaction time t1 = thinking distance
The area A1 is equal to the thinking distance, that is
the distance the vehicle travels in the time it takes the driver to
respond to a situation and starts to apply the brakes.
Right angled triangular area A2 = ½
x initial velocity v1 x braking time t2 = braking distance
The area A2 is the braking distance, that is the
distance the vehicle travels from its maximum initial speed, when
braking starts, until it
comes to a halt.
The total area = A1 + A2 = stopping
distance
The graph on the right of 1a shows a driver's slower reaction
and the vehicle is moving at a greater speed.
This means two factors have been
changed to emphasise how stopping distance is so easily and dramatically
increased.
So v2 > v1 and times t1 and t2 are both
increased, so both areas A1 and A2 are increased.
The purple shaded areas indicate the increase in
thinking distance A1 and braking distance A2.
This might mean lack of care and
attention e.g. tired and not concentrating on the speed limit.
Rectangular area A1 = initial
velocity v2 x reaction time t1 = thinking distance
Right angled triangular area A2 = ½
x initial velocity v2 x braking time t2 = braking distance
So, both area A1 and A2 are greatly
increased, increasing the likelihood of an accident if driving
carelessly!
The total area = A1 + A2 = stopping
distance, and much greater than before.
If you have followed the above
logical arguments, you should be able to interpret the graphs if only one
factors was changed.
Factors affecting the
thinking distance
Speed is the first obvious factor. The faster you are
going, the further you will travel for the same 'best' reaction time you can
manage, so the greater the thinking distance, which you can do nothing
about.
The longer your reaction time, the greater your
thinking distance. You can only keep this to a minimum by being fully alert
and able to respond as fast as your body is able to.
The effects of tiredness and alcohol will affect your
alertness and increase your response time and thinking distance.
There other factors too.
Are you on medication that
might affect your alertness?
Are you distracted by watching/thinking
about something else other than the road ahead?
Are you talking to somebody else
in the car, are the kids being silly?
Even using a mobile phone legally
with a handset, is still potentially a distraction.
Factors affecting the
braking distance
Again, speed is the first obvious factor. The faster
you are going the more kinetic energy has to removed from the car's kinetic
energy store. At a constant rate of braking force, it will take longer the
greater the speed, because more kinetic energy has to be converted to heat
energy in the brake pad and disc system.
This is shown on the right (brake pads P in contact with
disc D).
All of the factors discussed here become particularly
crucial in an emergency braking situation or you suddenly find
yourself too close to the car in front.
The greater your speed, the greater the stopping
distance and the greater distance you should allow between one vehicle
and another e.g. the two chevron distance for 70 mph you see on some
sections of a motorway.
However good your brakes are, its no
good being too close to another vehicle i.e. well within the stopping
distance, if you are to avoid an accident if the vehicle in front does
an emergency brake or the traffic head rapidly comes to halt!
Speed limits aren't simply about speed reduction,
they are also about reducing both the stopping distance where higher
speeds are considered hazardous for a particular section of road. This
for the safety of road users and pedestrians e.g. 20 mph in narrow
streets in built up areas where there are likely to be many people
walking and crossing roads.
Road
condition and the weather: The adverse condition of a road has already
been mentioned. With a dry road (and tyres in good condition) you will get
the maximum friction grip from the tyreroad surface contact on braking,
giving you the minimum distance travelled  the minimum thinking distance.
If the road is wet from rain, coated in snow or ice, the friction grip is
reduced (ice > snow >> standing water all leading to skidding on
braking). Modern tyres do very well on braking if the road is just a bit wet
and no obvious standing water  where can get 'aquaplaning'/'hydroplaning'
as you skid over a layer of water on the road surface. Leaves and split oil
also reduce the friction grip between tyre and road. All of these conditions
reduce the friction grip of tyre on road and increase the braking time and
braking distance
The condition of the tyres: Tyres are designed to
give the maximum road grip and expel water from under the tyres on wet
roads. If the tyres are worn (bald or little tread left), there is less grip
and the vital friction and water expelling function to slow the vehicle down
are reduced and so increasing the braking distance and the chance of
skidding. Also, the tyres should contain enough air to give the correct
operating pressure.
The effectiveness of the brakes: If the brakes are
not well maintained, braking function may be impaired. Brake pads might be
worn or a leak in the hydraulic brake system can be a source braking
impairment. Are the brakes balanced so that you slow down in a straight line
 this is point applies to the condition of the tyres too.
Some graphical
analysis of stopping distances, speed and kinetic energy of a moving car
see
calculations
Diagram KEY: KE = kinetic energy (J),
m = mass (kg), u = initial velocity (m/s),
v = final velocity (m/s), s = speed (m/s)
a = acceleration or
deceleration (m/s^{2}), W = work
done (J), F = force (N),
d = distance (m)
Graph
1b
The graph 1b above takes the thinking distance, braking
distance and stopping distance data and plots them against the typical speed of a road vehicle.
Obviously, all the distances increase with increase in
speed, but note two other very important points.
You should notice ...
(i) two of the graphs curve upwards, so there is a sort
of 'accelerating' effect of speed on the braking distance and overall
stopping distance (the latter is due to the increase in braking
distance).
Stopping distance and braking
distance are not proportional to speed, and crucially, the
braking distance is proportional to speed squared. This means
the stopping/braking distances increase faster than the increase in
speed.
e.g. doubling speed
quadruples the braking distance (2 ==> 2^{2} = 4) and
trebling speed increases the braking distance nine times (3 ==>
3^{2} = 9).
The thinking distance is
roughly proportional to speed, the graph is ~linear and does not
curve upwards. This is because your response time, if fully alert,
is pretty constant, so if your speed doubles, you just go twice as
far in the same response time.
(ii) and if you examine the graph or data carefully, you
can see that doubling the speed quadruples the braking distance.
This means by doubling your
speed, approximately quadruples the stopping distance, obviously
something you need to bear in mind the faster you drive. Doubling
speed quadruples braking distance and trebling speed increases it
nine times!
This is discussed further and is related to the formula
for kinetic energy KE = ½mv^{2}.
By doubling the speed, you quadruple
the kinetic energy of the car, hence you have quadrupled the kinetic
energy to be removed by braking (because
KE
v^{2}). See graphs 2 and 3 and notes below.
More on the physics of braking
The
mechanical process of braking primarily relies on friction between the brake
pad and a steel disc (shown on the right). When you press the brake pedal
the hydraulic
system pushes pads onto the surface of the disc causing work
to be done due to the resistive forces between the surfaces.
The resulting friction effect transfers energy from
the car's kinetic energy store to the thermal energy store of the braking
system which is eventually dissipated to the environment's energy store. The
friction does cause the brakes to heat up  the brake pads and disc must be
able to withstand a high temperature  both are made of high melting alloys.
So when the work is done between the brakes and the wheel
discs kinetic energy is converted to thermal/heat energy. The faster a
vehicle is going, the greater its kinetic energy store and more work must be
done to bring the car to a halt. It also means a greater force must be
applied to bring the vehicle to a halt within a certain braking/stopping
distance.
The greater the braking force, the greater the deceleration.
Big decelerations can be dangerous because the brakes may overheat affecting
their action AND there is a much greater chance of skidding, particularly if
the road surface is slippery due to conditions already described.
To put the point about kinetic energy into context, study graph 2 below.
Graph
2
Graph 2 shows how the kinetic energy of a
road vehicle (e.g. a car of 1200 kg) varies with its speed.
You can see that by doubling the speed,
you quadruple the kinetic energy of the car, hence you have quadrupled the
kinetic energy to be removed by braking.
This is because KE = ½mv^{2}. Its the speed^{2}
term that gives this crucial mathematical importance.
Assuming uniform deceleration and uniform
decrease in the rate of reducing kinetic energy, means the braking
distance is a function of kinetic energy and speed^{2}. See
graph 3 now.
Graph
3
Graph 3 shows the linear relationship between the kinetic
energy of the car and braking distance (using the UK Highway Code data and a 1200 kg car).
This is a result of KE = ½mv^{2}
and the braking distance data
assumes uniform deceleration and uniform decrease in the rate of reducing
kinetic energy due to the friction of the brakes.
As already mentioned, the braking
distance increases faster than the speed.
The total work done to stop a road
vehicle is equal to the initial maximum kinetic energy of the vehicle.
Work done to halt vehicle = total
KE of vehicle = braking force x braking distance
W = F x d =
KE = ½mv^{2}
(in a nutshell !)
W = work in J to come to a halt,
and all of the work is done by the brakes (assuming no skidding) via
friction from the vehicles KE store to the thermal energy store of
the brakes and environment
F = braking force in N (assumed
to be constant for the vehicle brakes),
d = braking distance in m, m =
mass of vehicle in kg, v = speed of vehicle in m/s
If you skid on a dry road,
the rubber left on the road tells you the tyres were doing a bit
of braking work too!
If we assume a constant braking force
(maximum push on brake pedal) and since the kinetic energy of the car is
proportional to speed^{2}, then the braking distance is
proportional to the initial kinetic energy of the car.
That's what the work done
equation says for a constant braking force:
KE
BD and so does the graph.
An extra consequence: If your car is full
of people or a lorry is fully loaded, then the kinetic energy at a given
speed is greater than if the vehicle only contained the driver.
Therefore, with extra mass in the vehicle, extra distance should be allowed
for your braking distance because of the extra kinetic energy.
Examples of typical
masses for road vehicles:
cars 1000  1500 kg; large
van/single decker bus ~9 000 10 000 kg; loaded lorry ~30 000  40 000
kg.
Q2 Using the above
kinetic energy formula:
A small domestic 1000 kg car (1 tonne)
with two axles at 60 mph (26.84 m/s)
will have a kinetic energy = ½ x 1000
x 26.84^{2} =
3.6 x 10^{5} J (360 kJ, 3 s.f.)
A heavy articulated goods vehicle of 6
axles may weigh, with a full load, up to 43 000 kg (43 tonnes) at 60 mph
(26.84 m/s)
will have a kinetic energy = ½ x
43,000 x 26.84^{2} =
1.55 x 10^{8} J
(15 500 kJ, 3 s.f.)
Now, both of these vehicles have to be
able to stop in the same safe distance in an emergency.
The two axle car will have four sets
of brake pads.
The six axle goods vehicle will have
twelve sets of brake pads, three times as many as the car.
This means to stop in the same safety
distance, the braking force exerted by each set of pads in the goods
vehicle must be much greater than for the car.
At 50 mph (22.37 m/s) suppose the
safe braking distance is 38 m.
We can then calculate the total
braking force needed to stop in three seconds.
(i) for both vehicles deceleration
a = ∆v / ∆t = 22.37 / 3 =
7.457 m/s^{2}
(ii) F = ma
from Newton's 2nd Law, force in newtons, mass in kg, deceleration in
metres per second^{2}
For the car: F
= 1000 x 7.457 = 7 460 N
(3 s.f.),
that is
1865 N braking force per set of brake pads.
For goods
vehicle: F = 43 000 x 7.457 =
321 000 N
(3 s.f.).
that is
26750 N braking force per set of brake pads.
This means the
heavy goods vehicle brake pads must generate over 14 x the braking force
of the car.
(For those expert
in road vehicle physics, I do appreciate these are simplified
calculations)
Health and safety issues related to collisions
involving road vehicles (motor bikes, cars, lorries, buses etc.)
You can build safety features into the design
of road vehicles, and, where appropriate, safety clothing.
What you are trying to in most cases is to
slow down the deceleration  increasing the collision time or absorbing the
kinetic energy of any rapid deceleration and in doing so minimise the force a
person's body experiences. A rapid impact produces an extreme deceleration 
much more so than even emergency braking.
Its all about minimising injury to
people in a rapid change of motion situation.
In terms of physics, its all about
absorbing impact energy and increasing the deceleration time  minimising
the a in F = ma!
From Newton's 2nd law of motion: F =
ma, so for a given mass m, if you can make
a the deceleration smaller,
the decelerating force F
is also reduced and minimises body impact and injury.
Wearing a seat belt reduces the impact of
the deceleration. On collision or in emergency braking, the seat belt
stretches, increasing your deceleration time and decreasing the force your
body experience against the seat belt.
Fast acting air bags, cushion your body
from a violent impact, they increase deceleration times and reduce the force
your body experiences.
A car body can have crumple zones in the
front and back to absorb the kinetic energy of the impact and by increasing
the deceleration time, thereby decreasing the force your body experiences.


The photographs of a (faked) moderately
violent crash of the car into a brick wall gives you an idea of what a
'crumple zone' is all about.
You would see similar damage to the rear
of your car if someone runs into the back of you.
Helmets worn by cyclists or motor bike
riders (motorcyclists) have an inner lining of foam (or other energy
absorbing material) to cushion the head on impact. The foam
increases the time before your head stops moving due to the impact. The
smaller deceleration decreases the force of impact your head experiences.

Everything is designed
with safety (and comfort) in mind. The main
safety features of the crash helmet are the hard outer shell and the
impact energy absorbing liner. BUT, even the comfort/fit padding 'foam'
will absorbed impact energy too.
Image
from the
CALIFORNIA MOTORCYCLIST SAFETY PROGRAM
and supported by the California Highway Patrol
The scheme advises motorcycle riders with helmets
that do not match all the safety design features illustrated, should
change helmet! 
When out walking, came across a
motorcycle couple who kindly allowed me to take photographs. Both survived a
serious accident, but once the crash helmet has been in an impact situation,
it must be replaced. You can clearly see all the features described in the
diagram above.
So, teenage motorcyclists, buy the safest
helmet, it might cost more, but without the best helmet, it might cost you
even more.
Research is always going on to develop
new materials to increase the performance of safety features, whether it be
car bodies or helmets.
Some advanced
GCSE calculations on braking force and kinetic energy
Diagram KEY: KE = kinetic energy (J),
m = mass (kg), u = initial velocity (m/s),
v = final velocity (m/s), s = speed (m/s)
a = acceleration or
deceleration (m/s^{2}), W =
work done (J), F = force (N),
d = distance (m)
Q1
Suppose a car of 1200 kg is travelling at 18 m/s (~40 mph) and has to do an
emergency stop with a hazard 30 m ahead.
(a) Calculate the deceleration of
the car and (b) the braking force involved.
(a) First use the motion equation v^{2}  u^{2}
= 2ad to calculate the deceleration.
where v = final velocity, u = initial velocity, a =
acceleration (∆v/∆t), d = distance travelled
Assuming uniform deceleration and v = 0 (comes to
halt), u = 18 m/s, d = 30 m
v^{2}  u^{2} = 2ad, 0  18^{2}
= 2 x a x 30
60a = 324, therefore a = 324/60 =
5.4 m/s^{2}
(note the negative sign for deceleration)
(This is easier to do if your
given the braking time, so can just use a = ∆v / ∆t, which I did in the previous section comparing a car and
a heavy goods vehicle, and called it
Q2)
(b) You then use Newton's 2nd Law equation
F = ma,
where F = decelerating braking force, m = mass of car,
a =
deceleration of car = change in speed / time taken
Substituting into the equation (and you can ignore
the ve acceleration sign here, but NOT above)
F = ma = 1200 x 5.4 =
6480 N
Comment: That's why your body is thrown forward. The
deceleration is just over half the value of the acceleration you
experience due to the Earth's gravitational field. If you are
involved in a high speed impact the force can be much greater and
hence destructive on you and the car!
See section on
safety features of road transport
Q2 See
Graph 3 braking
distance versus kinetic energy
Q3 Suppose
a car travelling at 30 m/s (~70 mph) has to make an emergency stop to avoid
a hazard.
If the mass of the car is 1500 kg, the
braking force of the car is 6000 N and the tired driver's
reaction time is 1.5 seconds, calculate the following:
(a) Calculate the thinking
distance of the driver (s = speed (m/s), d  distance (m), t = time
(s))
s = d / t, d = s x t = 30 x 1.5 =
45 m = thinking
distance
(b) Calculate the initial kinetic
energy of the car (m = mass of car in kg, v = speed of car (m/s)
KE = ½mv^{2} = 0.5 x 1500
x 30^{2} = 675000 =
6.75 x 10^{5} J =
initial KE of car
(c) Calculate the braking distance
to halt the car (W = braking work done (J), d = braking
distance (m)
Work done in braking the car must
equal the kinetic energy of the car (see
Graph 3
discussion)
W = F x d = KE = ½mv^{2}
= 6.75 x 10^{5} J
W = F x d, d = W / F = 6.75
x 10^{5} / 6000 =
113 m = braking distance (3 s.f.)
(d) Calculate the stopping distance
of the car
stopping distance = thinking
distance + braking distance
= 45 + 113 =
158 m = stopping distance
Q4 See
reaction
time experiment
Q5 A 1500 kg 4x4 car travelling at 18.0
m/s (~40 mph) veers off the road, without slowing until hitting and
demolishing a brick wall.
If it took 0.200 seconds to demolish
the wall, calculate the following ...
(a) What is the initial kinetic
energy of the car?
KE = ½mv^{2} = 0.5 x 1500
x 18^{2} = 243 000 =
2.43 x 10^{5}
J
(b) What work is done on the wall
and car in bringing the car to a halt?
2.43 x 10^{5}
J because all the kinetic energy of the car has to be
removed.
(c) What happens to the kinetic
energy of the car after impact?
The kinetic energy store of
the car is reduced to zero and the energy is converted into heat
(by compression or friction) and some sound energy (which will end
up as heat too). So the thermal energy store of the wall, car and
surrounding air is increased.
(d) Calculate the rate of
deceleration
Deceleration = change in speed /
time taken = ∆v / ∆t = (0  18) / 0.2 =
90 m/s^{2}
(e) What is the
decelerating force acting on the car?
From Newton's
2nd Law: F (N) = m (kg) x a (m/s^{2})
Decelerating
force = 1500 x 90 = 135 000 =
1.35 x 10^{5}
N
Q6 ?
Simple reaction
time experiments  but can be accompanied by some moderately
complicated calculations!
Your reaction time to a situation may be typically 0.2 to
0.8 seconds when fully alert. However your reaction time can be affected by
tiredness, feeling unwell, drugs, alcohol, in other words anything that
affects the speed of your brain function.
See
An introduction
to the nervous system including the reflex arc
You can conduct quite simple experiments to test your
reaction time to a particular situation. However, since the reaction time is
too short, a stopwatch is no good, but there are ways of measuring your
reaction time indirectly by making other measurements from which you can
calculate your reaction time.
(a) Computer screen  where you respond as quickly as possible to
something appearing on the screen.
In this situation, the computer
software generates something up on the screen and automatically times
your response by monitoring your contact with the keyboard or by
clicking the mouse.
I've quickly written an extremely
simple computer programme to test your response to a X appearing on
the screen.
Response time
test: It probably only works on Microsoft platforms, and
maybe not all of them?
Your antivirus protection might
query it, because it is a .exe file, but its written with
compiled BBC BASIC and should not pose any threat. Unfortunately I
never learned to write in a multiplatform professional computer
programming language, but I'm not exactly short of website projects!
(b) A simple physical response test  the
falling
ruler drop test
You get someone to hold a ruler vertically, with
thumb and first finger, above someone else's hand, who is ready to catch
it with their thumb and first finger.
First image on the right. The
ruler should be held at the top of the scale and steady hands from
both people.
The catching person should have
the middle of their thumb and finger adjacent to zero on the cm
scale  squat down to make sure you are reading the scale
horizontally.
Then, without warning, the person holding the ruler,
lets go of it. The second person has to react as fast as possible and
catch the dropped ruler between their thumb and first finger.
Second image on the right. The
longer the distance, the slower your reaction time!
When caught, you then read how far
the ruler as fallen by taking the reading, to the nearest centimetre,
from where the middle of their thumb and finger are.
You repeat the experiment a number
of times to get an average, but its not a particularly accurate
experiment.
You need to have steady hands and not
let the ruler wobble about or fall at an angle other than vertical. You
should also use the same ruler and the same people dropping the ruler
and catching it (fair test criteria), though, obviously, you can compare
one person's results with another.
The slower your response time, the
further the ruler falls before being caught. You might repeat the
experiment by having some background distractions  a group of people
talking nearby, or somebody trying to engage you in conversation or
music.
Q4 You can then do some 'nifty'
calculations to actually obtain a real response time  so you
using indirect data to get the response time.
It involves a two stage calculation.
Suppose the ruler is caught after an
average fall of 25 cm.
(i) You use the equation v^{2}
 u^{2} = 2ad,
to calculate the final velocity (more
calculations using this equation)
v = final velocity (m/s), u =
initial velocity (m/s), a = acceleration = 9.8 m/s^{2}
(gravitational acceleration),
and d = distance fallen (m)
Since u = 0 and d = 25/100 = 0.25
m
v^{2}  0 = 2 x 9.8 x
0.25 = 4.9
v = √4.9 = 2.214 m/s (its
not this accurate, but we'll leave the s.f. until the end)
(ii) We can now use the acceleration
formula to calculate the response time.
a = ∆v / ∆t,
where a = acceleration (9.8 m/s^{2}), ∆v = change in
velocity (m/s) and ∆t = response time
Therefore :
9.8 = 2.214 /
∆t, ∆t = 2.214 / 9.8 =
0.23 s
(2 s.f.)
So, on
average, the response time was about a quarter of a second.
INDEX
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