Download Science and the Road

Science and the Road
K M Cheng
Department of Physics, CUHK
Time Allocation: 6 hours
1
Content
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Measuring motion
Speeding up and slowing down
Tyre/road friction（摩擦力）
Stopping a car
The speed formula
Collision
Travelling round a bend
Driving safety
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Measuring motion

Speed（速率） tells us how fast an object
moves.
total distance travelled
Average speed 
total time taken
s 

v 

t 

The speed at a particular instant of time is
found by taking the average speed over a very
short time interval.

Velocity（速度） tells us the speed of the
object as well as its direction of travel.
3

An object accelerates when its speed increases
and decelerates when its speed decrease.
Acceleration（加速度）is the rate of change
of velocity.

For motion in a straight line,
change in speed
Average accelerati on 
time for the change
v 

a 

t 

When t becomes very small, the average
acceleration approaches the acceleration at a
particular time instant.
4

For uniformly accelerating motion in a straight
line:
uv
v
2
and
v u
a
t
where
u  initial velocity,
v  final velocity,
v  average velocity,
a  acceleration,
t  time.
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
From
these
two
expressions, the
equations of motion can be deduced:
three
v  u  at
1 2
s  ut  at
2
v 2  u 2  2as
Note that s is the displacement（位移）
which is the distance travelled along the straight
line.
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Speeding up and slowing down
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We may change the motion of a moving object
by pushing or pulling it. Scientists call this pull
or push to be a force（力）. The unit of force
is newton (N - 牛頓).

Newton formulated three laws of motion
describing the properties of force. In Newton’s
first law, it states that:
A body continues in a state of rest or uniform motion
in a straight line, unless an unbalanced force acts on it.
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
What does the term unbalanced force really
mean?
10 N
10 N
These forces balance. The
block does not move.
20 N
10 N
These forces does not balance. The
block moves as if it was being pulled
by a force of 10 N to the right.
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Newton's first law of motion tells us that all
objects resist change in their state of motion. All
objects have this tendency, i.e. they have inertia
（慣性）.
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The mass（質量）of an object is a measure of
its inertia. An object with a larger mass has more
inertia and thus has larger tendency to resist
changes in its state of motion. The mass of an
object is usually given in unit of g (gram) or kg
(kilogram).
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If an object is dropped, it falls to the ground at
increasing speed. The free falling object（自由
落 體 ） accelerates due to the pull of the
Earth’s gravity（重力）. When air resistance is
negligible, all free falling objects fall with the
same uniform acceleration (approximately)
known as the acceleration due to gravity g.
The value of g is independent of the mass of
the body, and is usually taken as 10 m/s2.
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
A car will eventually stop if the engine is turned
off. It is due to the friction（摩擦力）which
slows down the cars’ motion.

Friction arises whenever an object slides or
tends to slide over another object. It always acts
against the motion.
PUSH
To Move
FRICTIONAL FORCE to Resist Motion
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Friction is due to the irregularities on a surface.
Even a smooth surface has these “peaks and
valleys”. If two surfaces rub over each other, the
irregularities on the two surfaces would crush
together, resulting in an opposing force.
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What does friction depend on?

Friction does not depend on the area of contact,
but depends only on the nature of the contact
surface.
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Friction is much reduced if the contact surface is
lubricated with water or oil, and if rollers and
bearings are placed between the contact surface.
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The greater is the weight of the object, the greater
is the friction.
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Tyre/road friction
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According to Newton’s Third Law, the car is driven by
the forward frictional push of the ground on tyres.
If we apply the brakes (very hard), the wheels are
locked and thus prevent from rotating.
As a result, the car skids and decelerates. The
decelerating force is actually the sliding friction（滑動
摩擦）– a backward push of the ground on tyres.
Anti-lock braking system (ABS - 防鎖死煞車系統 ): a
system on motor vehicles which prevents the wheels
from locking while braking.
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Tyre/road friction

On a level road, this force is given by
F   mg
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where  is the coefficient of tyre/road friction and mg
is the weight of the car.
The  value changes only slightly with speed.
However, if the road surface is wet, the  value
depends significantly on tyre conditions, speed,
weight of vehicle and degree of wetness.
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Tyres and tread patterns
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Various tread patterns are designed to provide a firm grip on the
road surface
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Groove on the thread are for pushing out water on a wet road so
that the tyres and the road are still in direct contact and are not
lubricated
Tread pattern on a car tyre
(Wikimedia commons)
Tread pattern on a mountain
bicycle tyre
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Tyres and tread patterns
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Worn-out tyres are dangerous and illegal
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By law, tread depth must not be less than 1 mm over
three-quarters of the tread area
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A penny can be used to check tyre tread
(Penny test, USA):
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Take a penny and put Abe's head into one of
the grooves of the tyre tread
If part of his head is covered by the tread,
you're driving with the legal amount of tread
If you can see all of Abe's head, it's time to
replace the tyre
(Wikimedia commons)
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Slick tyre
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A type of tire that has no tread pattern
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Used mostly in auto racing
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By eliminating any grooves cut into the tread, such tyres provide the
largest possible contact patch to the road
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Can be very dangerous if the racing track is wet!
(Wikimedia commons)
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The skid marks
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When a car skids, particularly on a tarmac road surface, clearly
visible skid marks are usually left on the road

The speed of a vehicle prior to skidding can be estimated from the
skid mark length and the coefficient of friction（摩擦係數 ）
between the tyres and the road surface
(Wikimedia commons)
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Skidding over different surfaces

Very often, a skid mark may extend over different
surfaces, e.g., tarmac（柏油碎石）, concrete
（混凝土）, gravel（砂礫）, grass, etc.

In these cases, the speed of a vehicle prior to
skidding can be estimated from the skid mark
lengths on different surfaces and the coefficient
of frictions between the tyres and different
surfaces
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Skidding on slopes

When the road surface is not level, the skidding
distance is affected. It will be shorter for going
uphill and longer for going downhill.

One may define the gradient e（斜率）of a
non-level road by
H
e
L
H
L
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Skidding on slopes

If the coefficient of frictions of a level surface is
, then the effective value for an up-slope or
down-slope of gradient e and the same road type
is approximately given by
Up-slope:
us    e
Down-slope:
ds    e
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Skidding on slopes
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Proof of the formulae:
F   N  mg sin 
 mg cos    tan  
 mg cos    e
 mg    e
if  is small
N = mgcos
mgsin
N
H
mg

Force diagram for the case of down-slope
L
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Stopping a car
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When a driver sees a hazard, he hits the brake to
stop the car. The time taken for the driver to
react is called the reaction time (or thinking
time). Reaction time varies from person to
person but is, on average, about 0.2 s. The
distance travelled by a vehicle during reaction
time is called reaction distance, i.e.
reaction distance  reaction time
 average speed during this time
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Measuring reaction time by the ruler drop
technique
Your partner holds a ruler between your fingers. The
ruler is released suddenly. As soon as you see it, close
your finger to catch the ruler. Record how far the ruler
drops before you catch it. The faster you react, the less
the ruler will drop before being caught.
Remark: the reaction times are much longer in the case of
unexpected events – in the range 1.5 to 2.5 seconds!
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Once the driver applied the brakes, the car
slows down. The distance that the car moves
during braking is called the braking distance
(can be estimated by the skid mark length).
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Stopping distance is the distance taken to
stop a car, i.e.
stopping distance  reaction distance  braking distance
And
the time taken to stop a car is called the
stopping time.
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Example
Stop signal
Starts to brake
Speed
12 m/s
Slowing down
Stopped
0 m/s
0.75 s
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2.5 s
time
Thinking time = 0.75 s
Distance Travelled during thinking time = 12 m/s  0.75 s = 9 m
Distance Travelled during braking = ½  2.5 s  12 m/s = 15 m
Total stopping distance = 9 m + 15 m = 24 m
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
The stopping distance depends on the following
factors:
 the
speed of the car
 the
driver’s reaction time
 the
road surface
 the
condition of the car’s brakes and tyres
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The speed formula
Suppose a vehicle of mass m travel on a level road at a
speed u prior to skidding. The kinetic energy Ek of the
vehicle is given by
1
Ek  mu 2
2
Ek  0
Ek 
u
u=0
s
1
mu 2
2
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The speed formula
After skidding a distance s, the vehicle stops. The
decelerating force is the tyre/road friction and is given
by
F   mg
The work done W against friction is
W  Fs   mgs
u
u=0
F 0
s
F   mg
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The speed formula
When the vehicle stops, its kinetic energy is reduced to
zero. The change of kinetic energy is equal to the work
done against the friction, i.e.
1
mu 2   mgs
2
2
u
 s
2 g
OR
u  2 gs
So the skid-to-stop distance depends only on the
coefficient of tyre/road friction () and the speed of
the vehicle (u) prior to skidding.
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Some Learning Points
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In braking , a car’s kinetic energy (KE) must be
dissipated (usually as heat)
KE = work done in stopping the car = mgs
Braking distance is proportional to the speed
squared
The effective coefficient of friction is not the
same for all cars
The coefficient of friction decrease by about
40% in wet conditions (assuming the tyres are in
good condition)
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Collision
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Traffic accidents almost always involve collisions
of one kind or another
The principle of conservation of linear
momentum（動量守恆）can be applied:
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The total momentum of the objects before collision is
equal to the total momentum after collision, provided
that there is no external force acting on the colliding
objects. (Newton’s First Law)
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Collision
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The momentum of an object is defined as the
product of its mass m and velocity v. For two
colliding vehicles of masses mA and mB travelling
along the same straight path, their velocities are
uA and uB before impact and vA and vB after
impact. Then the total momentum before and
after impact is related
mAuA  mBuB  mAv A  mBvB
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Collision (remarks)
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Momentum is a vector quantity. If the collision occurs at
an angle, the vector sum of the momentum before impact
is equal to that after impact.
Vehicles exert force on one another on impact (Newton’s
Third Law). The force of impact causes damage to the
vehicles and changes their velocities (and hence momenta).
In fact, the force of impact, F, is related to the momentum,
P, by
Newton’s Second Law!
dP  mv  mu 
F


dt 
t

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Travelling round a bend
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A vehicle travelling
round a bend on a level
road can be viewed as
moving along a circular
path
The centripetal force
（向心力）is pointing
towards the centre of
circular motion
v
r
Centre of circular path
Centripetal force
mv 2

r
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Travelling round a bend

The centripetal force is provided by the sideway
friction between the tyres and the road surface:
F   mg

Equating the two forces, we get
mv 2
  mg
r
 v   gr

v is called the critical curve speed for the bend
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Travelling round a bend
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When the speed of the vehicle is smaller than the critical
curve speed of a bend, the vehicle has no difficulty in
negotiating the bend.

When the vehicle is just at the critical curve speed, it is
travelling at the limit of adhesion to the road. It cannot
brake or steer onto a tighter course to avoid an
unexpected hazard without risking side-slipping.

When the speed of the vehicle is greater than the critical
curve speed, the frictional force is not large enough to
provided the necessary centripetal force. As a result, the
vehicle side-slips.
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Travelling round a bend

If the bend is banked, the vehicle may negotiate a bend
at a higher speed because a component of the normal
reaction contributes to the centripetal force as shown:
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Travelling round a bend

When a bicycle (or motorcycle) travel round a bend, it
is necessary to lean it into the turn. Otherwise the
centrifugal forces will throw the bicycle (or motorcycle)
over on its side.
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Driving safety

Major reasons of car accidents [3]:

Following too closely to vehicle in front.

Losing control of vehicle.
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Careless lane changing.

Improper or illegal turn.

Starting negligently.
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
Many people are injured or even killed by
traffic accidents in each year. Whether as
pedestrian, driver or passenger, we should be
aware of road safety to avoid traffic accidents.

Almost all modern cars are equipped with some
safety features which can help to protect the
driver and passengers during an impact (see
next page figure).
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Don’t drink and drive!
Alcohol Affects Your Driving Ability

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Impairs judgment of speed and distance
Slows down reaction time
Affects the co-ordination of the body's
movements
Blurs vision
Gives a false sense of confidence
http://www.td.gov.hk/road_safety/drink_driving/index.htm
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Don’t drink and drive!
The effect of blood alcohol content (BAC)
No. of standard
drinks
BAC
(mg/100 ml)
Effects
1
20 - 30
5
100 to 150
noticeable effects (on
perception etc.)
intoxication
24
240 to 360
unconsciousness
36
360 to 480
coma or death
BAC < 50 ml/100 ml under the law in Hong Kong!
http://www.sciencetechnologyaction.com/lessons/53/NCT.pdf
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References:
1.
2.
3.
4.
5.
6.
G. Bethell and D. Coppock, Physics first. Oxford: Oxford
University Press, 1999.
G. Alderton, D. Berrington, and M. Brimicombe, Revise for
GCSE Science: Salters. Oxford: Heinemann, 1999.
http://www.roadsafety.gov.hk/eng/index.html
P. K. Tao, The Physics of Traffic Accident Investigation: Oxford
University Press, 1987
http://www.td.gov.hk.
http://www.sciencetechnologyaction.com/lessons/53/NCT.pdf
46
Example 1
A car skids with all four wheels locked and leaves
skid marks of 19.3 m on a tarmac road surface,
5.6 m on a concrete pavement and 15.4 m on
grass. The coefficients of friction of the tarmac,
concrete pavement and grass are 0.74, 0.82 and
0.46 respectively. Estimate the speed of the car at
the start of skidding.
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Example 1
The kinetic energy of the skidding car is equal
to the total work done against friction over
different surfaces. Hence
1
2
mu  1mgs1  2 mgs2  3 mgs3
2
where 1, 2, 3 are the coefficients of friction
of the tarmac, concrete pavement and grass
respectively.
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Example 1
The speed of the car at the start of skidding is
given by
u  2 1 gs1  2 2 gs2  2 3 gs3
 22.6 ms
1
 81.2 kmh
1
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Example 2
A car skids with all four wheels locked for 38.5 m
and then runs into a tree. The impact speed of the
car is estimated from the damage to be 40 kmh-1.
The coefficient of tyre/road friction is found to be
0.76. Estimate the speed of the car prior to
skidding.
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Example 2
The car decelerated on skidding. The decelerating
force F is given by
F  mg
If the average deceleration over a skidding
distance is a, then
F  ma   mg  a    g
(Newton’s Second Law)
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Example 2
Applying the equation of uniform acceleration,
v 2  u2  2as  2 gs

u  v 2  2  gs

u
 40 3.6
1
2
 2  0.76  9.81  38.5
 26.4 ms  95.1 kmh
1
52
Example 3
A car skidded 12.6 m before hitting a parked van
on its side. The two vehicles became locked
together and skidded a distance of 3.6 m before
coming to a stop. The coefficient of tyre/road
friction was found to be 0.71. Total mass of the
car and its passengers was 1,260 kg and total mass
of the van and its load was 2,100 kg. What was
the speed of the car before impact? What was its
speed at the start of skidding?
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Example 3
Since the car and the van were locked together
after impact, they had a common post-impact
velocity v. From the skid mark, we get
v  2  gs  2  0.71  9.81  3.6
1
 7.08 ms  25.5 kmh
1
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Example 3
Applying the principle of conservation of
momentum, we have
mAuA  mBuB  mAvA  mBvB   mA  mB  v
 1260uA  2100  0  1260  2100  7.08
 uA
1260  2100


 7.08  18.9 ms1  68.0 kmh 1
1260
The pre-impact speed of the car
55
Example 3
The speed of the car at the start of skidding is
u  u A2  2  gs

18.9 
2
 2  0.71  9.81  12.6
 23.1 ms1  83.1 kmh 1
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