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Transport

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Title: Transport


1
Transport
by S. Reid
5.1 On the move 5.2 Forces at work 5.3 Movement
means energy
2
Transport
by S. Reid
5.1 On the move 5.2 Forces at work 5.3 Movement
means energy
3
5.1 On the Move
This is sometimes written as ms-1.
Average speed can be calculated from the formula
4
To measure an average speed you must make certain
measurements
The distance travelled must be measured and the
time taken to cover that distance must also be
measured. These values should then be used in
the equation
To find the average speed.
5
Example 1. If a baby can crawl over a distance of
20 metres in a time of 10 seconds,
what is its average speed?
Example 2 A sprinter in training has an average
speed of 5m/s. How long does it take this
sprinter to complete a 100m race?
6
The instantaneous speed is how fast that object
is travelling at one particular point in
time. To measure the instantaneous speed a very
short period of time must be used
If the driver of a car glances at their
speedometer they are checking their instantaneous
speed.
The average speed of an object is not always the
same as the instantaneous speed of the same
object..
7
Take the example of a car travelling through a
busy town centre to reach the dual carriageway.
When the car is in town it should be travelling
at a maximum of 30 m.p.h and the car will have to
stop at various points such as traffic lights.
When the car is stopped it has an instantaneous
speed of 0 m.p.h. but it will have an average
speed over the journey of maybe 15 m.p.h.
8
When the car is on the dual carriageway it will
probably be travelling at a constant speed of 70
m.p.h.
This means that the instantaneous speed is always
70 m.p.h. and for the time that the car is on the
motorway the average speed is also 70 m.p.h.
9
The acceleration of an object is defined as the
rate of change of velocity of that object. i.e.
acceleration is a measure of how much an object
changes speed per unit time. As a formula
final speed (m/s)
initial speed (m/s)
time (s)
acceleration (m/s2)
N.B units of acceleration sometimes written as
ms-2
10
Example 1 If a bus can accelerate from rest to
60 m/s in a time of 7 seconds. Calculate the
acceleration of the bus . Solution u0 v60
m/s t7s a?
11
Example 2 A sports car may be able to accelerate
from rest to 100 m.p.h. in 10 seconds. Find the
acceleration of the car. Solution u0 v100
m.p.h t10s a?
The units of acceleration are different because
speeds are measured in m.p.h not m/s.
12
Speed - time graphs
A speed - time graph that shows a straight line
of positive gradient shows the motion of an
accelerating object.
A speed - time graph that shows a straight line
of zero gradient shows the motion of an object
travelling at a constant speed.
A speed - time graph that shows a straight line
of negative gradient shows the motion of an
decelerating object.
13
Speed - time graphs
A speed - time graph that shows a straight line
of positive gradient shows the motion of an
accelerating object.
A speed - time graph that shows a straight line
of zero gradient shows the motion of an object
travelling at a constant speed.
A speed - time graph that shows a straight line
of negative gradient shows the motion of an
decelerating object.
14
You must be able to use speed - time graphs to
find accelerations. Example
u10 m/s
At time 0s, the speed, u10 m/s
15
You must be able to use speed - time graphs to
find accelerations. Example
u10 m/s
At time 9s, the speed, v64 m/s
v64 m/s
16
You must be able to use speed - time graphs to
find accelerations. Example
u10 m/s v64 m/s
a 6ms-2
17
You can also calculate distance from a speed -
time graph. The distance travelled is equal to
the area under the graph. e.g.
3
1
4
2
Step 1 split graph into manageable areas.
18
You can also calculate distance from a speed -
time graph. The distance travelled is equal to
the area under the graph. e.g.
3
1
1
4
2
Area 1 0.5 x 10 x 5 25
19
You can also calculate distance from a speed -
time graph. The distance travelled is equal to
the area under the graph. e.g.
3
1
4
2
2
Area 1 0.5 x 10 x 5 25
Area 2 5 x 2 10
20
You can also calculate distance from a speed -
time graph. The distance travelled is equal to
the area under the graph. e.g.
3
3
1
4
2
Area 3 7 x 12 84
Area 1 0.5 x 10 x 5 25
Area 2 5 x 2 10
21
You can also calculate distance from a speed -
time graph. The distance travelled is equal to
the area under the graph. e.g.
3
1
4
4
2
Area 1 0.5 x 10 x 5 25
Area 3 7 x 12 84
Area 4 0.5 x 4 x 12 24
Area 2 5 x 2 10
22
You can also calculate distance from a speed -
time graph. The distance travelled is equal to
the area under the graph. e.g.
3
1
4
2
Area 1 25
Area 2 10
Area 3 84
Area 4 24
Total Area distance 25 10 84 24 143 m
23
Return to start of Transport Advance to next
section
24
Return to start of Transport Advance to next
section
25
5.2 Forces at Work
A force can have any of the following
effects change the speed of an object, change
the direction of an object or, change the shape
of an object.
Take the example of someone punching someone else
in the face.
26
When the punch lands
The persons head will move speed and direction
has changed. The persons face will also change
shape. Maybe permanently! These effects are all
due to the force being applied by the punch.
27
One way of measuring a force in the laboratory is
to use a Newton balance. (Sometimes called a
spring balance)
The balance has a spring inside which stretches
more when a heavier object is placed on the end.
To use a Newton balance the object should be
placed on the hook at the bottom of the
balance. The balance should then be held steady
and the reading read from the scale on the
balance.
This reading would be the objects weight. Weight
is a force and is therefore measured in Newtons
(N).
28
The weight of an object is a measure of the
Earths pull on the object.
Mass is a measure of how much substance makes up
a body or object. Mass is measured in kilograms
(kg).
Weight is a force and is measured in Newtons (N).
W mg
Where W weight, (N) m mass, (kg) g
gravitational field strength, (N/kg)
The value of g on Earth is 10 N/kg
29
Gravitational field strength, g varies from
planet to planet.
g on the moon is less than it is on the
Earth. This means that an object will have a
smaller weight on the moon than it does on Earth.
If a tin of beans was placed inside a rocket on
Earth and the rocket was launched from Earth on a
journey to the moon.
The tin of beans would have weight on Earth as
the rocket gets further from Earth and goes into
deep space, the weight of the beans would
decrease until the weight zero. As the rocket
approaches the moon the weight would increase
again but the final weight on the moon would be
less than the weight on Earth. Note that the mass
of the tin of beans does not change at all.
30
Friction is a force which always acts against the
direction of motion.
Direction of motion
Direction of motion
friction
Friction is due to the rough surface of the tyres
rubbing against the rough road surface.
Friction is due to air resistance as the
helicopter passes through the air. As the speed
increases so does the air friction.
friction
31
Friction is sometimes deliberately increased For
example A racing bike does not want to skid off
of the track when it takes a corner the tread
on the tyre causes friction between the bike and
the road and stops it skidding.
A sky-diver does not want to hit the Earth at a
large speed so when they approach the ground they
open a parachute. When the parachute opens, air
friction is increased and the falling person
slows down.
32
Friction is sometimes deliberately decreased For
example A ten pin bowling alley is made as
smooth as possible to make the bowls roll as
smoothly and as quickly as possible.
We use oil in engines to reduce the wear and tear
on the rubbing pieces of metal. If oil is not
used the engine will seize and stop moving.
33
2N
Mass
6N
6N
If two forces of the same size act in opposite
directions on the same object we say the forces
are balanced. These forces are then exactly the
same as no forces at all!
Mass
This is applied in Newtons First Law which
states An object will continue to travel in a
straight line at a constant speed unless an
unbalanced force acts upon it
2N
34
An example of Newtons first law. If a
passenger in a car is not wearing a seat belt and
the car crashes, the passenger would continue to
travel forward in a straight line at a constant
speed. (The same speed that the car was
travelling) If they had been wearing a seatbelt,
the belt would have exerted a retarding
(unbalanced) force on the passenger and would
have prevented them from leaving the seat.
35
The acceleration of an object depends on two main
factors The Mass of the object and The Force
applied to the object
If the Mass is doubled, the acceleration is
halved. If the Mass is trebled, the acceleration
is divided by 3. If the Mass is halved, the
acceleration doubles. If the Force is doubled,
the acceleration doubles. If the Force is
trebled, the acceleration trebles. If the Force
is halved, the acceleration is also halved.
36
These rules all obey Newtons Second Law which is
normally remembered as
Fma
Force (N)
acceleration (ms-2)
mass (kg)
Example 1 Calculate the force applied to a toy
car of mass 0.1kg which accelerates at
2ms-2. Soln.
F ma F 0.1 x 2 F 0.2 N
37
Example 2 Calculate the acceleration of a small
car (mass 150 kg) that can produce an engine
force of 1000N.
Soln.
F ma 1000 150 x a
a 6.67 ms-2
38
Example 3 What is the mass of a car if it has a
maximum acceleration of 20 ms-2 and the
engine can produce a maximum force of 3500 N?
F ma 3500 m x 20
m 175 kg
39
Example 4 Find the value of friction in a
vertical water chute if a boy of mass 50 kg
accelerates at a rate of 8 ms-2. Soln.
F ma W-Ff 50 x 8
(Ff friction) mg Ff 50 x 8 (50 x 10)
Ff 400 500 Ff 400 Ff 500
400 Ff 100 N
40
Return to start of Transport Advance to next
section
41
Return to start of Transport Advance to next
section
42
5.3 Movement Means Energy
There are 8 different forms of energy Electrical
Heat Kinetic Potential Chemical Light Sound
and Nuclear
43
The main energy transformations for a vehicle are
as follows For an accelerating vehicle Chemical
energy (Fuel) gt Kinetic energy. For a vehicle
moving at a constant speed Chemical energy (Fuel)
gt Heat energy (due to friction) For a braking
vehicle Kinetic energy gt Heat energy For a
vehicle climbing a hill Chemical energy gt
Potential Energy For a vehicle coming down a
hill Potential energy gt Kinetic energy
44
Work done is a measure of the energy
transferred. The work done will double if the
force applied doubles, The work done will also
double if the distance doubles. This leads to
the equation EW Fd
Distance travelled (m)
Work done (J)
Force (N)
45
Example How much work is done by a manual worker
in lifting a 2.5kg sledgehammer through a height
of 1.5m? Soln. Ew Fd Ew mg x d Ew (2.5 x
10) x 1.5 Ew 37.5 J
46
  • Work done is measured in joules so it has to be
    ENERGY.
  • Example.
  • A dog drags a large bone 2.3m in a time of 12s,
    if the dog exerts a force of 3.7N on the bone
    find
  • How much work the dog does, and
  • What power the dog develops.
  • Soln.

a) Ew Fd Ew 3.7 x 2.3 Ew 8.51 J
b)
47
A change in gravitational potential energy is
work done against/by gravity. A moving objects
Kinetic Energy (Ek) will increase if either its
mass or its speed increases.
Speed (m/s)
Kinetic Energy (J)
mass (kg)
48
Example How much kinetic energy does a helicopter
of 6000kg have if it is travelling at
150ms-1? Soln.
Ek 0.5 x 6000 x1502 Ek 3000 x 22,500 Ek
67.5 x 106 J
49
To calculate how much potential energy an object
has we use the equation Ep mgh
height (m)
Potential Energy (J)
mass (kg)
gravitational field strength (N/kg)
Example How much potential energy does a 100g
apple have when it is on a table 1.5m high?
Ep mgh Ep 0.1 x 10 x 1.5 Ep 1.5 J
50
To calculate how much potential energy an object
has we use the equation Ep mgh
height (m)
Potential Energy (J)
mass (kg)
gravitational field strength (N/kg)
Example How much potential energy does a 100g
apple have when it is on a table 1.5m high?
Ep mgh Ep 0.1 x 10 x 1.5 Ep 1.5 J
51
If an object travels either up or down a slope
Ek becomes Ep and vice versa. Example A
skateboarder leaves the top of a ramp (20m high),
calculate his speed at the bottom of the
ramp. Soln.
Initial Ep Final Ek mgh 0.5 mv2 gh 0.5
v2 10 x 20 0.5 x v2 200 0.5 v2 400 v2 v
20 m/s
52
Return to start of Transport
53
Return to start of Transport
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