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Stage 1 Physics
3.1 The Parameters of Motion
In analysing a motion, we wish to find information about the following parameters of the motion:
•
the time over which the motion of the body is being studied;
•
the displacement and/or the distance travelled by the body during this time interval;
•
the initial velocity of the body (i.e. the velocity of the body at the beginning of the time
interval);
•
the final velocity of the body (i.e. the velocity of the body at the end of the time interval); and
•
the acceleration of the body during the time interval.
Time
This refers to the duration of the constant acceleration, i.e. the time interval over which the
constant acceleration acts. The commonly used symbols for this are t or ∆t, and the standard
international (SI) unit for time is the second (s).
Displacement and Distance Travelled
The displacement of the body refers to its final displacement from its initial position during the
time interval in which the constant acceleration acts. Of course, if the body is just travelling in
one direction along a line, the distance travelled has the same numerical value as the displacement.
However, if the body reverses its direction of travel during the motion, then the displacement and
the distance travelled will have different values.
Consider the situation in Figure 3.1, which shows a body moving along a number line, as depicted.
The number line is marked in one-metre intervals. The body starts at point A and, in the first three
seconds, it moves to B. Then it reverses direction and, in the next four seconds, it moves to C. If
we are considering the motion in the first three seconds, the displacement of the body is 6m to the
right and the distance travelled is 6m. If we are considering the motion over the full seven
seconds, the displacement is 2m to the left and the total distance travelled is 14m.
−2
C
−1
0
A
1
2
Fig. 3.1
3
4
5
6
B
&
&
The usual symbols for these quantities are s or x. Thus, displacement is represented by s (s) or x
(x) and distance is represented by s or x. The SI unit for displacement and distance travelled is the
metre (m).
Velocity and Speed
The average velocity over some time interval ∆t is defined as the rate of change of displacement
with respect to time,
&
∆s
∆s
&
or
.
v avge =
v avge =
∆t
∆t
The average speed of a body over a given time interval equals the distance travelled, divided by
the time,
distance travelled
v avge =
.
∆t
The instantaneous velocity of a body is the limiting value of the average velocity as the time
interval gets very small (approaches zero). The instantaneous speed of a body is just the magnitude
of the average velocity.
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In our calculations, we will be mainly interested in the velocity of the body at the beginning of the
&
time interval over which the acceleration acts (i.e. the initial velocity vi or v i ), and the velocity of
&
the body at the end of this time interval (the final velocity v f or v f ). To avoid excessive use of
&
subscripts, it is a common practice to refer to initial velocity as u (or u); and to the final velocity
&
as v (or v). The SI unit of velocity and speed is metres per second (m/s or ms–1).
Acceleration
The average acceleration of a body over a time interval is calculated by the rate of change of
velocity with respect to time,
&
& &
∆v v f − v i
&
a avge =
=
.
∆t
∆t
If acceleration is constant, this formula also gives us the value of the instantaneous acceleration at
&
any time during this time interval. The common symbol for acceleration is a (or a). The SI unit
for acceleration is metres per second per second (m/s/s or ms–2).
3.2 Motion with Constant Velocity
This is a particular case of motion with constant acceleration, with acceleration being equal to
zero. This is a very simple motion. The initial velocity, the final velocity, the average velocity
over any time interval, and the instantaneous velocity at any time are all the same. The velocity of
the body v and its displacement s over any time interval ∆t are connected by the relationship:
&
s
* s
v=
or v =
.
∆t
∆t
In this particular case the velocity is in one direction only, and thus the distance travelled by the
body is the same as its displacement during the time interval. Since the only relevant direction in
this motion is the direction in which the body is travelling, this equation is often more simply
written as
s
v=
t
– where v is the speed of the body and s is the distance travelled during the time interval t.
Example 1
(1)
(2)
An athlete runs a distance of 800m in 1min, 47⋅6s at constant speed. At what speed is he running?
A ship is sailing at a constant speed of 25 knots. How far will it sail in 24h?
(1 knot = 0⋅5139ms–1)
Solutions
(1)
s = 800m; t = 1m, 47⋅6s = 107⋅6s (2)
v=
s
t
800
=
107 ⋅ 6
−1
≈ 7 ⋅ 435ms
v = 25 knots = 25×0⋅5139ms–1 ≈ 12⋅8475ms–1
t = 24h = 24×3600s = 86400s
s = vt
= 12 ⋅ 8475 × 86400
= 1,110,024m ≈ 1110km
Problem-Solving Method
1 If velocity is constant, there is only one formula that is applicable to the motion:
s
s
v = , or its rearrangements s = vt or t = .
t
v
2
Note that in both of the above examples, one of the parameters is given in non-standard units. To
avoid potential problems in your calculation, always convert all given data to SI units, immediately.
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Key Ideas3
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3.3 Motion with Constant Acceleration
Average Velocity
velocity
For all motions, whether acceleration is constant or not, the
average velocity is given by
v
displacement s
v avge =
= .
time
t
However, consider a motion with constant acceleration, as
depicted in Figure 3.2. The body starts with an initial velocity u
and accelerates for t seconds, finishing with a final velocity v.
The area of the shaded trapezium gives the displacement in this
time period.
u
Remember that the area of a trapezium is given by
§ a+b·
A=¨
¸h
© 2 ¹
0
0
– where a and b are the lengths of the parallel sides and h is the
distance between the two parallel sides.
§u+v·
Therefore, the displacement s = ¨
¸t
© 2 ¹
s u+v
v avge = =
.
and
t
2
Fig 3.2
t
time
Thus, when acceleration is constant, you can determine the average velocity in any time interval
just by finding the average of the initial and the final velocities in that time.
Newton’s Equations of Motion
In the late 17th century, Sir Isaac Newton, in his study of the motion of a body moving in a straight
line with constant acceleration, derived three equations that related the five parameters of this
motion. They are
v = u + at
s = ut + 12 at 2
v 2 − u 2 = 2as
where:
a = the acceleration of the body;
t = the time interval for which this acceleration acts;
u = the velocity of the body at the beginning of this time interval, i.e. the initial velocity of the
body;
v = the velocity of the body at the end of this time interval, i.e. the final velocity of the body;
s = the displacement of the body in this time interval.
It is possible to solve every problem involving the motion of a body in a straight line with constant
acceleration by using these three equations, together with the two formulae involving average
velocity given above, i.e.
u+v
v avge =
and s = v avge t .
2
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Derivation of equation 1
From the definition of acceleration:
i.e.
therefore
i.e.
∆v
∆t
v −u
a=
t
at = v − u
v = u + at
a=
Derivation of equation 2
s = v avge t
§u +v·
=¨
¸t
© 2 ¹
§ u + u + at ·
=¨
¸t
2
©
¹
1
= 2 (2u + at )t
this is true for all motions
acceleration is constant, ∴ v avge =
u+v
2
v = u + at (from equation 1)
s = ut + 12 at 2
Derivation of equation 3
s = v avge t
§v+u·
s =¨
¸t
© 2 ¹
§ v − u ·§ v + u ·
as = ¨
¸¨
¸t
© t ¹© 2 ¹
(v − u )(v + u )t
as =
2t
2
v − u2
as =
2
2
2
v − u = 2as
§v−u·
as a = ¨
¸
© t ¹
cancel the t ' s
Problem-Solving Using Newton’s Equations of Motion
Example 2
A car is moving with a speed of 3⋅0ms–1 and then speeds up uniformly for 8⋅0s with an acceleration of
4⋅5ms–2. Find
(1) the speed of the car at the end of this time;
(2) the distance that the car travels in this time.
(1)
u = 3⋅0ms–1, a = 4⋅5ms–2, t = 8⋅0s, v = ?
v = u + at
= 3 + 4 × 5.8
v = 39ms
−1
(2)
u = 3⋅0ms–1, a = 4⋅5ms–2, t = 8⋅0s, s = ?
s = ut + 12 at 2
= 3 × 8 + 12 × 4 ⋅ 5 × 64
= 24 + 144
s = 168m
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Note the following.
(1)
The simplest approach to these problems is to:
• list all the data given and also the quantity that you are asked to find;
• apply a formula that links the unknown parameter with the given parameters.
(2)
If the question states that the body is speeding up uniformly, it means that the acceleration is
constant, and hence all the equations given in this section are applicable.
(3)
This example is a simple and straightforward application in the direct use of the equations of
motion. In this case the diagram does not contribute significantly to the visualisation and solution of
the problems. However, in more difficult problems, the diagram significantly helps you to visualise
the problem. Get into the habit of sketching diagrams, even for simple problems.
Example 3
An aeroplane is landing on a runway with a speed of 75⋅0ms–1. The length of the runway is 1⋅20km. The
aeroplane comes to a halt just before the end of the runway. Find
(1) its acceleration if it slows down uniformly;
(2) the time taken for the aeroplane to stop.
(1)
u = 75⋅0ms–1, a = ?, v = 0, s = 1⋅2km = 1200m
v 2 − u 2 = 2as
0 − 75 2 = 2.a.1200
− 5625
a=
2400
−2
a ≈ −2 ⋅ 34ms
(2)
u = 75⋅0ms–1, v =0, s = 1200m,
a = 2⋅34375ms–2, t = ?
v = u + at
0 = 75 − 2 ⋅ 3438t
75
t=
2 ⋅ 3438
t = 32 ⋅ 0s
Note the following.
58
(1)
In this example we have vectors in two directions. By stating that the initial velocity u = 75⋅0ms–1,
we are saying that the direction of the motion of the aircraft is the positive direction (to the left on
the diagram). The acceleration of the aircraft is in the opposite direction to its velocity (as the
aircraft is slowing down). Thus, the acceleration of the aircraft must be negative. The negative sign
just means that the direction of the acceleration is opposite to the direction of the velocity (to the
right on the diagram).
(2)
The question states that the body is slowing down uniformly. This means that the acceleration is
constant, and hence all the equations given in this section are applicable.
(3)
The data in the question is all given to an accuracy of three significant figures; therefore the answers
should only be given to three significant figures. Any more figures than this implies a higher
accuracy than can possibly be justified from the data.
However, if you calculate a value – such as acceleration in this example – that you are going to use
later in the question, then keep a record of this value to one or two significant figures more than you
are using in your answers, and use this value in further calculations
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Example 4
A car starts from rest and accelerates uniformly for 10⋅0s, achieving a speed of 20ms−1. It then travels at
constant velocity for 2⋅0min and then brakes to a halt in 3⋅0s. What is the total distance travelled by the
car?
(1)
u = 0, t = 10s, v = 20ms−1
s = v avge t
(u + v )t
1
= 2 (0 + 20 ) × 10
=
1
2
(2)
t = 120s, v =20ms−1 = const
s = vt
= 20 × 120
= 2400m
= 100m
(3)
u = 20ms−1, v = 0, t = 3s
s = v avge t
= 12 (u + v)t
= 12 (20 + 0) × 3
= 30m
Thus, the total distance travelled by this car = 100 + 2400 + 30
= 2530m
Note: The equations of motion only apply for motion with constant acceleration. Thus we have to treat
this motion as the combination of three separate motions as follows (assuming the car is moving
from left to right):
• a motion with initial velocity of 0, and positive acceleration for 10 s;
• a motion with constant velocity of 20ms−1 for 120s;
• a motion with initial velocity of 20ms−1 and final velocity of 0, for 3s.
3.4 Motion in the Earth’s Gravitational Field
The ancient Greek philosopher Aristotle (384–322 BC) taught that
heavier objects dropped faster than lighter ones. This was
believed to be true for almost two millennia, until about the year
1600. At that time, the great Italian physicist and astronomer
Galileo Galilei (1564–1642) proved that all objects dropped near
the surface of the Earth fall at the same rate, regardless of their
mass. He showed that all objects experience the same constant
downward acceleration near the surface of the Earth. The measure
of this acceleration is 9⋅80ms−2. It is usually represented by the
symbol g.
• Any object dropped from some height near the surface of the
Earth experiences an acceleration of 9⋅80ms−2 downwards, in
Galileo Galilei (1564 – 1642)
the absence of air resistance. Thus it continues to accelerate
−1
down, its speed increasing by 9⋅80ms in every second until it hits the Earth.
• Any object projected vertically upwards near the surface of the Earth experiences a vertical
acceleration of 9⋅80ms−2 downwards. Thus it will slow down, its upward speed decreasing by
9⋅80ms−1 in every second until it has reduced to zero, at which stage the body has come to a
momentary halt, and then the body returns to Earth with a downward acceleration of 9⋅80ms−2.
Any body moving in the Earth’s gravitational field, which only experiences the gravitational
acceleration g = 9⋅80ms−2, is said to be in free-fall. It does not matter whether the body is moving
up or down; it is still described as free-falling. Gravitational acceleration g is often described as
free-fall acceleration. In all free-fall calculations, we neglect the effects of air resistance and
assume that the acceleration is constant for small heights above the surface of the Earth. Thus the
motion can be analysed as one-dimensional motion with constant acceleration, and all the
equations of motion, previously derived, will apply.
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Example 5
An object is dropped from a height of 40m.
(1) How long does it take to reach the ground?
(2) With what speed does it hit the ground?
(1)
u = 0, s = 40m, a = 9⋅80ms−2, t = ?
a=g
= 9.8ms-2
40m
(2)
s = ut + 12 at 2
t = 2⋅857s, u = 0, s = 40m, a = 9⋅80ms−2, v = ?
v 2 − u 2 = 2as
v 2 − 0 = 2 × 9 ⋅ 8 × 40
= 784
40 = 0 + 12 × 9 ⋅ 8 × t 2
40
4⋅9
≈ 8 ⋅163
t = 2 ⋅ 86s
t2 =
v ≈ 28 ⋅ 0ms
−1
Note: The use of the word “dropped” here implies that the initial velocity of the body is zero.
Because all the motion is downwards and the acceleration is downwards, we take this direction to
be positive. Thus velocity at any time, and displacement and acceleration, are all positive.
In part (2), we could have used v = u + at to determine the final speed of the body.
Example 6
A firework skyrocket is projected vertically upwards from ground level with
a speed of 49ms−1.
(1) How high will it rise?
(2) How long before its body falls back to the ground?
(3) What will be its speed when it returns to ground?
a=g
= −9.8ms-2
Note: In this case the motion is initially upwards and the acceleration is downward. We usually take the
upward direction to be positive and the downward direction to be negative. Thus acceleration is
negative throughout the motion, while velocity is positive on the upward path and negative when
the body is falling.
At the topmost point of its path, the velocity of the body is momentarily zero.
(1) Method 1. Find the displacement when its
velocity is equal to zero.
u = 49ms−1, a = −9⋅80ms−2, v = 0, s = ?
v 2 − u 2 = 2as
0 − 49 2 = 2 × −9 ⋅ 8 × s
− 2401
s=
− 19 ⋅ 6
= 122 ⋅ 5m
(1)
Alternative approach. Find the time taken
for it to reach the maximum height and then
find its displacement at that time.
v = u + at
0 = 49 − 9 ⋅ 8t
49
t=
9 ⋅8
= 5s
s = ut + 12 at 2
= 49 × 5 − 4 ⋅ 9 × 5 2
= 122 ⋅ 5m
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(2)
Method 1. Find the times when the
displacement of the body (from its original
position) is equal to zero.
(2)
u = 49ms−1, a = −9⋅80ms−2, s = 0, t = ?
s = ut + 12 at 2
Alternative approach. Find the time taken
for the body to rise to its maximum height (as
above) and then just double it. The time for it
to fall from its topmost point back to ground
is the same as the time taken to rise to its
maximum height, as shown below.
u = 0, s =122.5m, a = 9⋅8ms−2, t = ?
0 = 49t − 4 ⋅ 9t 2
t (49 − 4 ⋅ 9t ) = 0
s = ut + 12 at 2
49
4⋅9
= 0 or 10
time taken to return to ground = 10s
122 ⋅ 5 = 0 + 4 ⋅ 9t 2
122 ⋅ 5
t2 =
4⋅9
= 25
t = 5s
t = 0 or
Note that for this calculation all the motion is
downwards, and so we take the downward
direction to be positive.
(3)
Method 1. The body is free-falling from a
height of 122⋅5m, with an initial velocity = 0.
Thus, take the downward direction as
positive.
u =0, s = 122.5, t = 5s, a = 9⋅80ms−2, v = ?
v = u + at
= 0 + 9 ⋅8× 5
= 49ms
3.5
Applications:
−1
(3)
Alternative approach.
v 2 − u 2 = 2as
v 2 − 0 = 2 × 9 ⋅ 8 × 122 ⋅ 5
v 2 = 2401
v = 49ms
−1
Note that in the absence of air resistance, the
speed with which it returns to ground is the
same as the speed with which it is projected.
Traffic Flow & Accident Investigation
3.5.1 Traffic Flow
Cars moving along a road – accelerating, decelerating, stopping etc. – are an example of motion in
one dimension. Traffic engineers need to ensure that traffic runs as smoothly as possible with a
minimum of hold-ups. They need to research and make decisions on the following things.
Patterns of traffic flow
How many cars can we expect to have moving along a particular road
in a given time?
Timing of traffic lights
At a particular intersection, for how long – given the width of the roads
and the expected traffic density – for how long must the lights remain
green, in order to clear the traffic efficiently? Also, for how long must
the amber light be on, in order to enable cars to clear the intersection
before the red light comes on?
Sequencing of traffic
lights
If a group of cars is stopped at a red light at the beginning of a traffic
light sequence, and then moves off when the green light comes on,
when must the next set of lights change to green, in order for these cars
to be able to keep on moving without having to stop?
Speed limits in
different situations
What compulsory or advisory speed limits are suitable for the road
conditions at various places, taking into account the sharpness of
bends?
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