In this section we will…
…develop our understanding of using numbers
and equations to describe motion.
What equations to describe motion
have you come across before?
What do we understand about
'acceleration'?
acceleration (m s–2) is
v u
a
t
rate of change of velocity per unit time
vu
a
t
Rearrange this to give v =…
v = u + at
There are three equations
which together are known as
the equations of motion.
when acceleration is constant
(uniform)
and
motion is in a straight line
You need to be able to:
 select the correct formula
 identify symbols and units
 carry out calculations to solve
the problems of real-life
motion.
You need to be able to:
 carry out experiments to verify
the equations of motion
To do this fully, you might find
it an interesting challenge
to…
 understand where the
equations come from.
v = u + at
Label the formula using correct symbols and units
40
35
Describe the motion of this object
velocity (m s-1)
30
25
20
15
10
5
0
0
50
100
150
200
time (s)
250
300
350
40
How can we determine the displacement of the object?
35
velocity (m s-1)
30
25
20
15
10
5
0
0
50
100
150
200
time (s)
250
300
350
40
Area under the graph = 1500 + 4500 = 6000 m
35
velocity (m s-1)
30
25
20
Area 2 = ½bh
= ½ × (35 –- 5) × 300
= 4500 m
15
10
5
Area 1 = 5 × 300 = 1500 m
0
0
50
100
150
200
time (s)
250
300
350
40
35
v
velocity (m s-1)
30
25
20
Area 2 = ½bh
= ½ × (v – u) × t
Since v = u + at
and v – u = at
15
10
5
u0
Area 1 = ut
0
50
100
150
200
time (s)
250
t
300
350
40
35
Area under the graph = displacement s
v
velocity (m s-1)
30
25
20
Area 2 = ½bh
= ½ × (v – u) × t
= ½ × at × t
= ½ × at2
15
10
5
u0
Area 1 = ut
0
50
100
150
200
time (s)
250
t
300
350
40
35
s = ut + ½at2
v
velocity (m s-1)
30
25
20
Area 2 = ½bh
= ½ × (v – u) × t
= ½ × at × t
= ½ × at2
15
10
5
u0
Area 1 = ut
0
50
100
150
200
time (s)
250
t
300
350
Start with Equation 1
v = u + at
and square it
2
v
= (u +
2
at)
2
v
2
u
2
2
at
= + 2uat +
2
2
2
v = u + 2a(ut+ ½at )
2
v
=
2
u
+ 2as
What do we need to think about
when using the equations of
motion?
What do the following quantities
have in common?
velocity
displacement
acceleration
The sign convention
When dealing with vector quantities we
must have both magnitude and
direction.
When dealing with one-dimensional
kinematics (ie motion in straight lines) we
use + and – to indicate travel in opposite
directions.
The sign convention
Normally we use the following convention:
positive +
negative –
positive +
negative –
The sign convention
Normally we use the following convention
positive +
negative –
positive +
negative –
Take care – in some questions the sign
convention is reversed
v = u + at
What does a positive value of acceleration
mean?
Using the normal sign convention
–ve
+ve
–ve
+ve
Christine Arron is a 100-m
sprint athlete.
Immediately the starting
pistol is fired, Christine
accelerates uniformly from
rest, reaching maximum
velocity at the 50-m mark in
4.16 s.
Her maximum velocity is
10.49 m s–1.
Calculate her acceleration
over the first 50 m of the
race, showing full working.
–ve
+ve
Her acceleration is
2.52 m s–2.
In this case, acceleration is
a rate of change of velocity
with time, with which we are
familiar.
A positive value means, in
this case, increasing
velocity with time.
What else might it mean?
–ve
+ve
As she passes the finish
line, Christine begins to
slow down.
She comes to rest in 8.20 s
from a velocity of 9.73 m s–1.
Calculate her acceleration,
showing full working.
–ve
+ve
Her acceleration is
a = –1.19 m s–2.
Notice that the acceleration
has a negative value.
Explain this.
–ve
+ve
Now consider Christine
running in the opposite
direction.
Notice that the sign
convention remains the
same.
–ve
+ve
Immediately the starting
pistol is fired, Christine
accelerates uniformly from
rest, reaching maximum
velocity at the 50-m mark in
4.16 s.
Her maximum velocity is
–10.49 m s–1 (why is it
negative?).
Calculate her acceleration
over the first 50 m of the
race, showing full working.
–ve
+ve
Her acceleration is
–2.52 m s–2.
What does the negative
mean?
–ve
+ve
As she passes the finish line,
Christine begins to slow
down.
She comes to rest in 8.20 s
from a velocity of –9.73 ms–1.
Calculate her acceleration,
showing full working.
–ve
+ve
Her acceleration
is a = 1.19 m s–2.
Notice that the
acceleration has a
negative value.
Explain this.
Equation 1 and the sign convention
A positive value means gaining speed while
moving in the positive direction.
–ve
+ve
Initial velocity
0m
s–1
Final velocity
+10.49 m
s–1
Acceleration
a
v  u 10.49  0

 2.52 m s 2
t
4.16
OR
A positive value means the object is losing
speed while moving in the negative direction.
–ve
+ve
Final velocity
Initial velocity
–9.73 m
s–1
0m
s–1
Acceleration
a
v  u 0  ( 9.73)

 1.19 m s 2
t
8.20
In summary:
A negative value means the object is gaining
speed while moving in the negative direction.
–ve
+ve
Initial velocity
Final velocity
0 m s–1
–10.49 m
s–1
Acceleration
a
v  u 10.49  0

 2.52 m s 2
t
4.16
OR
A negative value means the object is losing
speed while moving in the positive direction.
–ve
Initial velocity
+10.49 m s–1
+ve
Final velocity
0m
s–1
Acceleration
a
v  u 0  10.49

 2.52 m s 2
t
4.16
–ve
+ve
Usain Bolt is a Jamaican
sprinter and a three-times
Olympic gold medallist.
Immediately the starting
pistol is fired, Usain
accelerates uniformly from
rest. He reaches 8.70 m s–1
in 1.75 s. Calculate his
displacement in this time.
Step 1:
Write down the sign convention.
Step 2:
Write down what you know (think
suvat).
s
u
v
a
t
Step 3:
displacement
initial velocity
final velocity
acceleration
time
Any other information,
eg acceleration due to force of gravity?
Step 4:
Select formula – use data sheet.
Step 5:
Substitute values then rearrange
formula.
Step 6:
Write the answer clearly, including
magnitude and direction, and units.
–ve
+ve
Usain Bolt is a Jamaican
sprinter and a three-times
Olympic gold medallist.
Immediately the starting
pistol is fired, Usain
accelerates uniformly from
rest. He reaches 8.70 m s-1
in 1.75 s. Calculate his
displacement in this time.
–ve
+ve
s=?m
u = 0 m s–1
v = 8.70 m s–1
a=?
t = 1.75 s
v  u  at
1 2
s  ut  at
2
2
2
v  u  2as
-ve
+ve
s=?m
u = 0 m s–1
v = 8.70 m s–1
a=?
t = 1.75 s
1 2
s  ut  at
2
1
2
s  (0  1.75)  (  a  1.75 )
2
-ve
+ve
s=?m
u = 0 m s–1
v = 8.70 m s–1
a=?
t = 1.75 s
v  u  at
8.70  0  (a  1.75)
8.70  1.75a
8.70
–2
a
 4.97m s
1.75
-ve
+ve
s=?m
u = 0 m s–1
v = 8.70 m s–1
a=?
t = 1.75 s
1 2
s  ut  at
2
1
2
s  (0  1.75)  (  a  1.75 )
2
1
s  0  x 4.97  1.75 2
2
s  7.61 m
In the previous section we
developed…
…our understanding of using graphs to
describe motion
…our skills in interpreting graphs of motion
…our skills in describing motion using physics
terms correctly.
In this section we planned to…
…develop our understanding of using numbers
and equations to describe motion.
Next we will bring all of this together
and use…
…our understanding of using graphs to
describe motion
…our skills in interpreting graphs of motion
…our skills in describing motion using physics
terms correctly
… our understanding of using numbers and
equations to describe motion
for vertical motion
Everyday acceleration
What sort of accelerations do you
experience in everyday life?
How can this be investigated?
Everyday acceleration
Do you experience
accelerations only in the
horizontal?
Everyday acceleration
An accelerometer (a device which measures
acceleration in three dimensions) can be used
to investigate accelerations which you
experience in everyday life.
A stationary tennis ball
Describe its motion.
A tennis ball travelling vertically
upwards
Describe its motion.
Film the ball as it is thrown upwards
and use tracker.jar to analyse its motion.
Once you have done this, describe the
motion in detail using the words velocity,
acceleration and displacement.
A tennis ball
dropped from a height
Describe its motion.
Film the ball as it falls and use tracker.jar
to analyse its motion.
Once you have done this, describe the
motion in detail using the words velocity,
acceleration and displacement.
Two tennis balls
dropped from a height
Predict the motion.
Observe.
Explain!
Two tennis balls
dropped from a height
Was your initial prediction that
the two identical tennis balls
dropped from the same height
would hit the ground at exactly
the same time?
Two tennis balls
dropped from a height
What did you think when you
discovered that one ball had a
significantly greater mass than
the other?
What do you think should
have happened?
What did you observe?
The mass does not matter!
Both balls will hit the ground at
the same time when dropped
from the same height.
If you do not believe this,
tracker.jar will allow you to
analyse the motion.
© Erich Schrempp / Science Photo Library
© Nicola Jones
The elephant and the feather in free-fall
Suppose we can
switch off air
resistance.
Which will hit
the ground first?
The elephant and the feather with air
resistance
The force of gravity near
the Earth’s surface gives
all objects the same
acceleration.
So why doesn’t the
feather reach the ground
at the same time as the
elephant?
Dropping an elephant…
We commonly use
a negative to
indicate
downward motion.
but be warned – you may come across
questions in which the sign convention is
reversed.
Dropping an elephant in
the absence of air
resistance.
Dropping an elephant in the
absence of air resistance
Calculate speed, velocity, distance and displacement at 1-s
intervals.
Time
(s)
Speed
(m s–1)
Velocity
(m s–1)
Distance fallen (m)
Displacement (m)
0
0
0
0
0
1
2
3
4
5
Sketch graphs to show how the speed,
distance, velocity and displacement vary
with time during the free-fall.
Speed
Distance
Time
Velocity
Time
Displacement
Time
Time
Sketch graphs to show how the speed,
distance, velocity and displacement vary
with time during free-fall.
Speed
Distance
What is missing
from the graph
sketches?
Time
Velocity
Time
Displacement
Time
Time
Speed (m s–1)
0
Distance ( m)
Time (s)
Velocity (m s–1)
0
Time (s)
0
Displacement (m)
0
Time (s)
Time (s)
A tennis ball
dropped from a height and allowed
to bounce
Consider the ball being dropped, allowed
to bounce and return to its original height.
Sketch your predictions for speed–time,
velocity–time and acceleration–time
graphs.
Compare to the results from this
simulation.
http://www.helpmyphys
ics.co.uk/bouncingball.html
speed (m s–1 )
Describe the motion.
0
time (s)
velocity m s
–1

When dropped, the ball gains speed in
the negative direction hence the –ve
sign for acceleration.
time (s)
0
The ball then loses speed in the
positive direction, coming to rest at the
original height.
Does this happen in real life? Explain!
a ( m s–2)
0
Time (s)
Consider a tennis ball thrown
upwards and allowed to fall back to
its starting position.
A tennis ball
thrown upwards then allowed to fall
back to its starting position
Sketch the velocity, speed and
acceleration graphs to describe
its motion until it returns to its
starting position.
Virtual Higher
Experiments
→ Higher
Physics →
Mechanics
and
Properties of
Matter →
Activity 5b.
What is the
force acting
on a tennis
ball thrown
upwards?
Estimate the
initial acceleration
of a jumping
popper.
Calculate the initial
acceleration of a jumping
popper.
What
assumptions are
you making?
How could your
calculation be
improved?
Observe what
happens when
the groaning tube
is dropped.
Explain!
In the previous section we
developed…
…our understanding of using graphs to
describe motion
…our skills in interpreting graphs of motion
…our skills in describing motion using physics
terms correctly.
In this section we…
…developed our understanding of using
numbers and equations to describe motion.
Review your progress!