2.1b Mechanics
On the move
Breithaupt pages 112 to 129
April 11th, 2010
AQA AS Specification
Lessons
Topics
1 to 4
Motion along a straight line
Displacement, speed, velocity and acceleration.
v = Δs / Δt and a = Δv / Δt
Representation by graphical methods of uniform and non-uniform acceleration;
interpretation of velocity-time and displacement-time graphs for uniform and
non uniform acceleration; significance of areas and gradients.
Equations for uniform acceleration;
v = u + at , v2 = u2 + 2as
s = ½ (u + v) t , s = ut + ½ at2
Acceleration due to gravity, g; detailed experimental methods of measuring g are
not required.
5&6
Projectile motion
Independence of vertical and horizontal motion; problems will be soluble from
first principles. The memorising of projectile equations is not required.
Distance (x) and Displacement (s)
Distance (x)
– the length of the path moved by an object
– scalar quantity
– SI unit: metre (m)
Displacement (s)
– the length and direction of the straight line
drawn from object’s initial position to its final
position
– vector quantity
– SI unit: metre (m)
Speed (v)
average speed = distance change
time taken
vav = Δx / Δt
scalar quantity
SI unit: ms -1
Instantaneous speed (v) is the rate of
change of distance with time: v = dx / dt
Velocity (v)
average velocity = displacement change
time taken
vav = Δs / Δt
vector quantity
direction: same as the displacement change
SI unit: ms -1
Instantaneous velocity (v) is the rate of change of
displacement with time: v = ds / dt
Speed and Velocity Conversions
1 kilometre per hour (km h-1)
= 1000 m h-1
= 1000 / 3600 ms-1
1 km h-1 = 0.28 ms-1
and 1 ms-1 = 3.6 km h-1
Also:
100 km h-1 = 28 ms-1 = approx 63 m.p.h
Complete
distance
time
speed
60 m
3s
20 ms-1
1400 m
35 s
40 ms-1
300 m
0.20 s
1500 ms-1
80 km
2h
40 km h-1
150 x 10 6 km
88 min 20 s
3.0 x 108 ms-1
1 km
3.03 s
330 ms-1
Speed and Velocity Question
Two cars (A and B)
travel from Chertsey
to Weybridge by the
routes shown
opposite. If both cars
take 30 minutes to
complete their
journeys calculate
their individual
average speeds and
velocities.
car A: distance = 6km
Chertsey
displacement = 2km EAST
Weybridge
car B: distance = 4km
Car A
Car B
speed
= 6km / 0.5h
= 12km h-1
speed
= 4km / 0.5h
= 8km h-1
velocity
= 2km EAST / 0.5h
= 4km h-1 EAST
velocity
= 2km EAST / 0.5h
= 4km h-1 EAST
Acceleration (a)
average acceleration = velocity change
time taken
aav = Δv / Δt
vector quantity
direction: same as the velocity change
SI unit: ms -2
Instantaneous acceleration (a) is the rate of
change of velocity with time: a = dv / dt
Notes:
1. Change in velocity:
= final velocity (v) – initial velocity (u)
so: aav = (v – u) / Δt
2. Uniform acceleration:
This is where the acceleration remains
constant over a period of time.
3. Deceleration:
This is where the magnitude of the velocity
is decreasing with time.
Acceleration due to gravity (g)
An example of uniform acceleration.
In equations ‘a’ is substituted by ‘g’
On average at sea level:
g = 9.81 ms-2 downwards
g is often approximated to 10 ms -2
YOU ARE EXPECTED TO USE 9.81 IN
EXAMINATIONS!!
Question
Calculate the average acceleration of a car
that moves from rest (0 ms-1) to 30 ms-1 over
a time of 8 seconds.
aav = (v – u) / Δt
= (30 – 0) / 8
average acceleration = 3.75 ms-2
Complete
Velocity / ms-1
Time
/s
Acceleration
/ ms-2
15
3
Initial
0
Final
45
0
24
3
8
30
90
10
6
20
5
3
-5
0
60
- -60
20
-3
Distance-time graphs
The gradient of a distance-time graph is
equal to the speed
Displacement-time graphs
The gradient of a
displacement-time graph is
equal to the velocity
The graph opposite shows
how the displacement of an
object thrown upwards varies
in time.
Note how the gradient falls
from a high positive value to
zero (at maximum height) to a
large negative value.
Estimate the initial velocity of the object.
Initial gradient = (5 – 0)m / (0.5 – 0)s = 10 ms-1
Initial velocity = 10 ms-1
Question
Describe the motion shown by the displacement-time graph below:
s/m
C
O → A: acceleration from rest
D
A → B: constant velocity
B
E
B → C: deceleration to rest
C → D: rest (no motion)
D → E: acceleration from rest
back towards the
starting point
A
t/s
Velocity-time graphs
velocity
With velocity-time graphs:
gradient
= acceleration
a = (v – u) / t
The area under the ‘curve’
= displacement
s = [u x t] + [½ (v – u) x t]
Question 1
Describe the motion shown by the velocity-time graph below:
O → A: UNIFORM POSITIVE
acceleration from rest to
velocity v1.
v / ms-1
v1
A
B
A → B: constant velocity v1.
t/s
C
F
v2
D
E
B → C → D : UNIFORM NEGATIVE
acceleration from v1 to
negative velocity v2.
At C:
The body reverses direction
D → E: constant negative velocity v2.
E → F: NON-UNIFORM POSITIVE
acceleration to rest
Question 2
v / ms-1
12
T
4
-10
6
t/s
11
The graph shows the velocitytime graph of a car. Calculate
or state:
(a) the acceleration of the car
during the first 4 seconds.
(b) the displacement of the car
after 6 seconds.
(c) time T.
(d) the displacement after 11
seconds.
(e) the average velocity of the
car over 11 seconds.
Question 2
v / ms-1
12
T
4
-10
6
t/s
11
a) the acceleration of
the car during the first
4 seconds.
acceleration = gradient
= (12 - 0)ms-1 / (4 – 0)s
= 12 / 4
acceleration = 3 ms-2
Question 2
v / ms-1
12
area
A
area
B
4
-10
6
T
t/s
11
(b) the displacement of
the car after 6
seconds.
displacement = area
= area A + area B
= ½ (12 x 4) + (12 x 2)
= 24 + 24
displacement = 48 m
Question 2
v / ms-1
12
area
A
area
B
4
-10
6
T
t/s
11
(c) time T.
By similar triangles:
(T - 6):(11 - T) = 12:10
i.e. (T - 6) / (11 - T) = 12 / 10
(T - 6) / (11 - T) = 1.20
(T - 6) = 1.20 (11 – T)
T - 6 = 13.2 – 1.2T
2.2T = 19.2
T = 19.2 / 2.2
T = 8.73 seconds
Note: T can also be found by scale
drawing or by using the equations of
uniform acceleration (see later).
Question 2
v / ms-1
12
area
A
area
B area
C
4
6
T
t/s
11
area
D
-10
(d) the displacement after
11 seconds.
displacement = area
= area A + area B + area C
– area D
= 24 + 24 + ½ (12 x 2.73)
– ½ (10 x 2.27)
= 24 + 24 + 16.38 – 11.35
= 53.03
displacement = 53.0 m
Question 2
v / ms-1
12
area
A
area
B area
C
4
6
T
t/s
11
area
D
-10
(e) the average velocity
of the car over 11
seconds.
average velocity
= displacement / time
= 53.03 / 11
average velocity
= 4.82 ms-1
Question 3
Sketch the displacementtime graph for the car of
question 2.
displacement-time
co-ordinates:
t/s
s/m
0
0
4
24
6
48
8.73
64.4
11
53.0
s / ms-1
64
53
48
24
4
6
T 11
t/s
Question 4
Sketch displacement and velocity time graphs for a bouncing ball.
Take the initial displacement of the ball to be h at time t = 0.
Use the same time axis for both curves and show at least three
bounces.
h
displacement
velocity
time
gradients = - 9.8 ms-2
The equations of uniform
acceleration
v = FINAL velocity
u = INITIAL velocity
a = acceleration
t = time for the velocity
change
s = displacement during
the velocity change
v = u + at
v2 = u2 + 2as
s = ½ (u + v) t
s = ut + ½ at2
THESE EQUATIONS ONLY APPLY
WHEN THE ACCELERATION
REMAINS CONSTANT
Question 1
Calculate the final velocity of a car that
accelerates at 2ms -2 from an initial velocity
of 3ms -1 for 5 seconds.
v = u + at
v = 3 + (2 x 5)
= 3 + 10
final velocity = 13 ms-1
Question 2
Calculate the stopping distance of a car that
is decelerated at 2.5 ms -2 from an initial
velocity of 20 ms -1.
v2 = u2 + 2as
0 = 202 + (2 x - 2.5 x s)
0 = 400 + - 5s
- 400 = - 5s
- 400 / - 5 = s
stopping distance = 80 m
Question 3
A stone is dropped from the edge of a cliff. If
it accelerates downwards at 9.81 ms -2 and
reaches the bottom after 1.5s calculate the
height of the cliff.
s = ut + ½ at2
s = (0 x 1.5) + ½ (9.81 x (1.5)2)
s = ½ (9.81 x 2.25)
cliff height = 11.0 m
Question 4
Calculate the time taken for a car to
accelerate uniformly from 5 ms -1 to 12 ms -1
over a distance of 30m.
s = ½ (u + v) t
30 = ½ (5 + 12) x t
30 = 8.5 x t
30 ÷ 8.5 = t
time = 3.53 s
Question 5
A ball is thrown upwards against gravity with
an initial speed of 8 ms -1. What is the
maximum height reached by the ball?
v2 = u2 + 2as
where:
s = height upwards
u = 8 ms -1 upwards
v = 0 ms -1 (at maximum height)
a = - 9.81 ms -2 (acceleration is downwards)
Question 5 continued
v2 = u2 + 2as
0 = (8)2 + 2 (-9.81 x s)
0 = 64 - 19.62 x s
- 64 = - 19.62 x s
- 64 / - 19.62 = s
maximum ball height = + 3.26 m
Calculate the ? quantities
u / ms-1 v / ms-1 a / ms-2
2
14
0
16
0
4
6
t/s
0.75
?
16
0.4
15
-8
s/m
?
45
?
16
4?
20
Calculate the other quantities
u / ms-1 v / ms-1 a / ms-2
t/s
s/m
2
14
0.75
16
128
0
6
0.4
15
45
16
0
-8
2
16
4
6
0.5
4
20
Projectile motion
This is where a body is moving in two
dimensions. For example a stone being
thrown across a stretch of water has both
horizontal and vertical motion.
The motion of the body in two such
mutually perpendicular directions can be
treated independently.
Example 1
A stone is thrown
horizontally at a speed of
8.0 ms-1 from the top of a
vertical cliff.
If the stone falls vertically by
30m calculate the time
taken for the stone to reach
the bottom of the cliff and
the horizontal distance
travelled by the stone
(called the ‘range’).
Neglect the effect of air
resistance.
height
of fall
path of
stone
range
Example 1
Stage 1
Consider vertical motion only
s = ut + ½ at2
30 = (0 x t) + ½ (9.81 x (t)2)
30 = ½ (9.81 x (t)2)
30 = 4.905 x t2
t2 = 6.116
height
of fall
path of
stone
time of fall = 2.47 s
range
Example 1
Stage 2
Consider horizontal motion only
During the time 2.47 seconds the
stone moves horizontally at a
constant speed of 8.0 ms-1
speed = distance / time
becomes:
distance = speed x time
= 8.0 x 2.47
= 19.8
height
of fall
path of
stone
range = 19.8 m
range
Further Questions
(a) Repeat this example this
time for a cliff of height 40m
with a stone thrown
horizontally at 20 ms-1.
time of fall = 2.83 s
range = 56.6 m
(b) How would these values
be changed if air resistance
was significant?
time of fall - longer
range - smaller
height
of fall
path of
stone
range
Example 2
A shell is fired at 200 ms-1 at an angle of 30 degrees to the
horizontal. Neglecting air resistance calculate:
(a) the maximum height reached by the shell
(b) the time of flight
path of
(c) the range
shell
30°
maximum
height
range
Example 2
Stage 1 - Part (a)
Consider vertical motion only
At the maximum height, s
The final VERTICAL velocity, v = 0.
v2 = u2 + 2as
0 = (200 sin 30°)2 + (2 x - 9.81 x s) [upwards +ve]
0 = (200 x 0.500)2 + (-19.62 x s)
0 = (100)2 + (-19.62 x s)
0 = 10000 - 19.62s
- 10000 = - 19.62s
s = 10000 / 19.62
s = 509.7
maximum height = 510 m
Example 2
Stage 2 – Part (b)
Consider vertical motion only
v = u + at
0 = (200 sin 30°) + (- 9.81 x t)
0 = 100 - 9.81t
-100 = - 9.81t
t = 100 / 9.81
t = 10.19
Time to reach maximum height = 10.19 s
If air resistance can be neglected then this is also the time
for the shell to fall to the ground again.
Hence time of flight = 2 x 10.19
time of flight = 20.4 seconds
Example 2
Stage 3 – Part (c)
Consider horizontal motion only
During the time 20.38 seconds the shell moves horizontally
at a constant speed of (200 cos 30°) ms-1
speed = distance / time
becomes:
distance = speed x time
= (200 cos 30°) x 20.38
= (200 x 0.8660) x 20.38
= 173.2 x 20.38
= 3530
range = 3530 m (3.53 km)
Question
Repeat example 2 this time for a firing angle of 45°.
sin 45° = 0.7071; 200 x sin 45° = 141.4
maximum height = 1020 m
time of flight = 28.8 s
range = 4072 m (4.07 km)
Note: 45° yields the maximum range in this situation.
path of
shell
45°
maximum
height
range
Internet Links
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The Moving Man - PhET - Learn about position,
velocity, and acceleration graphs. Move the little man
back and forth with the mouse and plot his motion. Set
the position, velocity, or acceleration and let the
simulation move the man for you.
Maze Game - PhET - Learn about position, velocity, and
acceleration in the "Arena of Pain". Use the green arrow
to move the ball. Add more walls to the arena to make
the game more difficult. Try to make a goal as fast as
you can.
Motion in 2D - PhET - Learn about velocity and
acceleration vectors. Move the ball with the mouse or let
the simulation move the ball in four types of motion (2
types of linear, simple harmonic, circle). See the velocity
and acceleration vectors change as the ball moves.
Motion with constant acceleration - Fendt
Bouncing ball with motion graphs - netfirms
Displacement-time graph with set velocities - NTNU
Displacement & Aceleration-time graphs with set
velocities - NTNU
Displacement & Velocity-time graphs with set
accelerations - NTNU
Football distance-time graphs - eChalk
Motion graphs with tiger- NTNU
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Two dogs running with graphs- NTNU
Motion graphs test - NTNU
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BBC KS3 Bitesize Revision: Speed - includes formula
triangle applet
Projectile Motion - PhET- Blast a Buick out of a cannon!
Learn about projectile motion by firing various objects.
Set the angle, initial speed, and mass. Add air resistance.
Make a game out of this simulation by trying to hit a
target.
Projectile motion - with or without air drag - NTNU
Projectile motion - NTNU
Projectile motion x- with variable height of projection netfirms
Projectile Motion - Fendt
Projectile motion - Virgina
Golf stroke projectile challenge - Explore Science
Shoot the monkey - Explore Science
Canon & target projectile challenge- Sean Russell
Slug projectile motion game - 7stones
Bombs released from an aeroplane- NTNU
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Core Notes from Breithaupt pages 112 to 129
1.
2.
3.
4.
5.
6.
7.
Define what is meant by: (a)
displacement; (b) speed & (c)
velocity.
Explain the difference between
speed and velocity.
Define what is meant by
acceleration.
What is the difference between
UNIFORM and NON-UNIFORM
acceleration? Illustrate your
answer with sketch graphs.
State the equations of constant
acceleration. Explain what each of
the five symbols used means.
What is meant by ‘free fall’?
What is the average value of the
acceleration of free fall near the
Earth’s surface?
8. What information is given by the
gradients of (a) distance-time; (b)
displacement-time; & (c) velocity-time
graphs?
9. What information is given by the area
under the curves of (a) speed-time &
(b) velocity-time graphs?
10. List and explain the three principles
applying to the motion of all projectiles.
11. Repeat the worked example on page
127 this time with the object projected
horizontally at a speed of 20 ms-1 from
the top of a tower of height 40 m.
Notes from Breithaupt pages 112 & 113
1. Define what is meant by: (a) displacement; (b)
speed & (c) velocity.
2. Explain the difference between speed and
velocity.
3. A ball is thrown vertically upwards and then
caught when it falls down again. Sketch
distance-time and displacement-time graphs of
the ball’s motion and explain why these graphs
are different from each other.
4. Try the summary questions on page 113
Notes from Breithaupt pages 114 & 115
1. Define what is meant by acceleration.
2. What is the difference between UNIFORM and
NON-UNIFORM acceleration? Illustrate your
answer with sketch graphs.
3. Repeat the worked example on page 115 this
time with the vehicle moving initially at 12 ms-1
applying its brakes for 20s.
4. Try the summary questions on page 115
Notes from Breithaupt pages 116 to 118
1. State the equations of constant acceleration.
Explain what each of the five symbols used
means.
2. Show how the four equations of constant
acceleration can be derived from the basic
definitions of average speed and acceleration.
3. Repeat the worked example on page 117 this
time with the vehicle moving initially at 40 ms-1
applying its brakes over a distance of 80m.
4. Try the summary questions on page 118
Notes from Breithaupt pages 119 to 121
1. What is meant by ‘free fall’?
2. What is the average value of the acceleration
of free fall near the Earth’s surface?
3. Describe a method of finding the acceleration
due to gravity in the laboratory.
4. Repeat the worked example on page 121 this
time with the coin taking 1.2s to reach the
bottom of the well.
5. Try the summary questions on page 121
Notes from Breithaupt pages 122 & 123
1.
2.
3.
4.
5.
What information is given by the gradients of (a)
distance-time; (b) displacement-time; & (c) velocitytime graphs?
What information is given by the area under the curves
of (a) speed-time & (b) velocity-time graphs?
Use figures 1 & 2 to explain the differences between
distance & displacement-time graphs.
Repeat the worked example on page 123 this time for
a ball released from 2.0 m rebounding to 1.2 m.
Try the summary questions on page 123
Notes from Breithaupt pages 124 & 125
1. Repeat the worked example on page 124 this
time with the vehicle moving initially at
3.0 ms-1, 40m from the docking station. Also
the motors only stay on 3 seconds longer this
time.
2. Repeat the worked example on page 125 this
time with the ball being released 0.75 m above
the bed of sand creating an impression 0.030m
in the sand.
3. Try the summary questions on page 125
Notes from Breithaupt pages 126 & 127
1. List and explain the three principles applying to
the motion of all projectiles.
2. Repeat the worked example on page 127 this
time with the object projected horizontally at a
speed of 20 ms-1 from the top of a tower of
height 40 m.
3. Derive expressions for the x and y
co-ordinates of a body at time t after it has
been projected horizontally with speed U.
4. Try the summary questions on page 127
Notes from Breithaupt pages 128 & 129
1. Try the summary questions on page 129