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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 ?s / ?t and a ?v / ?t 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 objects initial position to its final

position - vector quantity
- SI unit metre (m)

Speed (v)

- average speed distance change
- time taken
- vav ?x / ?t
- 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 ?s / ?t
- 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 3 s 20 ms-1

1400 m 35 s 40 ms-1

300 m 0.20 s 1500 ms-1

80 km 2 h 40 km h-1

150 x 10 6 km 8 min 20 s 3.0 x 108 ms-1

1 km 3.03 s 330 ms-1

20

1400

0.20

40

8

20

3.03

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
- speed
- 6km / 0.5h
- 12km h-1
- velocity
- 2km EAST / 0.5h
- 4km h-1 EAST

- Car B
- speed
- 4km / 0.5h
- 8km h-1
- velocity
- 2km EAST / 0.5h
- 4km h-1 EAST

Acceleration (a)

- average acceleration velocity change
- time taken
- aav ?v / ?t
- 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) / ?t
- 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) / ?t
- (30 0) / 8
- average acceleration 3.75 ms-2

Complete

Velocity / ms-1 Velocity / ms-1 Time / s Acceleration / ms-2

Initial Final Time / s Acceleration / ms-2

0 45 15 3

0 24 3 8

30 90 10 6

20 5 3 - 5

0 - 60 20 - 3

45

3

30

- 5

- 60

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

O ? A acceleration from rest A ? B constant

velocity B ? C deceleration to rest C ? D

rest (no motion) D ? E acceleration from rest

back towards the starting point

Velocity-time graphs

- 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. A ? B constant velocity v1. 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

- The graph shows the velocity-time 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

- 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

- (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

v / ms-1

12

area A

area B

t / s

T

4 6 11

-10

Question 2

- (c) time T.
- By similar triangles
- (T - 6)(11 - T) 1210
- 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).

v / ms-1

12

area A

area B

t / s

T

4 6 11

-10

Question 2

- (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

v / ms-1

12

area A

area B

area C

t / s

T

4 6 11

area D

-10

Question 2

- (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 displacement-time 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

t / s s / m

0

4

6

8.73

11

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.

velocity

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 t / s s / m

2 14 0.75 ?

0 0.4 15 ?

16 0 - 8 ?

4 6 ? 20

16

45

16

4

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

128

6

2

0.5

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
- time of fall 2.47 s

height of fall

path of stone

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
- range 19.8 m

height of fall

path of stone

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
- (c) the 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.

Internet Links

- 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

- Two dogs running with graphs- NTNU
- Motion graphs test - NTNU
- 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

Core Notes from Breithaupt pages 112 to 129

- 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 Earths surface?

- What information is given by the gradients of (a)

distance-time (b) displacement-time (c)

velocity-time graphs? - What information is given by the area under the

curves of (a) speed-time (b) velocity-time

graphs? - List and explain the three principles applying to

the motion of all projectiles. - 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

- Define what is meant by (a) displacement (b)

speed (c) velocity. - Explain the difference between speed and

velocity. - A ball is thrown vertically upwards and then

caught when it falls down again. Sketch

distance-time and displacement-time graphs of the

balls motion and explain why these graphs are

different from each other. - Try the summary questions on page 113

Notes from Breithaupt pages 114 115

- Define what is meant by acceleration.
- What is the difference between UNIFORM and

NON-UNIFORM acceleration? Illustrate your answer

with sketch graphs. - Repeat the worked example on page 115 this time

with the vehicle moving initially at 12 ms-1

applying its brakes for 20s. - Try the summary questions on page 115

Notes from Breithaupt pages 116 to 118

- State the equations of constant acceleration.

Explain what each of the five symbols used means.

- Show how the four equations of constant

acceleration can be derived from the basic

definitions of average speed and acceleration. - 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. - Try the summary questions on page 118

Notes from Breithaupt pages 119 to 121

- What is meant by free fall?
- What is the average value of the acceleration of

free fall near the Earths surface? - Describe a method of finding the acceleration due

to gravity in the laboratory. - Repeat the worked example on page 121 this time

with the coin taking 1.2s to reach the bottom of

the well. - Try the summary questions on page 121

Notes from Breithaupt pages 122 123

- What information is given by the gradients of (a)

distance-time (b) displacement-time (c)

velocity-time 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

- 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. - 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. - Try the summary questions on page 125

Notes from Breithaupt pages 126 127

- List and explain the three principles applying to

the motion of all projectiles. - 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. - Derive expressions for the x and y

co-ordinates of a body at time t after it has

been projected horizontally with speed U. - Try the summary questions on page 127

Notes from Breithaupt pages 128 129

- Try the summary questions on page 129