2.1b mechanics on the move breithaupt pages 112 to 129 april 11 th, 2010

2.1b Mechanics On the moveBreithaupt pages 112 to 129

April 11th, 2010

AQA AS Specification

Lessons Topics

1 to 4 Motion along a straight lineDisplacement, 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 + 2ass = 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 motionIndependence 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 changetime taken

vav = Δx / Δtscalar quantitySI 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

Completedistance 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 QuestionTwo 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

car B: distance = 4km

displacement = 2km EASTChertsey

Weybridge

Car Aspeed

= 6km / 0.5h

= 12km h-1

velocity

= 2km EAST / 0.5h

= 4km h-1 EAST

Car Bspeed

= 4km / 0.5h

= 8km h-1

velocity

= 2km EAST / 0.5h

= 4km h-1 EAST

Acceleration (a)average acceleration = velocity change

time taken

aav = Δv / Δtvector quantitydirection: same as the velocity changeSI 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-2Initial Final

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 graphsThe gradient of a distance-time graph is equal to the speed

Displacement-time graphsThe gradient of a displacement-time graph isequal 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

QuestionDescribe the motion shown by the displacement-time graph below:

s / m

t / sA

B

C D

E

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 = accelerationa = (v – u) / t

The area under the ‘curve’ = displacements = [u x t] + [½ (v – u) x t]

velocity

Question 1Describe 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

v / ms-1

t / s

AB

C

D E

F

v1

v2

Question 2The 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.

v / ms-1

t / sT

12

-10

4 6 11

Question 2a) 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

v / ms-1

t / sT

12

-10

4 6 11

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

t / sT

12

-10

4 6 11

areaA

areaB

Question 2(c) time T.By similar triangles:(T - 6):(11 - T) = 12:10i.e. (T - 6) / (11 - T) = 12 / 10(T - 6) / (11 - T) = 1.20(T - 6) = 1.20 (11 – T)T - 6 = 13.2 – 1.2T2.2T = 19.2T = 19.2 / 2.2T = 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

t / sT

12

-10

4 6 11

areaA

areaB

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 3.27)= 24 + 24 + 16.38 – 11.35= 53.03displacement = 53.0 m

v / ms-1

t / sT

12

-10

4 6 11

areaA

areaB area

C

areaD

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

v / ms-1

t / sT

12

-10

4 6 11

areaA

areaB area

C

areaD

Question 3

Sketch the displacement-time graph for the car of question 2.

displacement-time co-ordinates:

t / s s / m

0

4

6

8.73

11

t / s s / m

0 0

4 24

6 48

8.73 64.4

11 53.0

s / ms-1

t / sT

64

24

4 6 11

4853

Question 4Sketch 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.

displacement

time

h

gradients = - 9.8 ms-2

velocity

The equations of uniform acceleration

v = u + at

v2 = u2 + 2as

s = ½ (u + v) t

s = ut + ½ at2

v = FINAL velocity

u = INITIAL velocity

a = acceleration

t = time for the velocity

change

s = displacement during

the velocity change

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 + 2as0 = 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) t30 = ½ (5 + 12) x t30 = 8.5 x t30 ÷ 8.5 = ttime = 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 + 2aswhere: s = height upwardsu = 8 ms -1 upwardsv = 0 ms -1 (at maximum height)a = - 9.81 ms -2 (acceleration is downwards)

Question 5 continued

v2 = u2 + 2as0 = (8)2 + 2 (-9.81 x s)0 = 64 - 19.62 x s- 64 = - 19.62 x s- 64 / - 19.62 = smaximum 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 1A 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.

path of stone

range

height of fall

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

path of stone

range

height of fall

Example 1

Stage 2Consider horizontal motion only

During the time 2.47 seconds the stone moves horizontally at a constant speed of 8.0 ms-1 speed = distance / timebecomes:distance = speed x time= 8.0 x 2.47= 19.8

range = 19.8 m

path of stone

range

height of fall

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 srange = 56.6 m

(b) How would these values be changed if air resistance was significant?time of fall - longerrange - smaller

path of stone

range

height of fall

Example 2A 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 path of

shell

range

maximum height

30°

Example 2Stage 1 - Part (a)Consider vertical motion only At the maximum height, s The final VERTICAL velocity, v = 0.v2 = u2 + 2as0 = (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 = - 19.62ss = 10000 / 19.62s = 509.7maximum height = 510 m

Example 2Stage 2 – Part (b)Consider vertical motion only v = u + at 0 = (200 sin 30°) + (- 9.81 x t)0 = 100 - 9.81t-100 = - 9.81tt = 100 / 9.81t = 10.19Time 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 2Stage 3 – Part (c)Consider horizontal motion onlyDuring the time 20.38 seconds the shell moves horizontally at a constant speed of (200 cos 30°) ms-1 speed = distance / timebecomes: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)

QuestionRepeat example 2 this time for a firing angle of 45°.sin 45° = 0.7071; 200 x sin 45° = 141.4maximum height = 1020 mtime of flight = 28.8 srange = 4072 m (4.07 km) Note: 45° yields the maximum range in this situation.

path of shell

range

maximum height

45°

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 1291. Define what is meant by: (a)

displacement; (b) speed & (c) velocity.

2. Explain the difference between speed and velocity.

3. Define what is meant by acceleration.

4. What is the difference between UNIFORM and NON-UNIFORM acceleration? Illustrate your answer with sketch graphs.

5. State the equations of constant acceleration. Explain what each of the five symbols used means.

6. What is meant by ‘free fall’?7. 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. What information is given by the gradients of (a) distance-time; (b) displacement-time; & (c) velocity-time graphs?

2. What information is given by the area under the curves of (a) speed-time & (b) velocity-time graphs?

3. Use figures 1 & 2 to explain the differences between distance & displacement-time graphs.

4. Repeat the worked example on page 123 this time for a ball released from 2.0 m rebounding to 1.2 m.

5. 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