As 21b Onthemove

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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 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 at 2

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 linedrawn from object’s initial position to its final

position – vector quantity

– SI unit: metre (m)



Velocity (v )

average velocity = displacement change

time taken

v av = 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 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 Chertseyto Weybridge by theroutes shown

opposite. If both carstake 30 minutes tocomplete their

journeys calculate

their individualaverage speeds andvelocities.

car A: distance = 6km

car B: distance = 4km

displacement = 2km EASTChertsey

Weybridge



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 changetime taken

a av = Δv / Δt

vector quantitydirection: same as the velocity change

SI unit: ms -2

Instantaneous acceleration (a ) is the rate ofchange of velocity with time: a = dv / dt



Notes:

1. Change in velocity:

= final velocity (v ) – initial velocity (u )

so: a av = (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 INEXAMINATIONS!!



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.

a av = (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 graphs

The gradient of a distance-time graph isequal to the speed



Displacement-time graphs

The gradient of adisplacement-time graph is

equal to the velocity

The graph opposite showshow the displacement of an

object thrown upwards variesin time.

Note how the gradient fallsfrom a high positive value tozero (at maximum height) to alarge 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

t / s A

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

= acceleration

a = (v – u) / t

The area under the ‘curve’

= displacement

s = [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 v 1 .

A → B: constant velocity v 1 .

B → C → D : UNIFORM NEGATIVEacceleration from v 1 to

negative velocity v 2 .

At C: The body reverses direction

D → E: constant negative velocity v 2 .

E → F: NON-UNIFORM POSITIVE

acceleration to rest

v / ms-1

t / s

AB

C

D E

F

v 1

v 2



Question 2The graph shows the velocity-

time graph of a car. Calculateor state:

(a) the acceleration of the carduring the first 4 seconds.

(b) the displacement of the carafter 6 seconds.

(c) time T .

(d) the displacement after 11seconds.

(e) the average velocity of thecar over 11 seconds.

v / ms-1

t / sT

12

-10

4 6 11



Question 2

a) the acceleration ofthe 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 ofthe car after 6

seconds.

displacement = area= area A + area B

= ½ (12 x 4) + (12 x 2)

= 24 + 24displacement = 48 m

v / ms-1

t / sT

12

-10

4 6 11

area

A

area

B



Question 2(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 – 1.2T2.2T = 19.2

T = 19.2 / 2.2

T = 8.73 seconds

Note: T can also be found by scaledrawing or by using the equations ofuniform acceleration (see later).

v / ms-1

t / sT

12

-10

4 6 11

area

A

area

B



Question 2(d) the displacement after11 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

t / sT

12

-10

4 6 11

area

A

area

B area

C

area

D



Question 2

(e) the average velocityof 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

area

A

area

B area

C

area

D



Question 3

Sketch the displacement-time graph for the car ofquestion 2.

displacement-time

co-ordinates:

t / s s / m

0

46

8.73

11

t / s s / m

0 0

4 246 48

8.73 64.4

11 53.0

s / ms-1

t / sT

64

24

4 6 11

48

53



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 threebounces.

displacement

time

h

gradients = - 9.8 ms-2

velocity



The equations of uniform

acceleration

v = u + at

v 2 = u 2 + 2as

s = ½ (u + v) ts = u t + ½ at 2

v = FINAL velocity

u = INITIAL velocity

a = acceleration

t = time for the velocitychange

s = displacement during

the velocity change

THESE EQUATIONS ONLY APPLY

WHEN THE ACCELERATION

REMAINS CONSTANT



Question 1

Calculate the final velocity of a car thataccelerates 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 thatis decelerated at 2.5 ms -2 from an initial

velocity of 20 ms -1.

v 2 = u 2 + 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. Ifit accelerates downwards at 9.81 ms -2 and

reaches the bottom after 1.5s calculate the

height of the cliff.s = ut + ½ at 2

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 toaccelerate 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 withan initial speed of 8 ms -1. What is the

maximum height reached by the ball?

v 2 = u 2 + 2as where:

s = height upwards

u = 8 ms -1 upwardsv = 0 ms -1 (at maximum height)

a = - 9.81 ms -2 (acceleration is downwards)



Question 5 continued

v 2 = u 2 + 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 twodimensions. For example a stone beingthrown across a stretch of water has both

horizontal and vertical motion.

The motion of the body in two such

mutually perpendicular directions can betreated independently.



Example 1 A stone is thrownhorizontally at a speed of8.0 ms-1 from the top of avertical cliff.

If the stone falls vertically by

30m calculate the timetaken for the stone to reachthe bottom of the cliff andthe horizontal distancetravelled by the stone

(called the ‘range’).Neglect the effect of airresistance.

path of

stone

range

height

of fall



Example 1

Stage 1

Consider vertical motion only

s = ut + ½ at 2

30 = (0 x t ) + ½ (9.81 x (t )2)

30 = ½ (9.81 x (t )2)

30 = 4.905 x t 2

t 2 = 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 thestone 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

path of

stone

range

height

of fall



Further Questions

(a) Repeat this example thistime for a cliff of height 40mwith a stone thrownhorizontally at 20 ms-1.

time of fall = 2.83 s

range = 56.6 m

(b) How would these valuesbe changed if air resistancewas significant?

time of fall - longerrange - smaller

path of

stone

range

height

of fall



Example 2

A shell is fired at 200 ms-1

at an angle of 30 degrees to thehorizontal. 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.

v 2 = u 2 + 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 = - 19.62ss = 10000 / 19.62

s = 509.7

maximum 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 = - 9.81t

t = 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 only

During the time 20.38 seconds the shell moves horizontallyat 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 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 thesimulation 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 formulatriangle 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

http://phet.colorado.edu/new/simulations/sims.php?sim=The_Moving_Man

http://subscription.echalk.co.uk/Science/chemistry/atomicStructure/atomicStructure.html

http://phet.colorado.edu/new/simulations/sims.php?sim=Maze_Game


http://phet.colorado.edu/new/simulations/sims.php?sim=Motion_in_2D


http://www.walter-fendt.de/ph11e/acceleration.htm


http://www.ngsir.netfirms.com/englishhtm/Kinematics.htm


http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=185.0





http://subscription.echalk.co.uk/Science/physics/sycd/distanceTimeGraphs/index2.htm





http://www.bbc.co.uk/schools/ks3bitesize/science/physics/forces_motion_7.shtml



http://phet.colorado.edu/new/simulations/sims.php?sim=Projectile_Motion





http://www.ngsir.netfirms.com/englishhtm/ThrowABall.htm


http://www.walter-fendt.de/ph11e/projectile.htm


http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/jarapplet.html


http://www.ionaphysics.org/lab/Explore/dswmedia/range.htm


http://www.ionaphysics.org/lab/Explore/dswmedia/monkey.htm


http://jersey.uoregon.edu/vlab/Cannon/index.html

http://www.7stones.com/Homepage/Publisher/Vec04.html







http://jersey.uoregon.edu/vlab/Cannon/index.html


http://www.ionaphysics.org/lab/Explore/dswmedia/monkey.htm


http://www.ionaphysics.org/lab/Explore/dswmedia/range.htm


http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/jarapplet.html






http://www.ngsir.netfirms.com/englishhtm/ThrowABall.htm














http://www.bbc.co.uk/schools/ks3bitesize/science/physics/forces_motion_7.shtml




















http://www.ngsir.netfirms.com/englishhtm/Kinematics.htm


http://www.walter-fendt.de/ph11e/acceleration.htm


http://phet.colorado.edu/new/simulations/sims.php?sim=Motion_in_2D


http://phet.colorado.edu/new/simulations/sims.php?sim=Maze_Game


http://phet.colorado.edu/new/simulations/sims.php?sim=The_Moving_Man



Core Notes from Breithaupt pages 112 to 129

1. Define what is meant by: (a)displacement; (b) speed & (c)velocity.

2. Explain the difference betweenspeed and velocity.

3. Define what is meant byacceleration.

4. What is the difference between

UNIFORM and NON-UNIFORMacceleration? Illustrate youranswer with sketch graphs.

5. State the equations of constantacceleration. Explain what each ofthe five symbols used means.

6. What is meant by ‘free fall’?

7. What is the average value of theacceleration of free fall near theEarth’s surface?

8. What information is given by the

gradients of (a) distance-time; (b)displacement-time; & (c) velocity-timegraphs?

9. What information is given by the areaunder 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 projectedhorizontally at a speed of 20 ms-1 fromthe 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 andvelocity.

3. A ball is thrown vertically upwards and thencaught when it falls down again. Sketchdistance-time and displacement-time graphs of

the ball’s motion and explain why these graphsare different from each other.

4. Try the summary questions on page 113




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.




Notes from Breithaupt pages 116 to 118

1. State the equations of constant acceleration.Explain what each of the five symbols usedmeans.

2. Show how the four equations of constantacceleration can be derived from the basicdefinitions of average speed and acceleration.


time with the vehicle moving initially at 40 ms-1

applying its brakes over a distance of 80m.




Notes from Breithaupt pages 119 to 121

1. What is meant by ‘free fall’?

2. What is the average value of the accelerationof free fall near the Earth’s surface?

3. Describe a method of finding the accelerationdue to gravity in the laboratory.


time with the coin taking 1.2s to reach thebottom of the well.





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 betweendistance & displacement-time graphs.

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





1. Repeat the worked example on page 124 thistime with the vehicle moving initially at3.0 ms-1, 40m from the docking station. Alsothe 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 abovethe bed of sand creating an impression 0.030m

in the sand.3. Try the summary questions on page 125




1. List and explain the three principles applying tothe motion of all projectiles.

2. Repeat the worked example on page 127 thistime with the object projected horizontally at a

speed of 20 ms-1

from the top of a tower ofheight 40 m.

3. Derive expressions for the x and y

co-ordinates of a body at time t after it hasbeen projected horizontally with speed U.


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