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# AQA GCSE Physics 2-1 Motion - PowerPoint PPT Presentation

AQA GCSE Physics 2-1 Motion. GCSE Physics pages 120 to 131. February 28 th , 2011. AQA GCSE Specification. DESCRIBING MOVEMENT 12.1 How can we describe the way things move? Using skills, knowledge and understanding of how science works:

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### AQA GCSE Physics 2-1Motion

GCSE Physics pages 120 to 131

February 28th, 2011

DESCRIBING MOVEMENT

12.1 How can we describe the way things move?

Using skills, knowledge and understanding of how science works:

• to construct distance-time graphs for a body moving in a straight line when the body is stationary or moving with a constant speed

• to construct velocity-time graphs for a body moving with a constant velocity or a constant acceleration

HT to calculate the speed of a body from the slope of a distance-time graph

HT to calculate the acceleration of a body from the slope of a velocity-time graph

HT to calculate the distance travelled by a body from a velocity-time graph.

Skills, knowledge and understanding of how science works set in the context of:

• The slope of a distance-time graph represents speed.

• The velocity of a body is its speed in a given direction.

• The acceleration of a body is given by:

acceleration = change in velocity / time taken for change

• The slope of a velocity-time graph represents acceleration.

• The area under a velocity-time graph represents distance travelled.

speed = distance

time

In physics speed is usually measured in:

metres per second (m/s)

also:

distance =

and:

time =

distance

distance

speed

speed

time

1 kilometre per hour (km/h)

= 1000 metres per hour

but 1 hour = 3600 seconds

therefore 1 km/h = 1000m ÷ 3600 s

1 km/h = 0.28 m/s

and 1 m/s = 3.6 km/h

Also: 100 km/h = approx 63 m.p.h

Calculate the speed of a car that covers 500m in 20s.

speed = distance

time

= 500m / 20s

= 25 m/s (about 60 mph)

Sound waves travel at about 340m/s through air. How far will a sound wave travel in one minute?

distance = speed x time

= 340 m/s x 1 minute

= 340 m/s x 60 seconds

= 20 400 m

(20.4 km or about 13 miles)

20

1400

0.20

40

20

8

3.03

fast

slow

time

The slope or gradientof a distance-time graph is increases with speed.

The slope or gradientof a distance-time graph is equal to the speed.

In the graph opposite:

slope = 150m / 10s

= 15 m/s

= speed

distance

bus

car

time

Question 1

Sketch on the same set of axes distance-time graphs for:

(a) a car moving at a steady speed,

(b) a bus moving at a steady speed greater than the car,

(c) a lorry increasing in speed from rest.

Describe the motion of the three lorries X, Y and Z shown in the graph below.

Moving quickest

speed = 45000m / 1800s = 25 m/s

Lorry Y:

speed = 36000m / 1800s = 20 m/s

Lorry Z:

Moving slowest

0 to 600s; speed = 10000m / 600s = 16.7 m/s

600 to 1200s; stationary

1200 to 1800s; speed = 16.7 m/s

average speed = 20000m / 1800s = 11.1 m/s

Speed is equal to ________ divided by time and can be measured in _________ per second.

A speed of 20 m/s is the same as ______ km/h which is approximately equal to ______ mph.

The _________ of a distance against time graph can be used to calculate ________. The greater the gradient of the line the __________ is the speed. The line will be ___________ when the speed is zero.

distance

metres

72

40

slope

speed

higher

horizontal

WORD SELECTION:

slope

speed

distance

horizontal

40

higher

72

metres

Distance-time graphsNotes questions from pages 120 & 121

• Copy the equation for speed, along with the units used, at the bottom of page 120.

• Copy Figure 2 on page 120 and explain how a distance-time graph can be used to find speed.

• Copy and answer question (a) on page 120.

• Copy Figure 3 on page 121 and then copy and answer questions (b), (c) and (d) on page 121.

• Copy the Key Points on page 121.

• Answer the summary questions on page 121.

It would not have been as steep.

25 m/s

600 s

11.1 m/s

Summary questions:

(a) Speed

(b) Speed, distance. (c) Speed.

(a) 30 m/s

(b) 9000 m

The velocity of a body is its speed in a given direction.

The airplane opposite may loop at a constant speed but its velocity changes as its direction of motion changes.

A stone dropped off the top of a cliff falls down by 20m in 2s. Calculate its average velocity (a) downwards and (b) horizontally.

(a) average speed downwards = 20m / 2s

= 10m/s

Therefore velocity downwards = 10 m/s

(b) average speed horizontally = 0m / 2s

= 0m/s

Therefore velocity horizontally = 0 m/s

acceleration = velocity change

time taken

acceleration is measured in:

metres per second squared (m/s2)

acceleration = velocity change

time taken

velocity change is measured in m/s

time taken is measured in s

therefore acceleration = m/s ÷ s

= m/s2

1. Speed and velocity:

Often, but not always, speed can be used in the equation.

2. Change in velocity:

= final velocity – initial velocity

3. Deceleration:

This is where the speed is decreasing with time.

4. Circular motion at a constant speed:

Acceleration is occurring because the direction of motion is continually changing and hence so is velocity.

Complete the table below for an airplane accelerating at 8m/s2.

Calculate the acceleration of a car that changes in velocity from 5m/s to 25m/s in 4 seconds.

acceleration = velocity change

time taken

= (25m/s – 5m/s) / 4s

= 20 / 4

acceleration = 5 m/s2

Calculate the final velocity of a train that accelerates at 0.3m/s2 for 60 seconds from an initial velocity of 5m/s.

acceleration = velocity change

time taken

becomes: velocity change = acceleration x time taken

= 0.3m/s2 x 60s

= 18m/s

therefore final train velocity = 5m/s + 18m/s

= 23 m/s

Calculate the deceleration of a car that slows down from 18m/s to rest in 3 seconds.

acceleration = velocity change

time taken

= (0m/s – 18m/s) / 3s

= -18 / 3 (notice minus sign)

acceleration = - 6 m/s2

and so deceleration = 6 m/s2

Note: Deceleration is the negative of acceleration.

45

3

30

- 5

- 60

Velocity is speed measured in a particular ______________. A person walking northwards will have _______ velocity in a westwards direction.

Acceleration is equal to ________ change divided by the time taken. Acceleration is measured in metres per second ______.

Deceleration occurs when a body is _________ down. It is possible for a body to be accelerating even when its ______ is not changing provided its direction is, for example: a body moving in a ________.

direction

zero

velocity

squared

slowing

speed

circle

WORD SELECTION:

speed

zero

slowing

direction

squared

circle

velocity

Velocity and acceleration Notes questions from pages 122 & 123

• Explain the difference between speed and velocity.

• Explain how it is possible to be moving with a constant speed while changing in velocity.

• Copy and answer question (a) on page 122.

• (a) What is acceleration? (b) Copy the equation for acceleration, along with the units used, on page 123.

• Repeat the worked example shown on page 123 but this time change the time taken to 9 seconds.

• Copy and answer question (b) on page 123.

• Explain the meaning of ‘deceleration’.

• Copy the Key Points on page 123.

• Answer the summary questions on page 123.

600 m

2.5 m/s2

Summary questions:

(a) Speed.

(b) Acceleration.

(c) Velocity.

2. (a) 20 m/s

(b) 2.5 m/s2

high acceleration

low acceleration

constant velocity or zero acceleration

deceleration

time

Velocity-time graphs

The slope of a velocity-time graph represents acceleration.

time

The area under a velocity-time graph represents distance travelled.

area equals distance travelled

15

10

5

1 2 3 4

time (s)

Question 1

Sketch the velocity time graph of a car accelerating from rest to 15m/s in 3 seconds and then remaining at a constant speed for one more second.

12

8

4

area

1 2 3 4

time (s)

Question 2

Calculate the acceleration and the distance travelled after 4 seconds from using the graph opposite.

= y-step ÷ x-step

= (12 - 0)m/s ÷ (4 – 0)s

= 12 / 4

acceleration = 3 m/s2

distance = area under the graph

= area of triangle

= ½ x base x height

= ½ x 4s x 12m/s

distance travelled = 24m

Calculate the acceleration and distance travelled using the graph shown below.

Acceleration equals the slope of the graph

= y-step ÷ x-step

= (16 - 4)m/s ÷ (10s)

= 12 / 10

Acceleration = 1.2 m/s2

Distance travelled:

This equals the area below the graph

= area of rectangle + area of triangle

= (10s x 4m/s) + (½ x 10s x (12 – 4)m/s)

= 40m + 40m

Distance travelled = 80m

Calculate the distance travelled over 15 seconds and the deceleration during the final five seconds using the graph below.

This equals the area below the graph

= area of rectangle + area of triangle

= (10s x 20m/s) + (½ x 5s x 20m/s)

= 200m + 50m

Distance travelled = 250m

Deceleration:

Acceleration equals the slope of the graph

= y-step ÷ x-step

= (- 20m/s) ÷ (5s)

= - 4 m/s2

but deceleration = negative of acceleration

Deceleration = 4 m/s2

More about velocity-time graphsNotes questions from pages 124 & 125and also page 123

• Copy Figure 4 from page 123 – graph X ONLY.

• Explain how acceleration can be found from a velocity-time graph and hence find the acceleration represented by graph X.

• Copy Figure 3 from page 125 and describe the motion shown by this graph. You should include a calculation of the deceleration caused by braking.

• Copy and answer question (b) on page 125.

• How can distance travelled be found from a velocity-time graph?

• Copy and answer question (c) on page 125.

• Copy the Key Points on page 125.

• Answer the summary questions on page 125.

Less steep

It would not be as steep

Greater

Summary questions:

1. 1.B 2.A 3.D 4.C

2. (a) A (b) C

Using graphs Notes questions from pages 126 & 127

• Copy and answer questions (a) and (b) on page 126.

• Copy Figure 3 on page 126 and copy the calculations of the acceleration and the distance travelled over the ten second period shown on page 127.

• Copy the Key Points on page 127.

• Answer the summary questions on page 127.

NOTE – Graph paper needed for Q2.

15 m/s

The speed decreased gradually and became constant.

Summary questions:

(a) The cyclist accelerates with a constant acceleration for 40s, and then decelerates to a standstill in 20s.

(b) (i) 0.2 m/s2

(ii) 160 m

2. (a) Graph

(b) 2 m/s2

(c) (i) 400m

(ii) 400 m

Measuring Velocity

Measuring Acceleration

Virtual Physics Laboratory SimulationsNOTE: Links work only in school

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.

Ladybug motion in 2D - PhET - Learn about position, velocity and acceleration vectors. Move the ladybug by setting the position, velocity or acceleration, and see how the vectors change. Choose linear, circular or elliptical motion, and record and playback the motion to analyze the behavior

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 AQA GCSE Bitesize Revision:

Speed, distance and time

Distance-time graphs

Velocity-time graphs

Acceleration

Distance-time graphs (higher)

Velocity-time graphs (higher)

Online Simulations

Transport issuesNotes questions from pages 128 & 129

• Answer questions 1 and 2 on page 128.

• (a) (i) 5 litres (ii) 4 litres

(b) £1.70

• Journey 1: 6.7 km/h

Journey 2: 4000 km/h

They were base on non-scientific evidence. It could be hearsay or even prejudice. It is unlikely that ALL cars exceeded the speed limit.

Difficult to prove a casual relationship here. Evidence would be needed from both before and after the installation of road bumps at both sites.

Time for cars to cover a suitable distance recorded. Speed = distance / time.

Survey should be carried out at sample times at regular intervals during the day, for example 10 minutes every hour. A fair test with valid results.

As many as possible, but at least 10 in each sample. A preliminary test could be used to determine the best number and how often.

No. Otherwise drivers might behave differently. This is part of the control in such surveys.