AQA GCSE Physics 2-1

Motion

GCSE Physics pages 120 to 131

February 28th, 2011

AQA GCSE SpecificationDESCRIBING MOVEMENT12.1 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 accelerationHT to calculate the speed of a body from the slope of a distance-time graphHT to calculate the acceleration of a body from the slope of a velocity-time graphHT 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

speed = distance

time

In physics speed is usually measured in:

metres per second (m/s)

also:

distance =

and:

time =speed time

distance

speed x time

distance speed

Speed Conversions1 kilometre per hour (km/h)= 1000 metres per hourbut 1 hour = 3600 secondstherefore 1 km/h = 1000m ÷ 3600 s1 km/h = 0.28 m/sand 1 m/s = 3.6 km/h

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

Question 1

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

speed = distance

time

= 500m / 20s

= 25 m/s (about 60 mph)

Question 2

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)

Completedistance time speed

60 m 3 s 20 m/s

1400 m 35 s 40 m/s

300 m 0.20 s 1500 m/s

80 km 2 h 40 km/h

150 x 10 6 km 8 min 20 s 3.0 x 108 m/s

1 km 3.03 s 330 m/s

20

1400

0.20

40

8 20

3.03

Distance-time graphs

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

dis

tan

ce

time

slow

fast

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

In the graph opposite:

slope = 150m / 10s

= 15 m/s

= speed

Question 1Sketch 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.

dis

tan

ce

time

car

bus

lorry

Question 2Describe the motion of the three lorries X, Y and Z shown in the graph below.

Lorry X:Moving quickestspeed = 45000m / 1800s = 25 m/s

Lorry Y:speed = 36000m / 1800s = 20 m/s

Lorry Z:Moving slowest0 to 600s; speed = 10000m / 600s = 16.7 m/s600 to 1200s; stationary1200 to 1800s; speed = 16.7 m/saverage speed = 20000m / 1800s = 11.1 m/s

Choose appropriate words to fill in the gaps below:

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.

higher 7240

distanceslope

metres

WORD SELECTION:

speed horizontal

higher

72

40

distance

slope

metres

speed

horizontal

Distance-time graphsNotes questions from pages 120 & 121

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

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

3. Copy and answer question (a) on page 120.4. Copy Figure 3 on page 121 and then copy and answer

questions (b), (c) and (d) on page 121.5. Copy the Key Points on page 121.6. Answer the summary questions on page 121.

Distance-time graphs ANSWERS

In text questions:

(a) It would not have been as steep.

(b) 25 m/s

(c) 600 s

(d) 11.1 m/s

Summary questions:

1. (a) Speed

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

2. (a) 30 m/s

(b) 9000 m

Velocity

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.

Question 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/sTherefore velocity downwards = 10 m/s(b) average speed horizontally = 0m / 2s= 0m/sTherefore velocity horizontally = 0 m/s

Acceleration

acceleration = velocity change

time taken

acceleration is measured in:

metres per second squared (m/s2)

Why is acceleration measured in 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

Other notes: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.

Question 1

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

time (s) 0 1 2 3 4

velocity (m/s) 0 8 16 24 32

Question 2

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

Question 3Calculate 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

Question 4Calculate 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.

CompleteVelocity (m/s) Time

(s)

Acceleration

(m/s2)Initial 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

Answers

Choose appropriate words to fill in the gaps below:

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 ________.

slowing circlezerospeed direction squared velocity

WORD SELECTION:

slowing

circle

zero

speed

direction

squaredvelocity

Velocity and acceleration Notes questions from pages 122 & 123

1. Explain the difference between speed and velocity.2. Explain how it is possible to be moving with a constant speed

while changing in velocity.3. Copy and answer question (a) on page 122.4. (a) What is acceleration? (b) Copy the equation for acceleration,

along with the units used, on page 123.5. Repeat the worked example shown on page 123 but this time

change the time taken to 9 seconds.6. Copy and answer question (b) on page 123.7. Explain the meaning of ‘deceleration’.8. Copy the Key Points on page 123.9. Answer the summary questions on page 123.

Velocity and acceleration ANSWERS

In text question:

(a) 600 m

(b) 2.5 m/s2

Summary questions:

1. (a) Speed.

(b) Acceleration.

(c) Velocity.

2. (a) 20 m/s

(b) 2.5 m/s2

Velocity-time graphs

velo

city

time

low acceleration

high

acc

eler

atio

nThe slope of a velocity-time graph represents acceleration.

deceleration

constant velocity or zero acceleration

area equals distance travelled

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

loci

tytime

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.

velocity (m/s)

time (s)

1 2 3 4

15

5

10

area

Question 2Calculate the acceleration and the distance travelled after 4 seconds from using the graph opposite.

velocity (m/s)

time (s)

1 2 3 4

12

4

8

acceleration = gradient

= 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

Question 3Calculate the acceleration and distance travelled using the graph shown below.

Acceleration:Acceleration equals the slope of the graph= y-step ÷ x-step= (16 - 4)m/s ÷ (10s)= 12 / 10Acceleration = 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 + 40mDistance travelled = 80m

Question 4Calculate the distance travelled over 15 seconds and the deceleration during the final five seconds using the graph below.

Distance travelled:This equals the area below the graph= area of rectangle + area of triangle= (10s x 20m/s) + (½ x 5s x 20m/s)= 200m + 50mDistance travelled = 250m

Deceleration:Acceleration equals the slope of the graph= y-step ÷ x-step= (- 20m/s) ÷ (5s)= - 4 m/s2

but deceleration = negative of accelerationDeceleration = 4 m/s2

More about velocity-time graphs Notes questions from pages 124 & 125

and also page 123

1. Copy Figure 4 from page 123 – graph X ONLY.2. Explain how acceleration can be found from a velocity-time graph

and hence find the acceleration represented by graph X.3. 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.

4. Copy and answer question (b) on page 125.5. How can distance travelled be found from a velocity-time graph?6. Copy and answer question (c) on page 125.7. Copy the Key Points on page 125.8. Answer the summary questions on page 125.

More about velocity-time graphs ANSWERS

In text questions:

(a) Less steep

(b) It would not be as steep

(c) 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

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

2. 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.

3. Copy the Key Points on page 127.4. Answer the summary questions on page 127.

NOTE – Graph paper needed for Q2.

Using graphs ANSWERS

In text questions:(a) 15 m/s(b) The speed

decreased gradually and became constant.

Summary questions:1. (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

Virtual Physics Laboratory SimulationsNOTE: Links work only in school

Graphs

Measuring Velocity

Measuring Acceleration

Online SimulationsThe 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)

Transport issues

Notes questions from pages 128 & 129

1. Answer questions 1 and 2 on page 128.

Transport issues

ANSWERS

1. (a) (i) 5 litres (ii) 4 litres

(b) £1.70

2. Journey 1: 6.7 km/h

Journey 2: 4000 km/h

How Science Works ANSWERS

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

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

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

d) 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.

e) 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.

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