Section 1 Equations of Motion

North Berwick High School

Higher Physics

Department of Physics

Unit 1 Our Dynamic Universe

Section 1 Equations of Motion

Note Making

Make a dictionary with the meanings of any new words.

Vectors and scalars

1. Describe the differences between vectors and scalars. 2. Make a table showing both including units. You must learn this. 3. Make a table showing the sign convention. 4. Describe a method of adding vectors (either scale diagrams or trig). Don't forget top-to-tail. Include question 2 on page 22 as an example.

Graphs 1. Draw the three sets of graphs on pages 4 - 6 of the booklet. Make sure that you label them accordingly and that you understand the differences between them. 2. Copy and answer question 5 on page 29 as an example.

Equations of Motion 1. Describe a method to measure acceleration using light gates. 2. Write down the three equations of motion. 3. Show the derivation of these equations. 4. Make a note of the techniques shown on page 15 and the example on pages 16 and 17. Section 1 Equations of Motion

Contents

Content Statements...... 1 The physics of motion ...... 2 Vectors and scalars ...... 2 Scalar quantities ...... 2 Vector quantities ...... 3 Vectors and scalars: why do they matter? ...... 3 The sign convention ...... 3 Graphing motion and interpreting graphs...... 4 Interpreting displacement–time graphs ...... 7 Interpreting velocity–time graphs ...... 7 Investigating acceleration ...... 8 The equations of motion ...... 9 Direction matters ...... 13 Explaining good technique using an example ...... 15 Example ...... 16 Problems ...... 18 Solutions ...... 34

Content Statements content Notes context a) Equations of motion Candidates should Light gates, motion sensors for objects with undertake experiments to and software/hardware to constant verify the measure displacement, acceleration in a relationships shown in the velocity and acceleration. straight line. equations. Using software to analyse

videos of motion.

b) Motion-time graphs Displacement-time graphs. Motion sensors (including for motion with Gradient is velocity. wireless sensors) to enable constant Velocity-time graphs. Area graphical representation of acceleration. under graph is motion. displacement. Investigate the variation of Gradient is acceleration. acceleration on a slope Acceleration-time graphs. with the Restricted to zero and angle of the slope. constant Motion of athletes and acceleration. equipment used in sports.

Graphs for bouncing objects and objects thrown

vertically

upwards.

c) Motion of objects Objects in freefall and the Investigate the initial with constant speed movement of objects on acceleration of an object or constant slopes should be projected acceleration. investigated. vertically upwards (e.g. popper toy)

Section 1 Equations of Motion

The physics of motion

Footballers, golfers, tennis players, runners, skiers – they all have something in common. They have the ability to make split-second decisions about how their actions will affect their performance: how the curve of a ball will affect whether they score that crucial penalty, whether a change of angle of the club will give them a hole-in-one. They are making use of the physics of motion. In the first section of ‘Our Dynamic Universe’ you will learn more about the motion of objects: from raindrops to roller coasters. You will be able to use the language of physics to describe and explain the motion of bouncing balls and sky divers. You will understand the principles of the physics of motion and explain it using words, diagrams, graphs and equations.

Vectors and scalars

Exploring the physics of motion, the terms acceleration, velocity and displacement have been used. What do these terms mean? Is displacement the same as distance? Is velocity the same as speed? Quantities can be defined as vectors or scalars. You will be familiar with working with scalar quantities but you probably did not realise it. You may also be familiar with some vector quantities and have performed simple vector additions.

Scalar quantities

A scalar quantity is defined only by its magnitude (i.e. its size). Examples of scalar quantities include:

Quantity Units Distance m Speed m s–1 Energy J Time s Mass kg

Vector quantities

A vector quantity is defined both by its magnitude (i.e. its size) and its direction. Examples of vector quantities include:

Quantity Units Displacement m Velocity m s–1 Acceleration m s–2 Force N Impulse N s Momentum kg m s–1 Weight N

Vectors and scalars: why do they matter?

In physics, we are studying the real, physical world and finding models which explain our observations of motion in this world. Vectors give us a method of describing motion in the real world.

So why does this matter? Consider a pilot coming in to land an aircraft in very windy conditions. As he tries to land the plane in the centre of the runway, he must take into account the speed of the wind and its direction, i.e. the velocity of the wind.

Consider the corrective action that will have to be taken by the pilot to land an aircraft in a cross wind – and realise the importance of understanding vectors!

The sign convention

Direction matters and so a system of defining direction is needed. Normally, we use upwards  positive downwards  negative right  positive left  negative

Graphing motion and interpreting graphs

Although the ability to plot a graph accurately is important, it is absolutely essential to be able to interpret graphs and to visualise the shape of a graph for a given motion.

For an object that is stationary, the displacement–time, velocity–time and acceleration–time graphs are shown below.

For an object that is moving with constant velocity, the displacement–time, velocity–time and acceleration–time graphs are shown below.

For an object that is accelerating uniformly, (i.e. moving with a constant acceleration) the displacement–time, velocity–time and acceleration–time graphs are shown below.

Each time you are presented with a graph, you should ask yourself two key questions:

1. Is there any significance to the gradient of the graph? 2. Is there any significance to the area under the graph?

Interpreting displacement–time graphs

The gradient of the displacement–time graph is the velocity of the object. Mathematically this is the equivalent of dividing the change in the y-axis quantity by the change in the x-axis quantity: in this case displacement divided by time. Displacement divided by time is rate of change of displacement, which is velocity.

The area under the graph does not give us any meaningful information. Mathematically this is the equivalent of multiplying the two quantities, in this case displacement and time.

Interpreting velocity–time graphs

The gradient of the velocity–time graph is the acceleration of the object: the steeper the line the greater the rate of change of velocity (or acceleration) of the object.

The area under the graph is the displacement.

Investigating acceleration

Acceleration is defined as the rate of change of velocity per unit time. From this definition, an equation can be written:

a = acceleration in metres per second per second v u (m s–2) a –1 t v = final velocity in metres per second (m s ) u = initial velocity in metres per second (m s–1) t = time in seconds (s)

The acceleration of an object can be investigated in a number of ways. You should give consideration to the different methods, and the advantages and disadvantages of each method for measurement of acceleration in different circumstances. Methods of measurement of acceleration include:

1. using light gates connected to a timing unit, with single-mask and double-mask set-ups

2. using a motion sensor, which measures displacement with time and from this calculates velocity and acceleration

3. using an accelerometer, which measures acceleration directly.

You should be able to predict, investigate and explain the displacement–time, velocity–time and acceleration–time graphs for a variety of motions, including trolleys on slopes. Consider the uncertainties in each piece of equipment and which is most appropriate for the measurements you are making. Evaluate your experimental set-up to be able to suggest problems and potential improvements.

Remember that displacement (s), velocity (u, v) and acceleration (a) are vector quantities. A positive and negative direction must be chosen and used consistently in your predictions and explanations.

The equations of motion

The equation which links acceleration, initial and final velocity, and time is the first of the equations of motion.

These equations are used to describe motion in a straight line with uniform acceleration. You must to be able to:

select the correct formula identify the symbols and units used carry out calculations to solve problems of real life motion carry out experiments to verify the equations of motion.

You should develop an understanding of how the graphs of motion can be used to derive the equations. This is an important part of demonstrating that you understand the principles of describing motion, and the link between describing it graphically and mathematically.

a = acceleration in metres per second per second (m s–2) v u v = final velocity in metres per second (m s–1) a –1 t u = initial velocity in metres per second (m s ) t = time in seconds (s) v = u + at Equation of motion 1

You now should be sufficiently familiar with graphing motion to be able to describe the motion represented by the graph below.

time (s)

This graph represents an object moving with a positive velocity of 5 m s–1, which is accelerating at a constant rate. After 300 s the object is moving with velocity of 35 m s–1. A constant acceleration means the velocity is increasing at a constant rate.

The displacement of the object can be determined by calculating the area under the graph. You may already be familiar with the idea of using the area under a speed–time graph to determine the distance travelled by an object.

Removing the numbers from the axis, we can work instead with the symbols which represent the quantities, ie final velocity (v), initial velocity (u) and time (t).

Notice that Equation 1 has been rearranged to give v – u = at and substituted into the equation above.

Adding the two areas under the graph gives us:

s = displacement in metres (m) s = ut + ½at2 u = initial velocity in metres per second (m s–1) t = time in seconds (s) a = acceleration in metres per second per second (m s –2)

s = ut + ½at2 Equation of motion 2

The third equation of motion is derived from substituting Equation 1 into Equation 2.

Equation 1 v = u + at square each side to give v2 = (u + at)2 v2 = u2 + 2uat + a2t2 v2 = u2 + 2a(ut + ½at2) substitute in Equation 2 v2 = u2 + 2as v2 = u2 + 2as Equation of motion 3

Direction matters

Christine Arron was a 100-m sprint athlete.

Immediately the starting pistol is fired, Christine accelerates uniformly from rest, reaching maximum velocity at a distance of 21.8 m in 4.16 s.

Her maximum velocity is 10.49 m s–1.

Calculate her acceleration over the first 21.8 m of the race, showing full working.

Using the normal sign convection in which right is positive and left is negative, by calculation her acceleration is +2.52 m s –2. In this case, the positive value means increasing velocity with time in the positive direction.

As she passes the finish line, Christine begins to slow down.

She comes to rest in 8.20 s from a velocity of 9.73 m s–1.

Calculate her acceleration, showing full working.

Using the normal sign convection in which right is positive and left is negative, by calculation her acceleration is –1.19 m s–2. In this case, the negative value means decreasing velocity with time in the positive direction.

Before continuing you should give some thought to what else a positive or negative value of acceleration might mean.

Consider Christine running in the opposite direction, where the sign convention remains the same.

What will a positive value of acceleration mean in this case? What about a negative value?

Immediately the starting pistol is fired, Christine accelerates uniformly from rest, in the opposite direction, reaching maximum velocity at a distance of 21.8 m in 4.16 s.

Her maximum velocity is –10.49 m s–1 (why is it negative?). Calculate her acceleration at a distance of 21.8 m of the race, showing full working.

Her acceleration is –2.52 m s–2. The negative value indicates that she is gaining speed in the negative direction (or slowing down in the positive direction).

As she passes the finish line, Christine begins to slow down. She comes to rest in 8.20 s from a velocity of –9.73 m s–1. Calculate her acceleration, showing full working.

Her acceleration is a = 1.19 m s–2. The positive value indicates that she is losing speed in the negative direction.

When using equations in relation to motion, you must understand what the values mean. Remember, equations are just one way of describing motion – you should develop a picture in your head of the actual motion being described by the mathematics.

Using the equations of motion: explaining good technique using an example

Step 1: Write down the sign convention you are using for the situation.

Step 2: Write down what you know – think s s u v a t

s sign convention (see step 1) s displacement u initial velocity v final velocity a acceleration t time

Step 3: Write down any other information you have, e.g. acceleration due to gravity.

Step 4: Write down your formulae and check against data sheet. Select the most appropriate formula to use.

Step 5: Substitute values then rearrange the formula.

Step 6: Write answer clearly using magnitude with units and direction (if appropriate).

Example

Usain Bolt is a Jamaican sprinter and a three-time Olympic gold medallist (correct 2011).

Immediately the starting pistol is fired, Usain accelerates uniformly from rest. He reaches 8.70 m s–1 in 1.75 s.

Calculate his displacement in this time.

Working s  positive and  negative s = ? m u = 0 m s–1 v = 8.70 m s–1 a = ? m s–2 t = 1.75 s

(u is an easy one to miss – the phrase to look for is ‘starting from rest’)

Formula

v u at 1 s ut at2 2 v2 u2 2as

s = (u + v) 2

Can this be done in one calculation? Is there one formula which links s, v, u and t but does not require a?

On this occasion two formulae will be required, the first to determine acceleration a and the second to calculate displacement s. v = u + at 8.70 = 0 + a × 1.75 8.70 = 1.75a a = 8.70 1.75 a = 4.97 m s–2

then

1 s ut at2 2 1 s (0 t) 4.97 1.752 2 s 0 7.61 s = 7.61m

You should ensure that you are familiar with typical everyday velocities and accelerations. This is key to understanding work in physics on motion. For example, what is a realistic top speed for a world-class sprinter?

What sort of accelerations do you experience in everyday life? Do you experience motion only in the horizontal? An accelerometer (a device which measures acceleration in three dimensions) can be used to explore the accelerations that you experience during everyday activities. Try it out – you might be surprised by the results!

Problems

Revision problems – Speed

1. The world downhill skiing speed trial takes place at Les Arcs every year. Describe a method that could be used to find the average speed of the skier over the 1 km run. Your description should include:

any apparatus required details of what measurements need to be taken an explanation of how you would use the measurements to carry out the calculations.

2. An athlete runs a 1500 m race in a time of 3 min 40 s. Calculate his average speed for the race.

3. It takes light 8·0 minutes to travel from the Sun to the Earth. How far away is the Sun from the Earth? (speed of light = 3·0 × 108 m s 1).

4. The distance between London and New York is 4800 km. A plane travels at an average speed of Mach 1·3 between London and New York. Calculate the time, to the nearest minute, for this journey. (Mach 1 is the speed of sound. Take the speed of sound to be 340 m s 1).

5. The graph shows how the speed of a girl varies with time from the instant she starts to run for a bus.

She starts from stand still at O and jumps on the bus at Q. Find:

a) the steady speed at which she runs b) the distance she runs c) the increase in the speed of the bus while the girl is on it d) how far the bus travels during QR e) how far the girl travels during OR.

6. A ground-to-air guided missile starts from rest and accelerates at 150 m s 2 for 5 s. What is the speed of the missile 5 s after launching?

7. An Aston Martin has an acceleration of 6 m s 2 from rest. What time does it take to reach a speed of 30 m s 1?

8. A car is travelling at a speed of 34 m s 1. The driver applies the brakes and the car slows down at a rate of 15 m s 2. What is the time taken for the speed of the car to reduce to 4 m s 1?

Revision problems – Acceleration

1. A skateboarder starting from rest goes down a uniform slope and reaches a speed of 8 m s 1 in 4 s.

(a) What is the acceleration of the skateboarder? (b) Calculate the time taken for the skateboarder to reach a speed of 12 m s 1.

2. In the Tour de France a cyclist is travelling at 16 m s 1. When he reaches a downhill stretch he accelerates to a speed of 20 m s 1 in 2·0 s.

(a) What is the acceleration of the cyclist down the hill? (b) The cyclist maintains this constant acceleration. What is his speed after a further 2·0 s? (c) How long after he starts to accelerate does he reach a speed of 28 m s 1?

3. A student sets up the apparatus shown to find the acceleration of a trolley down a slope.

Length of card on trolley = 50 mm

Time on clock 1 = 0·10 s (time taken for card to interrupt top light gate) Time on clock 2 = 0·05 s (time taken for card to interrupt bottom light gate) Time on clock 3 = 2·50 s (time taken for trolley to travel between top and bottom light gate)

Use these results to calculate the acceleration of the trolley.

Revision problems – Vectors

1. A car travels 50 km due north and then returns 30 km due south. The whole journey takes 2 hours.

Calculate:

(a) the total distance travelled by the car (b) the average speed of the car (c) the resultant displacement of the car (d) the average velocity of the car.

2. A girl delivers newspapers to three houses, X, Y and Z, as shown in the diagram.

Z N

40 m

30 m X

She starts at X and walks directly from X to Y and then to Z.

(a) Calculate the total distance the girl walks. (b) Calculate the girl’s final displacement from X. (c) The girl walks at a steady speed of 1 m s 1. (i) Calculate the time she takes to get from X to Z. (ii) Calculate her resultant velocity.

3. Find the resultant force in the following examples:

4. State what is meant by a vector quantity and scalar quantity. Give two examples of each.

5. An orienteer runs 5 km due south then 4 km due west and then 2 km due north. The total time taken for this is 1 hour.

Calculate the average speed and average velocity of the orienteer for this run.

6. A football is kicked up at an angle of 70° at 15 m s 1.

Calculate: (a) the horizontal component of the velocity (b) the vertical component of the velocity.

Section 1: Equations of motion

Equations of motion

1. An object is travelling at a speed of 8·0 m s 1. It then accelerates uniformly at 4·0 m s 2 for 10 s. How far does the object travel in this 10 s?

2. A car is travelling at a speed of 15·0 m s 1. It accelerates uniformly at 6·0 m s 2 and travels a distance of 200 m while accelerating. Calculate the velocity of the car at the end of the 200 m.

3. A ball is thrown vertically upwards to a height of 40 m above its starting point. Calculate the speed at which it was thrown.

4. A car is travelling at a speed of 30·0 m s 1. It then slows down at 1·80 m s 2 until it comes to rest. It travels a distance of 250 m while slowing down. What time does it take to travel the 250 m?

5. A stone is thrown with an initial speed 5·0 m s 1 vertically down a well. The stone strikes the water 60 m below where it was thrown.

Calculate the time taken for the stone to reach the surface of the water. The effects of friction can be ignored.

6. A tennis ball launcher is 0·60 m long. A tennis ball leaves the launcher at a speed of 30 m s 1.

(a) Calculate the average acceleration of the tennis ball in the launcher. (b) Calculate the time the ball accelerates in the launcher.

7. In an experiment to find g a steel ball falls from rest through a distance of 0·40m. The time taken to fall this distance is 0·29s.

What is the value of g calculated from the data of this experiment?

8. A trolley accelerates uniformly down a slope. Two light gates connected to a motion computer are spaced 0·50 m apart on the slope. The speeds recorded as the trolley passes the light gates are 0·20 m s 1 and 0·50 m s 1. (a) Calculate the acceleration of the trolley. (b) What time does the trolley take to travel the 0·5 m between the light gates?

9. A helicopter is rising vertically at a speed of 10·0 m s 1 when a wheel falls off. The wheel hits the ground 8·00 s later.

Calculate the height of the helicopter above the ground when the wheel came off. The effects of friction can be ignored.

10. A ball is thrown vertically upwards from the edge of a cliff as shown in the diagram.

The effects of friction can be ignored.

(a) (i) What is the height of the ball above sea level 2·0 s after being thrown? (ii) What is the velocity of the ball 2·0 s after being thrown? (b) What is the total distance travelled by the ball from launch to landing in the sea?

Motion – time graphs

1. The graph shows how the displacement of an object varies with time.

displacement against time

North 8 7 6 5 4 3 2

displacement /m 1 0 0 1 2 3 4 5 6 7 8 time / s

(a) Calculate the velocity of the object between 0 and 1 s. (b) What is the velocity of the object between 2 and 4 s from the start? (c) Draw the corresponding distance against time graph for the movement of this object. (d) Calculate the average speed of the object for the 8 seconds shown on the graph. (e) Draw the corresponding velocity against time graph for the movement of this object.

2. The graph shows how the displacement of an object varies with time.

displacement against time

North 4.5 4 3.5 3 2.5 2 1.5 1 displacement /m 0.5 0 0 1 2 3 4 5 6 time / s

(a) Calculate the velocity of the object during the first second from the start. (b) Calculate the velocity of the object between 1 and 5 s from the start. (c) Draw the corresponding distance against time graph for this object. (d) Calculate the average speed of the object for the 5 seconds. (e) Draw the corresponding velocity against time graph for this object. (f) What are the displacement and the velocity of the object 0·5 seconds after the start? (g) What are the displacement and the velocity of the object 3 seconds after the start?

3. The graph shows the displacement against time graph for the movement of an object.

displacement against time

North 2.5 2 1.5 1 0.5 0 -0.5 0 1 2 3 4 5 -1

displacement /m -1.5 South -2 -2.5 time / s

(a) Calculate the velocity of the object between 0 and 2 s. (b) Calculate the velocity of the object between 2 and 4 s from the start. (c) Draw the corresponding distance against time graph for this object. (d) Calculate the average speed of the object for the 4 seconds. (e) Draw the corresponding velocity against time graph for this object. (f) What are the displacement and the velocity of the object 0·5 s after the start? (g) What are the displacement and the velocity of the object 3 seconds after the start?

4. An object starts from a displacement of 0 m. The graph shows how the velocity of the object varies with time from the start.

North velocity against time velocity m s–1 8 7 6 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 time / s time/s

(a) Calculate the acceleration of the object between 0 and 1 s. (b) What is the acceleration of the object between 2 and 4 s from the start? (c) Calculate the displacement of the object 2 seconds after the start. (d) What is the displacement of the object 8 seconds after the start? (e) Sketch the corresponding displacement against time graph for the movement of this object.

5. An object starts from a displacement of 0 m. The graph shows how the velocity of the object varies with time from the start.

North velocity against time velocity m s–1 2.5 2 North 1.5 1 0.5 0 time/s -0.5 0 1 2 3 4 5 -1 -1.5 South -2 -2.5 time / s

(a) Calculate the acceleration of the object between 0 and 2 s. (b) Calculate the acceleration of the object between 2 and 4 s from the start. (c) Draw the corresponding acceleration against time graph for this object. (d) What are the displacement and the velocity of the object 3 seconds after the start? (e) What are the displacement and the velocity of the object 4 seconds after the start? (f) Sketch the corresponding displacement against time graph for the movement of this object.

6. The velocity-time graph for an object is shown below.

velocity against time

15 velocity 10 -1 /m s 5 0 time/s -5 0 1 2 3 4 5 6 7 -10 time / s

A positive value indicates a velocity due north and a negative value indicates a velocity due south. The displacement of the object is 0 at the start of timing.

(a) Calculate the displacement of the object: (i) 3 s after timing starts (ii) 4 s after timing starts (iii) 6 s after timing starts.

(b) Draw the corresponding acceleration–time graph.

7. The graph shows how the acceleration a of an object, starting from rest, varies with time.

a -2 a/ms-2 m s 2

0 5 10 Time time/s / s

Draw a graph to show how the velocity of the object varies with time for the 10 seconds of the motion.

8. The graph shows the velocity of a ball that is dropped and bounces on a floor. An upwards direction is taken as being positive.

+ C Velocity O D 0 time - E B

(a) In which direction is the ball travelling during section OB of the graph? (b) Describe the velocity of the ball as represented by section CD of the graph. (c) Describe the velocity of the ball as represented by section DE of the graph. (d) What happened to the ball at the time represented by point B on the graph? (e) What happened to the ball at the time represented by point C on the graph? (f) How does the speed of the ball immediately before rebound from the floor compare with the speed immediately after rebound? (g) Sketch a graph of acceleration against time for the movement of the ball.

9. A ball is thrown vertically upwards and returns to the thrower 3 seconds later. Which velocity-time graph represents the motion of the ball?

10. A ball is dropped from a height and bounces up and down on a horizontal surface. Which velocity-time graph represents the motion of the ball from the moment it is released?

-1 -1 -1 v/ms v/ms v/ms - - - v / m s v / m s v / m s 1 1 1

t / s t / s t / s A B C -1 vv/ms / m s- -1 - v/msv / m s 1 1 0 t / s

D E t / s

11. Describe how you could measure the acceleration of a trolley that starts from rest and moves down a slope. You are provided with a metre stick and a stopwatch. Your description should include:

(a) a diagram (b) a list of the measurements taken (c) how you would use these measurements to calculate the acceleration of the trolley (d) how you would estimate the uncertainties involved in the experiment.

12. Describe a situation where a runner has a displacement of 100 m due north, a velocity of 3 m s 1 due north and an acceleration of 2 m s 2 due south. Your description should include a diagram.

13. Is it possible for an object to be accelerating but have a constant speed? You must justify your answer.

14. Is it possible for an object to move with a constant speed for 5 seconds and have a displacement of 0 m? You must justify your answer.

15 Is it possible for an object to move with a constant velocity for 5 s and have a displacement of 0 m? You must justify your answer.

Solutions

Revision problems – Speed

2. 6·8 m s 1

3. 1·4 × 1011 m

4. 181 minutes

5. (a) 5 m s 1 (b) 35 m (c) 10 m s 1 (d) 100 m (e) 135 m

6. 750 m s 1

7 5 s

8. 2 s

Revision problems – Acceleration

1. (a) 2 m s 2 (b) 6 s

2. (a) 2·0 m s 2 (b) 24 m s 1 (c) 6·0 s

3. 0·20 m s 2

Revision problems – Vectors

1. (a) 80 km (b) 40 km h 1 (c) 20 km north (d) 10 km h 1 north

2. (a) 70 m (b) 50 m bearing 037 (c) (i) 70 s (ii) 0·71 m s 1 bearing 037

3. (a) 6·8 N bearing 077 (b) 11·3 N bearing 045 (c) 6·4 N bearing 129

5. Average speed = 11 km h 1 Average velocity = 5 km h 1 bearing 233

6. (a) 5·1 m s 1 (b) 14·1 m s 1

Section 1: Equations of motion

Equations of motion

1. 280 m

2. 51·2 m

3. 28 m s 1

4. 16·7 s

5. 3·0 s

6. (a) 750 m s 2 (b) 0·04 s

7. 9·5 m s 2 or N kg 1

8. (a) 0·21 m s 2 (b) 1·4 s

9. 234 m

10. (a) (i) 21·4 m (ii) 15·6 m s 1 downwards (b) 34·6 m

Motion–time graphs

1. (a) 2 m s 1 due north (b) 0 m s 1 (d) 0·75 m s 1

2. (a) 4 m s 1 due north (b) 1·0 m s 1 due south (d) 1·6 m s 1 (f) displacement 2 m due north, velocity 4 m s 1 due north (g) displacement 2 m due north, velocity 1 m s 1 due south

3. (a) 1 m s 1 due north (b) 2 m s 1 due south (d) 1.5 m s 1 (f) displacement 0·5 m due north, velocity 1 m s 1 due north (g) displacement 0, velocity 2 m s 1 due south

4. (a) 2 m s 2 due north (b) 0 m s 2 (c) 4 m due north (d) 32 m due north

5. (a) 1 m s 2 due north (b) 2 m s 2 due south (d) displacement 3 m due north, velocity 0 m s 1 (e) displacement 2 m due north, velocity 2 m s 1 due south

6. (a) (i) 17·5 m due north (ii) 22·5 m due north (iii) 17·5 m due north

9. D. Note that in this question, downwards is taken to be the positive direction for vectors.

10. A. Note that in this question, upwards is taken to be the positive direction for vectors.