Physics · 1 Motion, forces and energy

1.2 Motion

Before you start

Unit conversions: $1$ km $= 1000$ m, $1$ minute $= 60$ s, $1$ hour $= 3600$ s.
From 1.1: a scalar has size only; a vector has size and direction.
The gradient of a straight line on a graph is the change in the vertical value divided by the change in the horizontal value.
Area of a rectangle (base × height) and of a triangle (½ × base × height).
Rearranging an equation with three quantities, e.g. getting $t$ out of $v = s / t$ .
Weight is the downward pull of gravity on an object, and when the forces on an object balance, its speed does not change.

Syllabus objectives covered (13)

Define speed as distance travelled per unit time; recall and use the equation v = s / t
Define velocity as speed in a given direction
Recall and use the equation average speed = total distance travelled / total time taken
Sketch, plot and interpret distance–time and speed–time graphs
Determine, qualitatively, from given data or the shape of a distance–time graph or speed–time graph when an object is: (a) at rest (b) moving with constant speed (c) accelerating (d) decelerating
Calculate speed from the gradient of a straight-line section of a distance–time graph
Calculate the area under a speed–time graph to determine the distance travelled for motion with constant speed or constant acceleration
State that the acceleration of free fall g for an object near to the surface of the Earth is approximately constant and is approximately 9.8 m / s2
Define acceleration as change in velocity per unit time; recall and use the equation a = Δv / Δt
Determine from given data or the shape of a speed–time graph when an object is moving with: (a) constant acceleration (b) changing acceleration
Calculate acceleration from the gradient of a speed–time graph
Know that a deceleration is a negative acceleration and use this in calculations
Describe the motion of objects falling in a uniform gravitational field with and without air/ liquid resistance, including reference to terminal velocity

Part 1 of 8

Speed is distance travelled per unit time

Speed answers one question: how much distance does the object cover in each unit of time? A speed of $5$ m/s means the object covers $5$ metres in every second.
That gives you the definition. Speed is distance travelled per unit time. If a question says Define speed, write those words.
The equation uses the syllabus letters: $v$ for speed, $s$ for distance and $t$ for time. There is no formula sheet in Paper 4, so you must recall this equation.
$v = \frac{s}{t}$
Work in metres and seconds and the speed comes out in m/s. If the distance is in km or the time is in minutes or hours, convert first: km × $1000$ , minutes × $60$ , hours × $3600$ . If you divide km by minutes, your answer is in km per minute, not m/s.
You also need the equation the other way round. Multiply both sides by $t$ to get the distance. Then divide both sides by $v$ to get the time. Doing the same thing to both sides keeps the equation true.
$s = v t t = \frac{s}{v}$
Real journeys do not happen at one speed. A bus speeds up, slows down and stops. For a whole journey you use average speed: the one steady speed that would cover the same distance in the same time.
$average speed = \frac{total distance travelled}{total time taken}$
Total time means all of the time, including time spent stopped. The clock does not stop when the bus does.
The trap here is to add the speeds of the stages and divide by the number of stages. That works only if every stage lasts the same time. Drive $1200$ m at $20$ m/s, which takes $60$ s, then $1200$ m at $10$ m/s, which takes $120$ s. The average speed is $2400 \div 180 = 13.3$ m/s, not $15$ m/s, because you spend twice as long at the slower speed.

Part 2 of 8

Velocity is speed in a given direction

From 1.1 you know that a scalar has size only and a vector has size and direction. Speed is a scalar. Velocity is the vector that goes with it.
Velocity is defined as speed in a given direction. A velocity has two parts: a size with a unit, and a direction, e.g. $15$ m/s due north.
Two cars on a straight road, one going north at $15$ m/s and one going south at $15$ m/s, have the same speed but different velocities. They are not doing the same thing, and velocity is the quantity that shows the difference.
Velocity can change while speed stays the same. A car that goes round a roundabout at a steady $8$ m/s keeps changing direction, so its velocity keeps changing although its speed does not.
Along a straight line you show direction with a sign. Choose one direction as positive and keep it for the whole question. Motion the other way then has a negative velocity, e.g. $- 3.0$ m/s.
This matters later in this lesson. Acceleration is defined using velocity, not speed. And the steady speed a falling object reaches is called its terminal velocity, because it has a direction: downwards.

Part 3 of 8

Distance–time graphs: the gradient is the speed

A distance–time graph has time on the horizontal axis and distance on the vertical axis. Each point tells you how far the object has travelled by that time.
The gradient of the line is the change in distance divided by the change in time. That is exactly the speed. So on a distance–time graph, steepness means speed: the steeper the line, the faster the object.
A horizontal line means the distance is not changing. The object is at rest.
A straight sloping line means equal distances in equal times. The object moves at constant speed. A steeper straight line means a greater constant speed.
A curve means the speed is changing. If the curve gets steeper, the speed is increasing: the object is accelerating. If the curve gets less steep, the speed is decreasing: the object is decelerating.
graph diagram — not drawn yet
Distance–time graph with axes labelled time / s and distance / m, showing in order: a straight line rising from the origin, a horizontal section, a steeper straight rising line, a curve that gets steeper, and a curve that gets less steep and levels off; label the sections constant speed, at rest, greater constant speed, accelerating, decelerating.
Steepness is speed: flat is at rest, straight is constant speed, a curve is changing speed.
To calculate the speed from a straight section, take the two points at the ends of the section. Read the time and the distance at each point. Then divide the change in distance by the change in time.
$speed = gradient = \frac{Δ s}{Δ t}$
The trap here: one distance reading divided by its own time reading gives the average speed since the start, not the speed of that section. The two are the same only when the section starts at the origin. For any later section, use the changes.
You can decide the motion from a table of data without a graph. Equal increases in distance in equal time intervals mean constant speed. No increase means at rest. Increases that get bigger mean accelerating; increases that get smaller mean decelerating.
Sketch means show the shape, not plot exact points. Label both axes, start at the right place, and give each section the right shape (horizontal, straight or curved) in the right order. Plot means mark the given points accurately on a grid.

Part 4 of 8

Speed–time graphs: the same shapes mean different things

A speed–time graph has time on the horizontal axis and speed on the vertical axis. Each point tells you how fast the object is moving at that instant.
This is where the shapes can mislead you. A horizontal line on a speed–time graph means the speed is not changing: constant speed, not at rest. At rest is a horizontal line along the time axis, where the speed is zero.
A straight line sloping upwards means the speed increases by the same amount every second. The object is accelerating, and the acceleration is constant.
A straight line sloping downwards means the speed decreases by the same amount every second. The object is decelerating at a constant rate.
A curved line means the acceleration is changing. If the curve rises but gets less steep, the object is still speeding up, but by less each second: its acceleration is decreasing.
graph diagram — not drawn yet
Speed–time graph with axes labelled time / s and speed / m/s, showing in order: a straight line rising from the origin, a horizontal section, a rising curve that gets less steep, a horizontal section, and a straight line falling to zero speed followed by a section along the time axis; label them constant acceleration, constant speed, changing (decreasing) acceleration, constant speed, constant deceleration, at rest.
On this graph a flat line is steady motion; only a line along the time axis is at rest.
With data instead of a graph, compare the speed changes in equal time intervals. Speeds of $0$ , $3$ , $6$ , $9$ m/s at one-second intervals rise by $3$ m/s every second: constant acceleration. Speeds of $0$ , $4$ , $7$ , $9$ m/s rise by $4$ , then $3$ , then $2$ m/s: changing acceleration, and here the acceleration is decreasing.
Before you answer any graph question, read the label on the vertical axis. The same straight sloping line means constant speed on a distance–time graph and constant acceleration on a speed–time graph.

Part 5 of 8

Acceleration is change in velocity per unit time

Acceleration is defined as change in velocity per unit time. It tells you how quickly the velocity changes, not how fast the object moves.
Δ means 'change in', and a change is always the final value minus the initial value. You must recall this equation for Paper 4.
$a = \frac{Δ v}{Δ t}$
The unit comes from the equation: velocity in m/s divided by time in s gives m/s². An acceleration of $2.0$ m/s² means the velocity increases by $2.0$ m/s in every second.
On a speed–time graph, the gradient is the change in speed divided by the change in time. For motion in one direction along a straight line, the change in speed is the change in velocity, so the gradient is the acceleration. Take the two points at the ends of a straight section and divide the speed change by the time change.
When an object slows down, the final speed is smaller than the initial speed. So $Δ v$ is negative, and so is $a$ . A deceleration is a negative acceleration. On a speed–time graph it is a line that slopes downwards, with a negative gradient.
There are two correct ways to say the same thing: 'an acceleration of $- 3.0$ m/s²' and 'a deceleration of $3.0$ m/s²'. Do not write 'a deceleration of $- 3.0$ m/s²'. The two negatives cancel, and that describes speeding up.
Keep the sign through a calculation. A car at $20$ m/s decelerates at $4.0$ m/s². To find the time it takes to stop, rearrange to $Δ t = Δ v / a$ and put the minus sign on $a$ , because the car slows down.
$Δ t = \frac{0 - 20}{- 4.0} = 5.0 s$
If you use $+ 4.0$ m/s² instead, you get $- 5.0$ s. A negative time tells you that a sign has gone wrong.

Part 6 of 8

The area under a speed–time graph is the distance

Start with constant speed. Distance = speed × time. On a speed–time graph, the speed is the height of the line and the time is the width of the section, so speed × time is the area of the rectangle under the line.
The same idea works for every section: the area between the line and the time axis is the distance travelled. The units agree, because (m/s) × s = m.
For constant acceleration from rest, the area is a triangle, so use ½ × base × height. Here is why that is right: the speed rises steadily from $0$ to its final value, so the average speed in that section is half the final speed, and half the final speed × time is the area of the triangle.
$distance = \frac{1}{2} \times time \times final speed$
If the speed rises from one value to another, not from zero, the area is a trapezium. Split it into a rectangle (the starting speed × time) and a triangle on top (½ × time × increase in speed), then add the two.
For a journey with several sections, find the area of each section separately, write each one down, and add them at the end. That way your working is visible even if one area goes wrong.
Read the scales before you multiply. If the time axis is in minutes and the speed is in m/s, convert the time to seconds first. If you do not, your distance in metres is $60$ times too small.
This works only on a speed–time graph. On a distance–time graph you read the distance straight off the vertical axis; the area under that line is not the distance.

Part 7 of 8

Free fall: the acceleration g

Near the surface of the Earth, an object that falls with no air resistance accelerates at the acceleration of free fall, $g$ . The value of $g$ is approximately constant, and it is approximately $9.8$ m/s².
So in free fall the speed increases by about $9.8$ m/s every second. You must be able to state this value. The front of Paper 4 also tells you to take the acceleration of free fall as $9.8$ m/s².
Without air resistance, every object falls with the same acceleration, whatever its mass. In a tube with the air pumped out, a feather and a coin released together reach the bottom together.
Because the acceleration is constant, the speed–time graph for free fall from rest is a straight line from the origin, with a gradient of $9.8$ m/s². The area under it gives the distance fallen, as in the previous section.

Part 8 of 8

Falling through air or liquid: terminal velocity

With air resistance, two forces act on a falling object. Its weight acts downwards and stays the same. Air resistance acts upwards and increases as the speed increases. The difference between them is the resultant force, and the resultant force is what makes the object accelerate.
At the instant the object is released from rest, its speed is zero, so there is no air resistance. Its acceleration is $g$ .
As the speed increases, air resistance increases. The resultant downward force (weight − air resistance) gets smaller, so the acceleration gets smaller. The speed still increases, but by less each second.
Eventually air resistance equals the weight. The resultant force is zero, so the acceleration is zero and the object falls at a constant speed. This constant speed is its terminal velocity.
graph diagram — not drawn yet
Speed–time graph with axes labelled time / s and speed / m/s, showing two lines from the origin: a straight line labelled 'no air resistance, gradient 9.8 m/s²', and below it a curve that starts along the straight line, gets less steep and becomes horizontal, with the horizontal level labelled 'terminal velocity'.
Both start with the same gradient, g. With air resistance the gradient falls to zero.
So the speed–time graph for a fall through air is a curve: steepest at the start, then less steep, then horizontal at the terminal velocity. A curved speed–time graph means changing acceleration. Here the acceleration decreases from $g$ to zero.
When a skydiver opens her parachute, air resistance suddenly becomes greater than her weight. The resultant force is now upwards, so she decelerates. As she slows down, air resistance decreases until it equals her weight again. She then falls at a new, lower terminal velocity.
An object falling through a liquid, e.g. a small steel ball dropped into a tall tube of oil, follows the same pattern: its acceleration decreases until it falls at a terminal velocity. The resistance of the oil is much larger than the resistance of air, so the ball reaches a much lower terminal velocity, after a much shorter fall. At terminal velocity the upward forces on it balance its weight.

Worked examples

Example 1. A cyclist rides $3.6$ km in $10$ minutes. She waits at traffic lights for $5.0$ minutes, then rides a further $1.8$ km in $5.0$ minutes. Calculate (a) her speed during the first stage, in m/s, and (b) her average speed for the whole journey.

Show the working

Convert to metres and seconds first, so that the answers come out in m/s. $3.6$ km $= 3600$ m and $10$ minutes $= 600$ s.
(a) Use $v = s / t$ for the first stage only.
$v = \frac{3600}{600} = 6.0 m/s$
(b) Average speed needs the total distance and the total time. Total distance: $3600 + 1800 = 5400$ m.
The total time includes the wait at the lights: $600 + 300 + 300 = 1200$ s. If you leave out the $300$ s wait, you get $5400 \div 900 = 6.0$ m/s, which is wrong.
Now divide.
$average speed = \frac{5400}{1200} = 4.5 m/s$
Check that it makes sense. She rides at $6.0$ m/s in both stages ( $1800 \div 300 = 6.0$ m/s too), but she is stopped for a quarter of the time. So her average speed must be less than $6.0$ m/s. It is.

Answer: (a) $6.0$ m/s; (b) $4.5$ m/s.

Example 2. The distance–time graph of a toy car is made of three straight sections joining these points: ( $0$ s, $0$ m), ( $4.0$ s, $10$ m), ( $7.0$ s, $10$ m) and ( $10$ s, $25$ m). Describe the motion in each section, and calculate the speed in each section where the car moves.

Show the working

Sketch the graph first. Three straight sections mean three stretches of constant speed, and one of those speeds may be zero.
graph diagram — not drawn yet
Distance–time graph with time 0 to 10 s on the horizontal axis and distance 0 to 25 m on the vertical axis, with straight lines joining (0, 0), (4.0, 10), (7.0, 10) and (10, 25).
The middle section is flat, and the last section is steeper than the first.
Section 1, $0$ to $4.0$ s: a straight sloping line, so constant speed. Its gradient is the speed.
$v = \frac{10 - 0}{4.0 - 0} = 2.5 m/s$
Section 2, $4.0$ to $7.0$ s: the distance stays at $10$ m. A horizontal line on a distance–time graph means the car is at rest.
Section 3, $7.0$ to $10$ s: straight and steeper than section 1, so a greater constant speed. Use the changes in distance and time between the ends of this section.
$v = \frac{25 - 10}{10 - 7.0} = \frac{15}{3.0} = 5.0 m/s$
Notice what $25 \div 10 = 2.5$ m/s would give you: the average speed for the whole $10$ s, not the speed in section 3.

Answer: $0$ to $4.0$ s: constant speed of $2.5$ m/s. $4.0$ to $7.0$ s: at rest. $7.0$ to $10$ s: constant speed of $5.0$ m/s.

Example 3. A car travelling at $10$ m/s accelerates uniformly to $24$ m/s in $7.0$ s. The driver then brakes, and the car decelerates uniformly to rest in $8.0$ s. Calculate (a) the acceleration while the car speeds up, (b) the acceleration while it brakes, and (c) the total distance travelled in these $15$ s.

Show the working

Sketch the speed–time graph: a straight line from $10$ m/s at $0$ s up to $24$ m/s at $7.0$ s, then a straight line down to $0$ m/s at $15$ s. Both sections are straight, so both accelerations are constant.
graph diagram — not drawn yet
Speed–time graph with time 0 to 15 s and speed 0 to 25 m/s: a straight line from (0, 10) to (7.0, 24), then a straight line from (7.0, 24) to (15, 0); the area under the first section is divided by a dashed line at 10 m/s into a rectangle and a triangle, and the area under the second section is shaded as a triangle.
The first section is a trapezium, split into a rectangle and a triangle; the second section is a triangle.
(a) Use $a = Δ v / Δ t$ , with final speed minus initial speed.
$a = \frac{24 - 10}{7.0} = \frac{14}{7.0} = 2.0 {m/s}^{2}$
(b) Use the same equation. The final speed is now $0$ , so $Δ v$ is negative.
$a = \frac{0 - 24}{8.0} = - 3.0 {m/s}^{2}$
The minus sign tells you that the car slows down. You can write $- 3.0$ m/s² or 'a deceleration of $3.0$ m/s²'.
(c) The distance is the area under the graph. The first section does not start at zero speed, so it is a trapezium. Split it into a rectangle of height $10$ m/s and a triangle of height $24 - 10 = 14$ m/s.
$10 \times 7.0 + \frac{1}{2} \times 7.0 \times 14 = 70 + 49 = 119 m$
The braking section is a triangle with a base of $8.0$ s and a height of $24$ m/s.
$\frac{1}{2} \times 8.0 \times 24 = 96 m$
Add the two sections.
$119 + 96 = 215 m$

Answer: (a) $2.0$ m/s²; (b) $- 3.0$ m/s², i.e. a deceleration of $3.0$ m/s²; (c) $215$ m.

Example 4. A stone is dropped from rest from a high bridge. Air resistance is small enough to ignore. Calculate (a) its speed after $1.5$ s and (b) the distance it falls in that time.

Show the working

With no air resistance, the stone falls with the acceleration of free fall, $g = 9.8$ m/s², and this acceleration is constant.
(a) Rearrange $a = Δ v / Δ t$ to $Δ v = a Δ t$ . The stone starts from rest, so the change in speed is equal to its final speed.
$Δ v = 9.8 \times 1.5 = 14.7 m/s$
(b) The speed–time graph is a straight line from the origin to $14.7$ m/s at $1.5$ s. The distance is the area of the triangle under it.
$\frac{1}{2} \times 1.5 \times 14.7 = 11.025 m$
The data have 2 significant figures, so give $11$ m.
Do not use $s = v t$ with $v = 14.7$ m/s. The stone moves that fast only at the very end, so $14.7 \times 1.5 = 22$ m is twice the true distance.

Answer: (a) $14.7$ m/s ( $15$ m/s to 2 significant figures); (b) $11$ m.

Where people go wrong

Mistake: Reading a horizontal line on a speed–time graph as 'at rest'.

Instead: A horizontal line on a speed–time graph means constant speed. At rest is a line along the time axis, at zero speed. A horizontal line on a distance–time graph means at rest.

The two graphs plot different quantities. A flat line means the plotted quantity does not change: distance not changing means stopped; speed not changing means steady motion.

Mistake: Finding average speed by adding the speeds of the stages and dividing by the number of stages, or by leaving the stopped time out of the total time.

Instead: Average speed = total distance travelled ÷ total time taken, and the total time includes every stop.

A stage that lasts longer counts for more. A simple mean of the speeds treats a one-minute stage and a one-hour stage as equal.

Mistake: Calculating the speed in a later section of a distance–time graph by dividing one distance reading by one time reading.

Instead: Divide the change in distance by the change in time between the two ends of that section.

One reading divided by its own time gives the average speed since the start, which includes the earlier sections.

Mistake: Writing 'a deceleration of $- 3.0$ m/s²', or dropping the minus sign of a negative acceleration part way through a calculation.

Instead: Write either 'acceleration $= - 3.0$ m/s²' or 'deceleration $= 3.0$ m/s²', and keep the sign of $a$ in every later step.

A deceleration is already a negative acceleration. A second minus sign reverses the meaning, and a lost sign gives impossible results such as a negative time.

Mistake: Saying that at terminal velocity no forces act on the object, or that gravity has stopped acting.

Instead: The weight still acts. Air resistance has increased until it equals the weight, so the resultant force is zero.

Zero acceleration needs a zero resultant force, not zero forces. The weight of an object near the Earth does not change as it falls.

Check yourself

A train travels $270$ km in $1.5$ hours. Calculate its speed in m/s.
Answer
$270$ km $= 270 000$ m and $1.5$ h $= 5400$ s, so $v = s / t = 270 000 \div 5400 = 50$ m/s.
Two cyclists ride past each other on a straight path. One rides east at $6.0$ m/s and the other rides west at $6.0$ m/s. Do they have the same speed? Do they have the same velocity?
Answer
The same speed, $6.0$ m/s. Different velocities: velocity is speed in a given direction, and their directions are opposite. If east is positive, the velocities are $+ 6.0$ m/s and $- 6.0$ m/s.
State the motion shown by each of these: (a) a distance–time graph that is a curve getting steeper; (b) a speed–time graph that is a horizontal line at $5$ m/s; (c) a distance–time graph that is a horizontal line.
Answer
(a) Accelerating. (b) Moving at a constant speed of $5$ m/s. (c) At rest.
Runner A's speed, measured every second from the start, is $0$ , $2.5$ , $5.0$ , $7.5$ m/s. Runner B's speed at the same times is $0$ , $4.0$ , $6.0$ , $7.0$ m/s. Which runner has a constant acceleration? Describe the other runner's acceleration.
Answer
Runner A: her speed increases by $2.5$ m/s every second, so her acceleration is constant at $2.5$ m/s². Runner B's speed increases by $4.0$ , then $2.0$ , then $1.0$ m/s, so her acceleration changes: it decreases.
A car moving at $18$ m/s brakes with a deceleration of $3.0$ m/s². Write its acceleration with the correct sign, and calculate how long it takes to stop.
Answer
$a = - 3.0$ m/s². $Δ t = Δ v / a = (0 - 18) \div (- 3.0) = 6.0$ s.
A sprinter accelerates uniformly from rest to $6.0$ m/s in $3.0$ s. Use the speed–time graph of this motion to find her acceleration and the distance she covers.
Answer
Acceleration = gradient $= 6.0 \div 3.0 = 2.0$ m/s². Distance = area of the triangle $= \frac{1}{2} \times 3.0 \times 6.0 = 9.0$ m.
A steel ball and a table-tennis ball are released together from the same height inside a tall tube with all the air removed. Which reaches the bottom first? State the acceleration of each ball.
Answer
They reach the bottom together. With no air resistance, both accelerate at $g$ , approximately $9.8$ m/s², whatever their mass.
A skydiver falls at terminal velocity and then opens her parachute. Describe what happens to her speed, and explain why.
Answer
Her speed decreases, then becomes constant at a new, lower terminal velocity. The open parachute makes air resistance greater than her weight, so the resultant force is upwards and she decelerates. As she slows down, air resistance decreases until it equals her weight again.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A cyclist starts from rest. Her speed increases uniformly to $8.0$ m/s in $10$ s. She then rides at a constant $8.0$ m/s for $30$ s. She then brakes and slows down uniformly, and stops $4.0$ s later.

(a) Sketch the speed–time graph of her journey. Label the axes and mark the values of speed and time where the motion changes. 3 marks No calculator

Mark scheme

B1 straight line from the origin to 8.0 m/s at 10 s
B1 horizontal line at 8.0 m/s from 10 s to 40 s
B1 straight line from 8.0 m/s at 40 s down to zero speed at 44 s

(b) Calculate her acceleration during the first $10$ s. 2 marks

Mark scheme

C1 a = Δv / Δt, or 8.0 ÷ 10 seen
A1 0.80 m/s² (accept 0.8 m/s²); unit required

(c) Calculate her acceleration while she brakes. 2 marks

Mark scheme

C1 a = Δv / Δt, or (0 − 8.0) ÷ 4.0, or 8.0 ÷ 4.0 seen
A1 −2.0 m/s² (accept a deceleration of 2.0 m/s²; do not accept a deceleration of −2.0 m/s²); unit required

(d) Determine the total distance she travels. 3 marks

Mark scheme

C1 distance = area under the graph, or any one of 40 m, 240 m, 16 m seen
C1 40 + 240 + 16, or all three areas correct, or ½ × (30 + 44) × 8.0 seen (trapezium method)
A1 296 m (accept 300 m or 3.0 × 10² m); unit required

Question 2. A student walks from home to a shop along a straight road. Her distance–time graph is made of straight lines that join these points. Time / s: $0$ , $100$ , $200$ , $300$ , $400$ , $500$ . Distance from home / m: $0$ , $150$ , $300$ , $300$ , $600$ , $900$ .

(a) Identify the time interval during which the student is at rest. 1 mark No calculator

Mark scheme

B1 200 s to 300 s

(b) Calculate her speed between $300$ s and $500$ s. 2 marks

Mark scheme

C1 change in distance ÷ change in time, or (900 − 300) ÷ (500 − 300) seen
A1 3.0 m/s; unit required

(c) Calculate her average speed for the whole $500$ s. 2 marks

Mark scheme

C1 total distance ÷ total time, or 900 ÷ 500 seen
A1 1.8 m/s; unit required

Question 3. A skydiver steps out of a hot-air balloon that is not moving. She falls for some time before she opens her parachute.

(a) State the value of her acceleration at the instant she steps out of the balloon. 1 mark No calculator

Mark scheme

B1 9.8 m/s² (accept 9.81 m/s², or g with the value 9.8 m/s²); unit required

(b) Explain why her acceleration decreases as she falls, and why she eventually falls at a constant speed. 3 marks No calculator

Mark scheme

B1 air resistance increases as her speed increases
B1 resultant (downward) force decreases / weight − air resistance decreases, so acceleration decreases
B1 air resistance becomes equal to her weight, so resultant force is zero and acceleration is zero / she reaches terminal velocity

(c) Describe the shape of the speed–time graph of her fall, from the moment she steps out until she reaches terminal velocity. 2 marks No calculator

Mark scheme

B1 starts at the origin and curves, with a gradient that decreases / gets less steep
B1 becomes a horizontal line (constant speed), owtte

How this is examined

Finding the distance travelled from the area under a speed–time graph: a three-mark Calculate or Determine part after a speed–time graph or a description of a journey with two or three straight sections Distance = area under the graph, stated or used; each section's area found separately, with ½ × base × height for each triangle and a rectangle plus a triangle for each trapezium; the sum, with the unit m. A correct area for one section is working that a C mark can credit when the total is wrong.
Why a falling object stops speeding up and reaches terminal velocity: a three-mark Explain about a skydiver, or about an object that falls through a liquid Three linked points: air resistance increases as the speed increases; the resultant downward force (weight − air resistance), and so the acceleration, decreases; when air resistance equals the weight, the resultant force is zero, the acceleration is zero and the object falls at constant (terminal) velocity.
Acceleration and deceleration from $a = Δ v / Δ t$ or from the gradient: a two-mark Calculate part, with two speeds and a time read from a speed–time graph or given in the question The equation, or the substitution (final speed − initial speed) ÷ time, then the value with the unit m/s². For slowing down, either ' $- 2.0$ m/s²' or 'a deceleration of $2.0$ m/s²'; never a deceleration with a minus sign.
Reading the motion from the shape of a graph or from a data table: a one- or two-mark State, Describe or Identify part that points at one section of a distance–time or speed–time graph, or a Sketch worth one mark per section The correct motion named for the correct graph: horizontal on distance–time = at rest; horizontal on speed–time = constant speed; straight sloping line on speed–time = constant acceleration (or deceleration if it slopes down); curve on speed–time = changing acceleration. In a sketch, each section has the right shape and starts and ends at the right values.
Speed from a gradient, and average speed over a journey with a stop: two-mark Calculate parts after a distance–time graph or a table of distances and times For a section: change in distance ÷ change in time between the ends of that section. For the whole journey: total distance ÷ total time, with the stopped time included. Each answer in m/s.
The three definitions and the value of g: one-mark Define or State parts Speed: distance travelled per unit time. Velocity: speed in a given direction. Acceleration: change in velocity per unit time. The acceleration of free fall near the Earth's surface: approximately $9.8$ m/s², with the unit.

Words to know

Speed: Distance travelled per unit time, $v = s / t$ . A scalar, measured in m/s.
Velocity: Speed in a given direction. A vector.
Average speed: Total distance travelled ÷ total time taken, with any time spent stopped included.
Acceleration: Change in velocity per unit time, $a = Δ v / Δ t$ . Measured in m/s².
Deceleration: A negative acceleration: the object slows down.
Gradient: The change in the vertical quantity divided by the change in the horizontal quantity. On a distance–time graph it is the speed; on a speed–time graph it is the acceleration.
Acceleration of free fall, $g$: The acceleration of an object falling with no air resistance. Near the Earth's surface it is approximately constant and approximately $9.8$ m/s².
Air resistance: The force from the air that opposes the motion of an object through it. It increases as the speed increases.
Terminal velocity: The constant velocity a falling object reaches when the upward forces on it balance its weight, so the resultant force and the acceleration are zero.

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