IB DP Physics · Theme A · Space, time and motion

Kinematics.

A.1 · Describing motion, equations of motion, projectiles
Standard level and higher level

SL + HL · 9 hours PanPhyTrack ↗

The journey

Four parts.

Part 01

Describing motion

Displacement, velocity, acceleration

Part 02

Motion graphs

What the gradient and the area mean

Part 03

Equations of motion

Four equations for constant acceleration

Part 04

Projectiles

Two motions at once, then air resistance

Every slide is for SL and HL: A.1 has no additional higher level content.

A.1 KINEMATICS · SL + HL

Part 01 · SL + HL

Describing motion.

Where is it, how fast, and how is that changing?

A.1 KINEMATICS · SL + HL

Where is it?

Distance and displacement.

Position: where a body is, measured from a chosen origin
Distance: the length of the path traveled. A scalar
Displacement: the change in position. A vector, with a direction

ONE JOURNEY, TWO MEASURES

A.1 KINEMATICS · SL + HL

How fast?

Speed and velocity.

Speed: rate of change of distance. A scalar
Velocity: rate of change of displacement. A vector

\(v = \dfrac{\Delta s}{\Delta t}\)

One 400 m lap in 50 s Average speed \(= \dfrac{\text{distance}}{\text{time}} = \dfrac{400}{50} = \class{res}{8.0 \,\mathrm{m} \,\mathrm{s}^{-1}}\) Displacement \(= 0\) Average velocity \(= \dfrac{\Delta s}{\Delta t} = \dfrac{0}{50} = \class{res}{0}\)

A.1 KINEMATICS · SL + HL

Over a journey, or at a moment?

Average and instantaneous.

Average velocity: the whole displacement divided by the whole time
Instantaneous velocity: the velocity at one moment
On a displacement–time graph: a chord gives the average, a tangent the instantaneous

DISPLACEMENT AGAINST TIME

A.1 KINEMATICS · SL + HL

The story · Palo Alto, California, 1878

Background story · not assessed

Freezing a gallop.

A printed card of twelve photographs in three rows, eleven showing the silhouette of a galloping horse and rider at successive points of its stride against a wall marked with vertical lines, and the last showing the horse standing still — **The Horse in Motion**: the mare Sallie Gardner on Leland Stanford’s track, 19 June 1878

Do all four hooves of a galloping horse leave the ground at once? The photographs said yes
Eadweard Muybridge’s 12 cameras fired in turn: frames 27 inches and about \(\frac{1}{25}\) s apart
0.69 m in 0.04 s: an average speed of about 17 m s⁻¹, over a time short enough to be nearly instantaneous

Photographs: Eadweard Muybridge, Library of Congress, public domain.

A.1 KINEMATICS · SL + HL

How is the velocity changing?

Acceleration.

\(a = \dfrac{\Delta v}{\Delta t} = \dfrac{v - u}{t}\)

The rate of change of velocity. A vector
\(u\): initial velocity, \(v\): final velocity
Uniform acceleration: the same at every moment

A car goes from 12 to 30 m s⁻¹ in 6.0 s \(a = \dfrac{v - u}{t} = \dfrac{30 - 12}{6.0} = \class{res}{3.0 \,\mathrm{m} \,\mathrm{s}^{-2}}\) Then brakes from 30 to 10 m s⁻¹ in 4.0 s \(a = \dfrac{v - u}{t} = \dfrac{10 - 30}{4.0} = \class{res}{-5.0 \,\mathrm{m} \,\mathrm{s}^{-2}}\)

A.1 KINEMATICS · SL + HL

Summary of part 1

Five quantities.

Quantity	Scalar or vector?	Unit	Tells you
Distance	Scalar	m	How far along the path
Displacement	Vector	m	How far from the start, and which way
Speed	Scalar	m s⁻¹	How fast
Velocity	Vector	m s⁻¹	How fast, and which way
Acceleration	Vector	m s⁻²	How quickly the velocity changes

A.1 KINEMATICS · SL + HL

Part 02 · SL + HL

Motion graphs.

The gradient and the area tell the whole story

A.1 KINEMATICS · SL + HL

Graph 1

Displacement–time graphs.

Gradient = velocity
Straight line: constant velocity
Flat: at rest
Curve: the velocity is changing

DISPLACEMENT AGAINST TIME

A.1 KINEMATICS · SL + HL

Graph 2

Velocity–time graphs.

Gradient = acceleration
Area under the graph = displacement
Flat: constant velocity, not at rest
Below the time axis: moving the other way

VELOCITY AGAINST TIME

A.1 KINEMATICS · SL + HL

Linking the graphs

Three graphs, one motion.

UNIFORM ACCELERATION FROM REST

Area under an acceleration–time graph = the change in velocity.

A.1 KINEMATICS · SL + HL

When the acceleration changes

Non-uniform acceleration.

The velocity–time graph is a curve
Acceleration at one moment: gradient of the tangent
Displacement: still the area. Count squares

The equations of motion in part 3 do not apply.

VELOCITY AGAINST TIME

A.1 KINEMATICS · SL + HL

Worked example

Reading a journey.

First 4 s: gradient \(a = \dfrac{\Delta v}{\Delta t} = \dfrac{12 - 0}{4.0} = \class{res}{3.0 \,\mathrm{m} \,\mathrm{s}^{-2}}\) Last 3 s: gradient \(a = \dfrac{\Delta v}{\Delta t} = \dfrac{0 - 12}{3.0} = \class{res}{-4.0 \,\mathrm{m} \,\mathrm{s}^{-2}}\) Whole journey: area \(s = \text{area} = 24 + 72 + 18 = \class{res}{114 \,\mathrm{m}}\) Average speed \(= \dfrac{s}{t} = \dfrac{114}{13} = \class{res}{8.8 \,\mathrm{m} \,\mathrm{s}^{-1}}\)

A SHORT CAR JOURNEY

A.1 KINEMATICS · SL + HL

Quick check

Describing motion.

01A swimmer swims two 25 m lengths, there and back. Distance? Displacement?

50 m and zero

02What does the gradient of a displacement–time graph give?

Velocity

03What does the area under a velocity–time graph give?

Displacement

04A train goes from 8 to 20 m s⁻¹ in 30 s. Acceleration?

\(\frac{20 - 8}{30} = 0.40 \,\mathrm{m} \,\mathrm{s}^{-2}\)

05Velocity −6 m s⁻¹, acceleration −2 m s⁻². Speeding up or slowing?

Speeding up: same sign

06How do you find instantaneous velocity from a curved displacement–time graph?

Gradient of the tangent at that time

PRESS → TO REVEAL EACH ANSWER

Part 03 · SL + HL

Equations of motion.

Four equations for uniform acceleration

A.1 KINEMATICS · SL + HL

The story · Italy, published 1638

Background story · not assessed

Galileo slows down a fall.

Painted portrait of Galileo, an old man with a white beard looking upwards — **Galileo Galilei**, from a portrait by Justus Sustermans

A long wooden ramp with a groove along its top edge and five small brass bells mounted along it at increasing distances — A demonstration **inclined plane**, Florence, early 1800s. The ball rings each bell in turn, at **equal time intervals**

A falling body was too fast to time. So Galileo slowed it down: a bronze ball rolling down a gentle slope
His clock: he weighed the water that ran from a tank during each run
Twice the time: four times the distance
1638: published in his last book, Two New Sciences

Uniform acceleration, found by measuring.

Portrait: Justus Sustermans, public domain. Inclined plane: Museo Galileo, CC BY-SA 4.0.

A.1 KINEMATICS · SL + HL

From the velocity–time graph

Where the equations come from.

Gradient = acceleration \(a = \dfrac{v - u}{t}\) so \(\class{res}{v = u + at}\) Area = displacement: rectangle + triangle \(\class{res}{s = ut + \tfrac{1}{2}at^{2}}\) Area = displacement: a trapezium \(\class{res}{s = \dfrac{u + v}{2}t}\)

UNIFORM ACCELERATION FROM u TO v

A.1 KINEMATICS · SL + HL

The triangle

Its base is \(t\) and its height is the gain in velocity, \(v - u\), which is \(at\). Half base times height gives \(\frac{1}{2} \times t \times at = \frac{1}{2}at^{2}\).

The fourth equation

It has no \(t\) in it. From the first, \(t = \frac{v - u}{a}\). Put that into \(s = \frac{u + v}{2}t\): \(s = \frac{(v + u)(v - u)}{2a} = \frac{v^{2} - u^{2}}{2a}\), which rearranges to \(v^{2} - u^{2} = 2as\).

What the trapezium says

\(\frac{u + v}{2}\) is the average velocity when the velocity changes steadily. Displacement is average velocity × time.

You do not have to derive these in an examination: all four are in the data booklet. Knowing where they come from tells you when they can be used.

Uniform acceleration in a straight line

The four equations.

Equation	Leaves out	Use it when…
\(s = \dfrac{u + v}{2}t\)	\(a\)	the acceleration is not given or wanted
\(v = u + at\)	\(s\)	the displacement is not given or wanted
\(s = ut + \tfrac{1}{2}at^{2}\)	\(v\)	the final velocity is not given or wanted
\(v^{2} - u^{2} = 2as\)	\(t\)	the time is not given or wanted

Five quantities: \(s\), \(u\), \(v\), \(a\), \(t\). Know three, and you can find the other two.

A.1 KINEMATICS · SL + HL

Why three are enough

Each equation links four of the five quantities. With three known, the equation that contains those three and the one you want has a single unknown.

The condition

Every one of these was built on a straight velocity–time line. If the acceleration changes during the motion, split the motion into stages that each have a constant acceleration and treat them one at a time. The final velocity of one stage is the initial velocity of the next.

\(s\) is displacement, not distance. If a body turns round during the motion, \(s\) is still measured from the start in a straight line.

The data booklet prints the last equation as \(v^{2} = u^{2} + 2as\). It is the same equation, with \(u^{2}\) moved to the other side.

A method that always works

Solving a problem.

01 · Direction

Choose which way is positive

Anything pointing the other way is negative

02 · List

Write down \(s\), \(u\), \(v\), \(a\), \(t\)

Fill in the three you know and mark the one you want

03 · Choose

Pick the equation

The one without the quantity you neither know nor want

04 · Solve

Substitute, with signs

Then give the answer a unit and a direction

A.1 KINEMATICS · SL + HL

Worked example

A car brakes to a stop.

\(u = 30 \,\mathrm{m} \,\mathrm{s}^{-1}\) · \(v = 0\) · \(a = -5.0 \,\mathrm{m} \,\mathrm{s}^{-2}\) Distance: no \(t\), so \(v^{2} - u^{2} = 2as\) \(s = \dfrac{v^{2} - u^{2}}{2a} = \dfrac{0 - 30^{2}}{2(-5.0)} = \class{res}{90 \,\mathrm{m}}\) Time: \(v = u + at\) \(t = \dfrac{v - u}{a} = \dfrac{0 - 30}{-5.0} = \class{res}{6.0 \,\mathrm{s}}\)

Check: \(s = \dfrac{u + v}{2}t = 15 \times 6.0 = 90 \,\mathrm{m}\)
At twice the speed: four times the distance

A.1 KINEMATICS · SL + HL

Motion under gravity alone

Free fall.

With no air resistance, every body falls with the same acceleration
Near the Earth’s surface: \(\class{res}{g = 9.8 \,\mathrm{m} \,\mathrm{s}^{-2}}\), downward
It is the same on the way up, at the top and on the way down

FALLING FROM RESTEQUAL TIME STEPS

A.1 KINEMATICS · SL + HL

The story · The Moon, 2 August 1971

Background story · not assessed

A hammer and a feather on the Moon.

Portrait of astronaut David Scott in a white spacesuit, holding his helmet — **David Scott**, commander of Apollo 15

A blurred color television picture of an astronaut in a white spacesuit standing on the lunar surface in front of the lunar module — From the **television broadcast**: Scott watches the hammer and the feather reach the ground

The last minutes of the last moonwalk of Apollo 15
In one hand a 1.32 kg hammer. In the other a falcon feather
He lets go of both from about 1.6 m
They fall side by side and land together

With no air, mass makes no difference.

Photos: NASA and NASA Johnson Space Center, public domain.

A.1 KINEMATICS · SL + HL

Worked example

A ball thrown straight up.

Up is positive \(u = +14.7 \,\mathrm{m} \,\mathrm{s}^{-1}\) · \(a = -9.8 \,\mathrm{m} \,\mathrm{s}^{-2}\) Top: \(v = 0\), and \(v = u + at\) \(t = \dfrac{v - u}{a} = \dfrac{0 - 14.7}{-9.8} = \class{res}{1.5 \,\mathrm{s}}\) Height: \(v^{2} - u^{2} = 2as\) \(s = \dfrac{v^{2} - u^{2}}{2a} = \dfrac{0 - 14.7^{2}}{2(-9.8)} = \class{res}{11 \,\mathrm{m}}\) Back at the hand: \(t = \class{res}{3.0 \,\mathrm{s}}\), \(v = \class{res}{-14.7 \,\mathrm{m} \,\mathrm{s}^{-1}}\)

VELOCITY AGAINST TIME

A.1 KINEMATICS · SL + HL

Practical

Measuring \(g\).

Release a steel ball from rest
Time its fall through a measured height \(h\)
Repeat for a range of heights
\(h = \tfrac{1}{2}gt^{2}\): plot \(h\) against \(t^{2}\)

A straight line through the origin, gradient \(\tfrac{1}{2}g\).

FREE-FALL TIMERNOT TO SCALE

A.1 KINEMATICS · SL + HL

Why a graph and not one drop

A single height gives one value of \(g\) with no way of telling how good it is. A line through several points averages out the random scatter in the times, and the way the points sit about the line shows the quality of the data.

What a systematic error does

An electromagnet does not let go at once, so every time is too long by about the same amount. The points then no longer fall on a line through the origin. If instead \(h\) is measured to the wrong part of the ball, every height is wrong by the same amount: the line is shifted and has an intercept, but its gradient, and so \(g\), is unchanged.

The times are short, about 0.3 s for a 0.5 m drop, so they must be measured electronically. A hand-held stopwatch has a reaction-time error as large as the reading.

The PanPhyTrack ↗ does the same from a video of a falling ball, frame by frame.

Quick check

Equations of motion.

01When can the four equations be used?

Only when the acceleration is constant

02A stone is dropped from rest. How fast is it moving after 4.0 s?

\(9.8 \times 4.0 = 39.2 \,\mathrm{m} \,\mathrm{s}^{-1}\)

03How far has it fallen in that time?

\(\frac{1}{2} \times 9.8 \times 4.0^{2} = 78.4 \,\mathrm{m}\)

04Which equation has no \(t\) in it?

\(v^{2} - u^{2} = 2as\)

05A cyclist accelerates from rest at 1.5 m s⁻² for 8.0 s. Final velocity?

\(1.5 \times 8.0 = 12 \,\mathrm{m} \,\mathrm{s}^{-1}\)

06What is the acceleration of a ball at the top of its flight?

9.8 m s⁻² downward

PRESS → TO REVEAL EACH ANSWER

Part 04 · SL + HL

Projectile motion.

Two simple motions happening at once

A.1 KINEMATICS · SL + HL

The key idea

Two independent motions.

Horizontal: no force, so constant velocity
Vertical: gravity, so acceleration \(g\)
Neither affects the other

The only thing they share is the time.

RELEASED AT THE SAME MOMENT

A.1 KINEMATICS · SL + HL

Method

Keep two columns.

	Horizontal	Vertical
Acceleration	Zero	\(g = 9.8 \,\mathrm{m} \,\mathrm{s}^{-2}\), downward
Velocity	Stays at its starting value	Changes: \(v = u + at\)
Displacement	velocity × time	\(s = ut + \tfrac{1}{2}at^{2}\)

Find the time from one column, then use it in the other.

A.1 KINEMATICS · SL + HL

Worked example

Launched horizontally from a cliff.

44.1 m high Launched horizontally at 12 m s⁻¹ Vertical: \(u = 0\), so \(s = \tfrac{1}{2}gt^{2}\) \(t = \sqrt{\dfrac{2s}{g}} = \sqrt{\dfrac{2 \times 44.1}{9.8}} = \class{res}{3.0 \,\mathrm{s}}\) Horizontal: \(x = u_{x}t = 12 \times 3.0 = \class{res}{36 \,\mathrm{m}}\)

Velocity on landing Vertical: \(v_{y} = gt = 9.8 \times 3.0 = 29.4 \,\mathrm{m} \,\mathrm{s}^{-1}\) \(v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{12^{2} + 29.4^{2}} = \class{res}{32 \,\mathrm{m} \,\mathrm{s}^{-1}}\) \(\tan \theta = \dfrac{v_{y}}{v_{x}} = \dfrac{29.4}{12}\) so \(\class{res}{68^{\circ}}\) below horizontal

A.1 KINEMATICS · SL + HL

Resolving the launch velocity

Launched at an angle.

Horizontal: \(u \cos \theta\). It never changes
Vertical: \(u \sin \theta\). It changes at \(g\)
At the top the vertical velocity is zero
Landing at launch height: the path is symmetrical

COMPONENTS OF THE LAUNCH VELOCITY

A.1 KINEMATICS · SL + HL

Following the velocity

What changes, and what does not.

VELOCITY COMPONENTS AT EQUAL TIME STEPSNO AIR RESISTANCE

The acceleration is the same at every point: \(g\), straight down.

A.1 KINEMATICS · SL + HL

Worked example

20 m s⁻¹ at 30° above the horizontal.

Components Horizontal: \(u \cos \theta = 20 \cos 30^{\circ} = \class{res}{17.3 \,\mathrm{m} \,\mathrm{s}^{-1}}\) Vertical: \(u \sin \theta = 20 \sin 30^{\circ} = \class{res}{10.0 \,\mathrm{m} \,\mathrm{s}^{-1}}\) Time to the top: vertical \(v = 0\) \(t = \dfrac{u_{y}}{g} = \dfrac{10.0}{9.8} = \class{res}{1.02 \,\mathrm{s}}\)

Maximum height: \(v^{2} - u^{2} = 2as\) \(s = \dfrac{u_{y}^{2}}{2g} = \dfrac{10.0^{2}}{2 \times 9.8} = \class{res}{5.1 \,\mathrm{m}}\) Lands at launch height Time of flight \(= 2t = 2 \times 1.02 = \class{res}{2.04 \,\mathrm{s}}\) Range \(= u_{x} \times 2t = 17.3 \times 2.04 = \class{res}{35.3 \,\mathrm{m}}\)

A.1 KINEMATICS · SL + HL

Worked example

Thrown downwards from a height.

8.0 m s⁻¹ at 30° below horizontal 25 m above ground · down is positive Horizontal: \(u \cos \theta = 8.0 \cos 30^{\circ} = \class{res}{6.93 \,\mathrm{m} \,\mathrm{s}^{-1}}\) Vertical: \(u \sin \theta = 8.0 \sin 30^{\circ} = \class{res}{4.0 \,\mathrm{m} \,\mathrm{s}^{-1}}\)

Vertical velocity on landing \(v^{2} - u^{2} = 2as\) \(v^{2} = 4.0^{2} + 2(9.8)(25) = 506\) \(v = \class{res}{22.5 \,\mathrm{m} \,\mathrm{s}^{-1}}\) \(t = \dfrac{v - u}{a} = \dfrac{22.5 - 4.0}{9.8} = \class{res}{1.89 \,\mathrm{s}}\) Lands \(u_{x}t = 6.93 \times 1.89 = \class{res}{13.1 \,\mathrm{m}}\) away

A.1 KINEMATICS · SL + HL

Real projectiles · no calculations

Adding air resistance.

Fluid resistance acts against the velocity, and grows with speed
Lower maximum height
Shorter range and time of flight
Path not symmetrical: steeper coming down

THE SAME LAUNCH, WITH AND WITHOUT AIR

A.1 KINEMATICS · SL + HL

Horizontal motion

There is now a force against the sideways motion, so the horizontal velocity is no longer constant: it falls throughout the flight. The body covers less ground in each second than the one before, which shortens the range and makes the last part of the path steep.

Vertical motion

On the way up, resistance and gravity both act downward, so the body slows more quickly than at \(g\) and does not climb as high. On the way down, resistance acts upward against gravity, so it speeds up more slowly than at \(g\). It therefore takes longer to come down than it took to go up, and lands more slowly than it was launched.

These effects are examined in words and sketches only. Describe the change and give the reason: which way the resistance acts and what it does to each component.

Falling through a fluid

Terminal velocity.

From rest: the acceleration starts at \(g\)
As the speed rises, the resistance rises
So the acceleration falls
Resistance equals weight: no acceleration, constant terminal velocity

VELOCITY AGAINST TIME

A.1 KINEMATICS · SL + HL

The graph tells the same story

Its gradient is the acceleration. It begins at \(g\), when the body is hardly moving and there is no resistance. The curve then bends over as the resistance grows, and becomes flat when the two forces balance.

A skydiver

In free fall the terminal velocity is about 55 m s⁻¹. Opening the parachute suddenly makes the resistance far greater than the weight, so the acceleration is upward and the skydiver slows. As the speed drops, so does the resistance, until the forces balance again at a new, much lower terminal velocity of about 5 m s⁻¹.

Saying the resistance becomes greater than the weight as the body falls. It rises until it equals the weight and then stops rising, because the speed stops rising.

This ignores the upthrust of the fluid, which is tiny in air. In a liquid the terminal velocity is reached when drag and upthrust together equal the weight. The forces themselves, including the viscous drag on a small sphere, are in A.2.

Nature of science

Every model leaves something out.

THIS TOPIC ASSUMESIT STOPS WORKING WHEN

No air resistance

→

The body is light or fast: a shuttlecock, a raindrop

\(g\) is constant

→

The height is not small compared with the Earth: a satellite

The body is a point

→

Its size or its spin matters: a curving football

A model is good enough when its predictions agree with measurement to the accuracy you need.

A.1 KINEMATICS · SL + HL

Why simplify at all?

With air resistance included there are no simple equations: the acceleration changes from moment to moment and the path has to be worked out step by step by computer. Leaving it out gives an answer in three lines, and for a dense, slow body such as a shot or a dropped stone that answer is very close to what is measured.

Galileo did the same

His rolling ball was slowed a little by friction and by the air. He treated those as small disturbances to be reduced, with a hard ball and a smooth groove, and looked for the simple rule underneath. Deciding what can be ignored is part of doing physics.

When a question asks for an assumption, name the one that matters: “air resistance is negligible” for a projectile, “the acceleration is constant” for the equations of motion.

Quick check

Projectiles.

01What is the horizontal acceleration of a projectile (no air resistance)?

Zero

02A ball rolls off a table 1.25 m high. Time to reach the floor?

\(\sqrt{\frac{2 \times 1.25}{9.8}} = 0.51 \,\mathrm{s}\)

03It left the table at 2.0 m s⁻¹. How far out does it land?

\(2.0 \times 0.51 = 1.0 \,\mathrm{m}\)

04What is the velocity of a projectile at the top of its path?

\(u \cos \theta\), horizontal

05Give two effects of air resistance on the path.

Lower maximum height, shorter range

06What is true of the forces at terminal velocity?

Resistance equals weight: no resultant force

PRESS → TO REVEAL EACH ANSWER

Summary

Kinematics on one slide.

Describing motion

Vectors have direction

Displacement, velocity, acceleration
Average: over a journey. Instantaneous: at a moment

Motion graphs

Gradient and area

Gradient of \(s\)–\(t\) is \(v\); gradient of \(v\)–\(t\) is \(a\)
Area under \(v\)–\(t\) is displacement

Equations of motion

Constant acceleration only

Know three of \(s\), \(u\), \(v\), \(a\), \(t\)
Free fall: \(g = 9.8 \,\mathrm{m} \,\mathrm{s}^{-2}\)

Projectiles

Two columns, one time

Horizontal: constant velocity. Vertical: \(g\)
Air resistance: lower, shorter, not symmetrical

A.1 KINEMATICS · SL + HL

Summary

Key words.

DisplacementChange in positionA vector

VelocityRate of change of displacement

AccelerationRate of change of velocity

InstantaneousAt one momentGradient of a tangent

Uniform accelerationConstant in size and direction

Free fallGravity is the only force\(g = 9.8 \,\mathrm{m} \,\mathrm{s}^{-2}\)

ProjectileMoves freely under gravity after launch

RangeHorizontal distance to landing

Terminal velocityConstant velocity reached when resistance equals weight

A.1 KINEMATICS · SL + HL