2.1.1 Scalar and vector quantities

Scalar and vector quantities

Definition

Scalar quantity

A scalar quantity has magnitude but no specific direction, such as distance, speed, mass or energy.

Definition

Vector quantity

A vector quantity has both magnitude and a specific direction, such as displacement, velocity, acceleration or force.

A physical quantity needs a magnitude and a unit. Magnitude means its numerical size.
A scalar is completely described by its magnitude and unit, such as a mass of $kg70\ \text{kg}$ or an energy transfer of $kJ84\ \text{kJ}$ .
A vector also needs a direction, such as a force of $N5\ \text{N}$ downwards or a velocity of $m/s12\ \text{m/s}$ east.
A vector can be represented by an arrow. The arrow points in the vector’s direction, while its length can be drawn to scale to represent the magnitude.

Classifying quantities

The scalar quantities are distance, speed, mass and energy.
The vector quantities are displacement, velocity, acceleration, force, weight and momentum.
Distance is the total length of the path travelled, measured in $m\text{m}$ .
Displacement is the straight-line change in position from the starting point to the finishing point, stated with a direction and measured in $m\text{m}$ .
Speed describes how quickly distance is travelled and is measured in $m/s\text{m/s}$ .
Velocity is speed in a stated direction and is measured in $m/s\text{m/s}$ .
Mass measures the amount of matter in an object and is measured in $kg\text{kg}$ .
Weight is the gravitational force acting on an object, measured in $N\text{N}$ and directed towards the centre of the attracting body.
Acceleration, force and momentum are vectors because changing their direction changes the physical situation. Energy is a scalar because it has no direction.

Common Mistake

Do not use mass and weight as interchangeable terms. Mass is a scalar measured in $kg\text{kg}$ , while weight is a vector force measured in $N\text{N}$ .
Do not identify a vector only from its unit. Distance and displacement both use $m\text{m}$ , while speed and velocity both use $m/s\text{m/s}$ .

Direction changes vector results

Choose one direction as positive when motion is along a straight line. Quantities in the opposite direction are negative.
A velocity of $m/s-12\ \text{m/s}$ means a velocity of magnitude $m/s12\ \text{m/s}$ in the chosen negative direction. It does not mean a speed below zero.
Vectors in opposite directions can partly or completely cancel. Scalars do not cancel because they carry no direction.

Example

A walker travels $m30\ \text{m}$ east and then $m10\ \text{m}$ west.
The distance is $m30+10=40\ \text{m}$ because distance counts the whole route.
Taking east as positive, the displacement is $m30-10=20\ \text{m}$ east.

Speed and velocity

Definition

Velocity

Velocity is speed in a stated direction, measured in metres per second.

An object can have constant speed while its velocity changes. When it moves around a curve, the direction changes even if the magnitude of the velocity stays constant.
After one complete lap of a circular track, the distance is the circumference of the track but the displacement is zero because the finishing position is the starting position.

Example

A runner completes one $m400\ \text{m}$ lap in $s50\ \text{s}$ .
The average speed is $m/s\dfrac{400}{50}=8.0\ \text{m/s}$ .
The displacement is $m0\ \text{m}$ , so the average velocity is $m/s\dfrac{0}{50}=0\ \text{m/s}$ .

Exam technique

When asked to compare a scalar with a vector, state that both have magnitude and then add that only the vector has a specific direction.
For a displacement or velocity answer, include the direction. A magnitude and unit alone give only distance or speed.
Use signs consistently in one-dimensional calculations and state which direction you selected as positive.

Self review

What information describes a scalar quantity completely?
What additional information is required for a vector quantity?
Which listed motion quantities form the scalar and vector pairs?
Why can velocity change while speed remains constant?

2.2.1 Speed and distance–time graphs

Speed and average speed

Definition

Average speed

Average speed is the total distance travelled divided by the total time taken.

Definition

Instantaneous speed

Instantaneous speed is the speed of an object at a particular moment.

Speed is a scalar quantity because it has magnitude but no direction.
The equation for average speed is
$taken\text{average speed}=\frac{\text{distance travelled}}{\text{time taken}}$
Using symbols,
$v=stv=\frac{s}{t}$
where $v$ is average speed in $m/s\text{m/s}$ , $s$ is distance in $m\text{m}$ and $t$ is time in $s\text{s}$ .
The rearranged forms are $s = v t$ and $t=svt=\dfrac{s}{v}$ . Convert every measurement to compatible units before substitution.

Example

A cyclist covers $m750\ \text{m}$ in $s50\ \text{s}$ .
Substitution gives $v=75050v=\dfrac{750}{50}$ .
The average speed is $m/s15\ \text{m/s}$ .

Average speed for a whole journey

Average speed is not normally the arithmetic mean of two speeds. It depends on the total distance and total time.
Include time spent stationary because it contributes to the total journey time even though no distance is travelled during that interval.

Example

A bus travels $m600\ \text{m}$ in $s40\ \text{s}$ , waits for $s20\ \text{s}$ and then travels $m900\ \text{m}$ in $s60\ \text{s}$ .
The total distance is $m600+900=1500\ \text{m}$ .
The total time is $s40+20+60=120\ \text{s}$ .
The average speed is $m/s\dfrac{1500}{120}=12.5\ \text{m/s}$ .

Distance–time graphs

Definition

Gradient

The gradient of a graph is the change in the vertical-axis quantity divided by the change in the horizontal-axis quantity.

A distance–time graph shows distance on the vertical axis and time on the horizontal axis.
The gradient represents speed because
$gradient=ΔdistanceΔtime=speed\text{gradient}=\frac{\Delta \text{distance}}{\Delta \text{time}}=\text{speed}$
A horizontal line has zero gradient, so the object is stationary.
A straight sloping line has constant gradient, so the object moves at constant speed.
A steeper line represents a greater speed because more distance is covered in the same time.
A curve with an increasing gradient represents increasing speed. A curve that becomes less steep represents decreasing speed.

Calculating a graph gradient

Choose two points far apart on the straight section of the graph. If a best-fit line is present, the points should lie on that line and do not need to be plotted data points.
Draw a large gradient triangle and read the coordinate differences from the axes.
Calculate
$speed=d2−d1t2−t1\text{speed}=\frac{d_2-d_1}{t_2-t_1}$
Include the unit obtained from the graph axes, commonly $m/s\text{m/s}$ .

Example

A straight section passes through $m)(10\ \text{s},40\ \text{m})$ and $m)(30\ \text{s},140\ \text{m})$ .
The change in distance is $m140-40=100\ \text{m}$ .
The change in time is $s30-10=20\ \text{s}$ .
The speed is $m/s\dfrac{100}{20}=5.0\ \text{m/s}$ .

Common Mistake

Do not calculate a gradient using $d / t$ from one point unless the straight line passes through the origin.
Do not describe a horizontal section as moving at constant speed. Its speed is zero.
A distance–time graph does not show direction, so its distance values do not decrease.

Exam technique

For a graph description, link each shape to motion: horizontal means stationary, straight and sloping means constant speed, and changing gradient means changing speed.
Show the coordinate differences used for a gradient. A large triangle reduces the percentage uncertainty in reading the axes.
Keep units with all substitutions and quote a sensible number of significant figures.

Self review

What equation links average speed, distance and time?
What does the gradient of a distance–time graph represent?
What motion produces a horizontal section?
How does a curved distance–time graph show acceleration or deceleration?

2.2.2 Acceleration and the uniform-acceleration equation

Acceleration

Definition

Acceleration

Acceleration is the rate of change of velocity.

Definition

Deceleration

Deceleration is acceleration that reduces the magnitude of an object's velocity.

Acceleration is a vector quantity because velocity is a vector. A change in speed, direction or both produces acceleration.
The acceleration equation is
$a=v−uta=\frac{v-u}{t}$
where $a$ is acceleration in $m/s2\text{m/s}^2$ , $u$ is initial velocity in $m/s\text{m/s}$ , $v$ is final velocity in $m/s\text{m/s}$ and $t$ is time in $s\text{s}$ .
The unit $m/s2\text{m/s}^2$ means that velocity changes by that many metres per second during each second.
With one direction chosen as positive, a negative acceleration points in the negative direction. It represents slowing down only when velocity points in the positive direction.

Example

A car increases its velocity from $m/s6.0\ \text{m/s}$ to $m/s22\ \text{m/s}$ in $s4.0\ \text{s}$ .
Substitution gives $a=22−6.04.0a=\dfrac{22-6.0}{4.0}$ .
The acceleration is $m/s24.0\ \text{m/s}^2$ .

Rearranging the acceleration equation

Useful rearrangements are
$v = u + a t$
$u = v - a t$
$t=v−uat=\frac{v-u}{a}$
Keep the signs of velocities and acceleration during substitution. A negative value may carry essential directional information.

Example

A train travelling at $m/s25\ \text{m/s}$ slows uniformly at $m/s2-0.50\ \text{m/s}^2$ for $s30\ \text{s}$ .
Using $v = u + a t$ gives $v = 25 + (- 0.50) (30)$ .
The final velocity is $m/s10\ \text{m/s}$ .

Uniform acceleration without time

Definition

Uniform acceleration

Uniform acceleration is acceleration that remains constant.

When acceleration is uniform and time is not given, use
$v^2-u^2=2ax$
where $x$ is displacement in $m\text{m}$ . This equation applies only when acceleration is constant.
Useful rearrangements include
$x=v2−u22ax=\frac{v^2-u^2}{2a}$
$a=v2−u22xa=\frac{v^2-u^2}{2x}$
When solving for velocity, take the square root only after evaluating $v^2$ . Select the root that matches the stated direction of motion.

Example

A vehicle moving at $m/s20\ \text{m/s}$ comes to rest with uniform acceleration over $m50\ \text{m}$ .
Substitution gives $0^2-20^2=2a(50)$ .
Rearrangement gives $m/s2a=\dfrac{-400}{100}=-4.0\ \text{m/s}^2$ .

Interpreting acceleration

Positive and negative signs indicate direction relative to the chosen positive direction. They do not automatically mean speeding up and slowing down.
An object speeds up when its velocity and acceleration have the same sign.
An object slows down when its velocity and acceleration have opposite signs.
An object moving at constant speed in a circle accelerates because its direction, and therefore its velocity, continually changes.

Common Mistake

Do not use $m/s\text{m/s}$ as the unit of acceleration. The correct unit is $m/s2\text{m/s}^2$ .
Do not use $v^2-u^2=2ax$ when acceleration changes during the motion.
Do not discard a negative acceleration. Explain its direction or identify it as deceleration only when the motion direction makes that description correct.

Exam technique

Write the selected equation before substituting and keep all quantities in SI units.
Edexcel calculation marks commonly reward substitution, evaluation and the final unit, so show these stages clearly even when your calculator gives the answer directly.
Check whether the question gives distance or displacement and whether the signs make physical sense.

Self review

What does an acceleration of $m/s23\ \text{m/s}^2$ mean?
What equation links $u$ , $v$ , $a$ and $t$ ?
When may $v^2-u^2=2ax$ be used?
When does a negative acceleration make an object speed up?

2.2.3 Velocity–time graphs

Velocity–time graphs

Definition

Velocity-time graph

A velocity-time graph shows how an object's velocity changes with time.

Definition

Uniform acceleration

Uniform acceleration is acceleration that remains constant.

Velocity is plotted on the vertical axis and time on the horizontal axis. Values below the time axis represent motion in the chosen negative direction.
A horizontal line represents constant velocity and zero acceleration.
A straight sloping line represents uniform acceleration.
A curve represents changing acceleration because its gradient changes.
Crossing the time axis means the velocity is zero at that instant. If the graph continues across the axis, the object reverses direction.

Three velocity-time graphs showing constant velocity (horizontal line), uniformly accelerated motion (straight line sloping upwards from the origin), and uniformly decelerated motion (straight line sloping downwards to the time axis).

Gradient gives acceleration

The gradient of a velocity–time graph is
$gradient=ΔvΔt=a\text{gradient}=\frac{\Delta v}{\Delta t}=a$
A positive gradient gives acceleration in the positive direction and a negative gradient gives acceleration in the negative direction.
To calculate uniform acceleration, choose two points far apart on the straight section and use
$a=v2−v1t2−t1a=\frac{v_2-v_1}{t_2-t_1}$

Example

Velocity rises uniformly from $m/s4.0\ \text{m/s}$ at $s2.0\ \text{s}$ to $m/s16\ \text{m/s}$ at $s8.0\ \text{s}$ .
The velocity change is $m/s16-4.0=12\ \text{m/s}$ .
The time change is $s8.0-2.0=6.0\ \text{s}$ .
The acceleration is $m/s2\dfrac{12}{6.0}=2.0\ \text{m/s}^2$ .

Area gives displacement

Definition

Displacement from a velocity-time graph

Displacement is the signed area between a velocity-time graph and the time axis.

For uniform acceleration, split the region into rectangles and triangles, then calculate each area.
For a rectangle, $A = bh$ . For a triangle, $A=12bhA=\dfrac{1}{2}bh$ . Time provides the base and velocity provides the height, so the area unit is $s×m/s=m\text{s}\times\text{m/s}=\text{m}$ .
Area above the time axis is positive displacement. Area below the axis is negative displacement.
Total displacement is the signed sum of all areas. Total distance is the sum of the magnitudes of the separate areas.

A velocity-time graph showing uniform acceleration from an initial velocity u to a final velocity v over time t. The area under the graph is shaded to show it can be split into a rectangle (area u times t) and a triangle (area 0.5 times t times (v minus u)).

Example

A car accelerates uniformly from $m/s5.0\ \text{m/s}$ to $m/s15\ \text{m/s}$ in $s4.0\ \text{s}$ .
The area is a rectangle plus a triangle: $x=(5.0)(4.0)+12(4.0)(15−5.0)x=(5.0)(4.0)+\dfrac{1}{2}(4.0)(15-5.0)$ .
The displacement is $m20+20=40\ \text{m}$ .

Describing motion from the graph

State the velocity at the start of each section, whether it is constant or changing, and the direction of motion.
Use the sign and gradient separately. The sign of velocity gives the direction of motion, while the sign of the gradient gives the direction of acceleration.
An upward-sloping line below the axis can represent an object slowing down while travelling in the negative direction because its velocity moves towards zero.
A downward-sloping line below the axis can represent speeding up in the negative direction because the magnitude of velocity increases.

Common Mistake

Do not call every negative gradient deceleration. Compare the signs of velocity and acceleration.
Do not use the line length as distance. Calculate the area between the graph and the time axis.
Do not add areas below the axis as positive when finding displacement. Use signed areas.

Exam technique

For a comparison of accelerations, compare gradients and state which line is steeper.
For displacement, label the geometric shapes, show every area calculation and combine them with the correct signs.
When a graph crosses the axis, state that the object is momentarily at rest and then changes direction.

Self review

What does the gradient of a velocity–time graph represent?
What does the area between the line and time axis represent?
How is total distance found when part of the graph lies below the axis?
What happens when a velocity–time graph crosses the time axis?

2.2.4 Measuring speed, typical speeds and free-fall acceleration

Measuring speed

Definition

Average speed

Average speed is the total distance travelled divided by the total time taken.

Definition

Resolution

Resolution is the smallest change in a quantity that a measuring instrument can distinguish.

Every speed method measures a distance and the corresponding time, then uses
$speed=distancetime\text{average speed}=\frac{\text{distance}}{\text{time}}$
For a stopwatch method, mark two positions a measured distance apart. Time the object between the markers, repeat the measurement and calculate a mean.
A long measured distance reduces the percentage uncertainty in the distance and makes reaction-time uncertainty a smaller fraction of the measured time.

Using light gates

Attach a card of measured length to the moving object. Connect a light gate to a timer or data logger.
As the card passes through the beam, the logger measures the interruption time. The instantaneous speed at the gate is approximated by
$timev=\frac{\text{card length}}{\text{interruption time}}$
With two light gates, measure their separation and the travel time between them to calculate average speed over that interval.
Light gates remove the delay caused by starting and stopping a handheld stopwatch, so they reduce reaction-time uncertainty.

Other laboratory methods

A ticker timer makes dots on a moving tape at equal time intervals. Measure the distance across several intervals and divide by their total time.
If the mains frequency is $Hz50\ \text{Hz}$ , successive dots are separated by $s\dfrac{1}{50}=0.020\ \text{s}$ . Increasing gaps show increasing speed.
Video analysis uses a known scale in the frame and the recording frame rate. Measure the displacement across a known number of frames and divide by the elapsed time.
Motion sensors can record distance at short regular intervals. The gradient of a distance–time graph then gives speed.

Exam technique

A complete method names the equipment, states what distance and time are measured, gives the speed equation, repeats the readings and calculates a mean.
When explaining an improvement, link it to the relevant uncertainty. Light gates reduce reaction-time uncertainty, while a longer timing distance reduces percentage uncertainty.
Use consistency, reliability, precision or reduced random error when justifying repeats. Do not claim that repeats automatically remove systematic error.

Typical speeds

Typical values are estimates that provide an order-of-magnitude check rather than exact constants.
Walking is about $m/s1.5\ \text{m/s}$ , running about $m/s3\ \text{m/s}$ and cycling about $m/s6\ \text{m/s}$ .
A car in urban traffic is about $m/s13\ \text{m/s}$ , which is roughly $mph30\ \text{mph}$ , while motorway travel is about $m/s30\ \text{m/s}$ .
A train may travel at about $m/s50\ \text{m/s}$ and a cruising passenger aircraft at about $m/s250\ \text{m/s}$ .
A gentle breeze is about $m/s5\ \text{m/s}$ and a gale about $m/s20\ \text{m/s}$ .
Sound in air travels at about $m/s330\ \text{m/s}$ .
Useful conversions are $km/h1\ \text{m/s}=3.6\ \text{km/h}$ and $mph1\ \text{m/s}\approx2.2\ \text{mph}$ .

Example

A person walks for $min20\ \text{min}$ at an estimated $m/s1.5\ \text{m/s}$ .
The time is $s20\times60=1200\ \text{s}$ .
The estimated distance is $ms=vt=(1.5)(1200)=1800\ \text{m}$ , which is about $km2\ \text{km}$ .

Free-fall acceleration

Definition

Free fall

Free fall is motion in which gravity is the only force acting on an object.

Definition

Gravitational acceleration

Gravitational acceleration is the acceleration caused by a gravitational field.

Near the Earth’s surface, the acceleration in free fall is
$m/s2g=10\ \text{m/s}^2$
The acceleration is directed downwards towards the centre of the Earth. An object’s downward velocity therefore increases by about $m/s10\ \text{m/s}$ each second when air resistance is negligible.
All objects have the same free-fall acceleration at the same location, regardless of mass. A greater mass produces a proportionally greater weight, so $a = F / m$ remains the same.
A ball thrown upwards is still accelerating downwards. It slows while rising, has zero instantaneous velocity at its highest point, then speeds up downwards.
The symbol $g$ can also represent gravitational field strength in $N/kg\text{N/kg}$ . The numerical value near Earth is also about $10$ , but acceleration and field strength have different units.

Example

A ball is released from rest and falls freely for $s0.80\ \text{s}$ .
Using $v = u + a t$ with $u = 0$ and $m/s2a=10\ \text{m/s}^2$ gives $v = 0 + (10) (0.80)$ .
The speed after $s0.80\ \text{s}$ is $m/s8.0\ \text{m/s}$ .

Estimating accelerations

Use $a=v−uta=\dfrac{v-u}{t}$ with sensible estimates for the velocity change and time interval.
A train leaving a station may accelerate at about $m/s20.5\ \text{m/s}^2$ , a car pulling away at about $m/s22\ \text{m/s}^2$ and a sprinter at about $m/s25\ \text{m/s}^2$ .
Hard braking can produce a deceleration with magnitude approaching $m/s210\ \text{m/s}^2$ .
Check that the estimate matches the context and quote an appropriate unit and sensible precision.

Common Mistake

Do not state that heavier objects have a larger free-fall acceleration. Differences observed in air result from air resistance.
Do not describe $g$ as a force. Weight is the force, calculated using $W = m g$ .
Do not use a single short timing interval when a longer interval or electronic timer is available.

Self review

What two measurements are required in every speed method?
How does a light gate determine speed from a card?
What are suitable estimates for walking speed and the speed of sound in air?
What is the value and direction of free-fall acceleration near Earth?
Why do different masses have the same free-fall acceleration?

2.1.1 Scalar and vector quantities

Scalar and vector quantities

Definition

Scalar quantity

A scalar quantity has magnitude but no specific direction, such as distance, speed, mass or energy.

Definition

Vector quantity

A vector quantity has both magnitude and a specific direction, such as displacement, velocity, acceleration or force.

A physical quantity needs a magnitude and a unit. Magnitude means its numerical size.
A scalar is completely described by its magnitude and unit, such as a mass of $kg70\ \text{kg}$ or an energy transfer of $kJ84\ \text{kJ}$ .
A vector also needs a direction, such as a force of $N5\ \text{N}$ downwards or a velocity of $m/s12\ \text{m/s}$ east.
A vector can be represented by an arrow. The arrow points in the vector’s direction, while its length can be drawn to scale to represent the magnitude.

Classifying quantities

The scalar quantities are distance, speed, mass and energy.
The vector quantities are displacement, velocity, acceleration, force, weight and momentum.
Distance is the total length of the path travelled, measured in $m\text{m}$ .
Displacement is the straight-line change in position from the starting point to the finishing point, stated with a direction and measured in $m\text{m}$ .
Speed describes how quickly distance is travelled and is measured in $m/s\text{m/s}$ .
Velocity is speed in a stated direction and is measured in $m/s\text{m/s}$ .
Mass measures the amount of matter in an object and is measured in $kg\text{kg}$ .
Weight is the gravitational force acting on an object, measured in $N\text{N}$ and directed towards the centre of the attracting body.
Acceleration, force and momentum are vectors because changing their direction changes the physical situation. Energy is a scalar because it has no direction.

Common Mistake

Do not use mass and weight as interchangeable terms. Mass is a scalar measured in $kg\text{kg}$ , while weight is a vector force measured in $N\text{N}$ .
Do not identify a vector only from its unit. Distance and displacement both use $m\text{m}$ , while speed and velocity both use $m/s\text{m/s}$ .

Direction changes vector results

Choose one direction as positive when motion is along a straight line. Quantities in the opposite direction are negative.
A velocity of $m/s-12\ \text{m/s}$ means a velocity of magnitude $m/s12\ \text{m/s}$ in the chosen negative direction. It does not mean a speed below zero.
Vectors in opposite directions can partly or completely cancel. Scalars do not cancel because they carry no direction.

Example

A walker travels $m30\ \text{m}$ east and then $m10\ \text{m}$ west.
The distance is $m30+10=40\ \text{m}$ because distance counts the whole route.
Taking east as positive, the displacement is $m30-10=20\ \text{m}$ east.

Speed and velocity

Definition

Velocity

Velocity is speed in a stated direction, measured in metres per second.

An object can have constant speed while its velocity changes. When it moves around a curve, the direction changes even if the magnitude of the velocity stays constant.
After one complete lap of a circular track, the distance is the circumference of the track but the displacement is zero because the finishing position is the starting position.

Example

A runner completes one $m400\ \text{m}$ lap in $s50\ \text{s}$ .
The average speed is $m/s\dfrac{400}{50}=8.0\ \text{m/s}$ .
The displacement is $m0\ \text{m}$ , so the average velocity is $m/s\dfrac{0}{50}=0\ \text{m/s}$ .

Exam technique

When asked to compare a scalar with a vector, state that both have magnitude and then add that only the vector has a specific direction.
For a displacement or velocity answer, include the direction. A magnitude and unit alone give only distance or speed.
Use signs consistently in one-dimensional calculations and state which direction you selected as positive.

Self review

What information describes a scalar quantity completely?
What additional information is required for a vector quantity?
Which listed motion quantities form the scalar and vector pairs?
Why can velocity change while speed remains constant?

2.2.1 Speed and distance–time graphs

Speed and average speed

Definition

Average speed

Average speed is the total distance travelled divided by the total time taken.

Definition

Instantaneous speed

Instantaneous speed is the speed of an object at a particular moment.

Speed is a scalar quantity because it has magnitude but no direction.
The equation for average speed is
$taken\text{average speed}=\frac{\text{distance travelled}}{\text{time taken}}$
Using symbols,
$v=stv=\frac{s}{t}$
where $v$ is average speed in $m/s\text{m/s}$ , $s$ is distance in $m\text{m}$ and $t$ is time in $s\text{s}$ .
The rearranged forms are $s = v t$ and $t=svt=\dfrac{s}{v}$ . Convert every measurement to compatible units before substitution.

Example

A cyclist covers $m750\ \text{m}$ in $s50\ \text{s}$ .
Substitution gives $v=75050v=\dfrac{750}{50}$ .
The average speed is $m/s15\ \text{m/s}$ .

Average speed for a whole journey

Average speed is not normally the arithmetic mean of two speeds. It depends on the total distance and total time.
Include time spent stationary because it contributes to the total journey time even though no distance is travelled during that interval.

Example

A bus travels $m600\ \text{m}$ in $s40\ \text{s}$ , waits for $s20\ \text{s}$ and then travels $m900\ \text{m}$ in $s60\ \text{s}$ .
The total distance is $m600+900=1500\ \text{m}$ .
The total time is $s40+20+60=120\ \text{s}$ .
The average speed is $m/s\dfrac{1500}{120}=12.5\ \text{m/s}$ .

Distance–time graphs

Definition

Gradient

The gradient of a graph is the change in the vertical-axis quantity divided by the change in the horizontal-axis quantity.

A distance–time graph shows distance on the vertical axis and time on the horizontal axis.
The gradient represents speed because
$gradient=ΔdistanceΔtime=speed\text{gradient}=\frac{\Delta \text{distance}}{\Delta \text{time}}=\text{speed}$
A horizontal line has zero gradient, so the object is stationary.
A straight sloping line has constant gradient, so the object moves at constant speed.
A steeper line represents a greater speed because more distance is covered in the same time.
A curve with an increasing gradient represents increasing speed. A curve that becomes less steep represents decreasing speed.

Calculating a graph gradient

Choose two points far apart on the straight section of the graph. If a best-fit line is present, the points should lie on that line and do not need to be plotted data points.
Draw a large gradient triangle and read the coordinate differences from the axes.
Calculate
$speed=d2−d1t2−t1\text{speed}=\frac{d_2-d_1}{t_2-t_1}$
Include the unit obtained from the graph axes, commonly $m/s\text{m/s}$ .

Example

A straight section passes through $m)(10\ \text{s},40\ \text{m})$ and $m)(30\ \text{s},140\ \text{m})$ .
The change in distance is $m140-40=100\ \text{m}$ .
The change in time is $s30-10=20\ \text{s}$ .
The speed is $m/s\dfrac{100}{20}=5.0\ \text{m/s}$ .

Common Mistake

Do not calculate a gradient using $d / t$ from one point unless the straight line passes through the origin.
Do not describe a horizontal section as moving at constant speed. Its speed is zero.
A distance–time graph does not show direction, so its distance values do not decrease.

Exam technique

For a graph description, link each shape to motion: horizontal means stationary, straight and sloping means constant speed, and changing gradient means changing speed.
Show the coordinate differences used for a gradient. A large triangle reduces the percentage uncertainty in reading the axes.
Keep units with all substitutions and quote a sensible number of significant figures.

Self review

What equation links average speed, distance and time?
What does the gradient of a distance–time graph represent?
What motion produces a horizontal section?
How does a curved distance–time graph show acceleration or deceleration?

2.2.2 Acceleration and the uniform-acceleration equation

Acceleration

Definition

Acceleration

Acceleration is the rate of change of velocity.

Definition

Deceleration

Deceleration is acceleration that reduces the magnitude of an object's velocity.

Acceleration is a vector quantity because velocity is a vector. A change in speed, direction or both produces acceleration.
The acceleration equation is
$a=v−uta=\frac{v-u}{t}$
where $a$ is acceleration in $m/s2\text{m/s}^2$ , $u$ is initial velocity in $m/s\text{m/s}$ , $v$ is final velocity in $m/s\text{m/s}$ and $t$ is time in $s\text{s}$ .
The unit $m/s2\text{m/s}^2$ means that velocity changes by that many metres per second during each second.
With one direction chosen as positive, a negative acceleration points in the negative direction. It represents slowing down only when velocity points in the positive direction.

Example

A car increases its velocity from $m/s6.0\ \text{m/s}$ to $m/s22\ \text{m/s}$ in $s4.0\ \text{s}$ .
Substitution gives $a=22−6.04.0a=\dfrac{22-6.0}{4.0}$ .
The acceleration is $m/s24.0\ \text{m/s}^2$ .

Rearranging the acceleration equation

Useful rearrangements are
$v = u + a t$
$u = v - a t$
$t=v−uat=\frac{v-u}{a}$
Keep the signs of velocities and acceleration during substitution. A negative value may carry essential directional information.

Example

A train travelling at $m/s25\ \text{m/s}$ slows uniformly at $m/s2-0.50\ \text{m/s}^2$ for $s30\ \text{s}$ .
Using $v = u + a t$ gives $v = 25 + (- 0.50) (30)$ .
The final velocity is $m/s10\ \text{m/s}$ .

Uniform acceleration without time

Definition

Uniform acceleration

Uniform acceleration is acceleration that remains constant.

When acceleration is uniform and time is not given, use
$v^2-u^2=2ax$
where $x$ is displacement in $m\text{m}$ . This equation applies only when acceleration is constant.
Useful rearrangements include
$x=v2−u22ax=\frac{v^2-u^2}{2a}$
$a=v2−u22xa=\frac{v^2-u^2}{2x}$
When solving for velocity, take the square root only after evaluating $v^2$ . Select the root that matches the stated direction of motion.

Example

A vehicle moving at $m/s20\ \text{m/s}$ comes to rest with uniform acceleration over $m50\ \text{m}$ .
Substitution gives $0^2-20^2=2a(50)$ .
Rearrangement gives $m/s2a=\dfrac{-400}{100}=-4.0\ \text{m/s}^2$ .

Interpreting acceleration

Positive and negative signs indicate direction relative to the chosen positive direction. They do not automatically mean speeding up and slowing down.
An object speeds up when its velocity and acceleration have the same sign.
An object slows down when its velocity and acceleration have opposite signs.
An object moving at constant speed in a circle accelerates because its direction, and therefore its velocity, continually changes.

Common Mistake

Do not use $m/s\text{m/s}$ as the unit of acceleration. The correct unit is $m/s2\text{m/s}^2$ .
Do not use $v^2-u^2=2ax$ when acceleration changes during the motion.
Do not discard a negative acceleration. Explain its direction or identify it as deceleration only when the motion direction makes that description correct.

Exam technique

Write the selected equation before substituting and keep all quantities in SI units.
Edexcel calculation marks commonly reward substitution, evaluation and the final unit, so show these stages clearly even when your calculator gives the answer directly.
Check whether the question gives distance or displacement and whether the signs make physical sense.

Self review

What does an acceleration of $m/s23\ \text{m/s}^2$ mean?
What equation links $u$ , $v$ , $a$ and $t$ ?
When may $v^2-u^2=2ax$ be used?
When does a negative acceleration make an object speed up?

2.2.3 Velocity–time graphs

Velocity–time graphs

Definition

Velocity-time graph

A velocity-time graph shows how an object's velocity changes with time.

Definition

Uniform acceleration

Uniform acceleration is acceleration that remains constant.

Velocity is plotted on the vertical axis and time on the horizontal axis. Values below the time axis represent motion in the chosen negative direction.
A horizontal line represents constant velocity and zero acceleration.
A straight sloping line represents uniform acceleration.
A curve represents changing acceleration because its gradient changes.
Crossing the time axis means the velocity is zero at that instant. If the graph continues across the axis, the object reverses direction.

Gradient gives acceleration

The gradient of a velocity–time graph is
$gradient=ΔvΔt=a\text{gradient}=\frac{\Delta v}{\Delta t}=a$
A positive gradient gives acceleration in the positive direction and a negative gradient gives acceleration in the negative direction.
To calculate uniform acceleration, choose two points far apart on the straight section and use
$a=v2−v1t2−t1a=\frac{v_2-v_1}{t_2-t_1}$

Example

Velocity rises uniformly from $m/s4.0\ \text{m/s}$ at $s2.0\ \text{s}$ to $m/s16\ \text{m/s}$ at $s8.0\ \text{s}$ .
The velocity change is $m/s16-4.0=12\ \text{m/s}$ .
The time change is $s8.0-2.0=6.0\ \text{s}$ .
The acceleration is $m/s2\dfrac{12}{6.0}=2.0\ \text{m/s}^2$ .

Area gives displacement

Definition

Displacement from a velocity-time graph

Displacement is the signed area between a velocity-time graph and the time axis.

For uniform acceleration, split the region into rectangles and triangles, then calculate each area.
For a rectangle, $A = bh$ . For a triangle, $A=12bhA=\dfrac{1}{2}bh$ . Time provides the base and velocity provides the height, so the area unit is $s×m/s=m\text{s}\times\text{m/s}=\text{m}$ .
Area above the time axis is positive displacement. Area below the axis is negative displacement.
Total displacement is the signed sum of all areas. Total distance is the sum of the magnitudes of the separate areas.

Example

A car accelerates uniformly from $m/s5.0\ \text{m/s}$ to $m/s15\ \text{m/s}$ in $s4.0\ \text{s}$ .
The area is a rectangle plus a triangle: $x=(5.0)(4.0)+12(4.0)(15−5.0)x=(5.0)(4.0)+\dfrac{1}{2}(4.0)(15-5.0)$ .
The displacement is $m20+20=40\ \text{m}$ .

Describing motion from the graph

State the velocity at the start of each section, whether it is constant or changing, and the direction of motion.
Use the sign and gradient separately. The sign of velocity gives the direction of motion, while the sign of the gradient gives the direction of acceleration.
An upward-sloping line below the axis can represent an object slowing down while travelling in the negative direction because its velocity moves towards zero.
A downward-sloping line below the axis can represent speeding up in the negative direction because the magnitude of velocity increases.

Common Mistake

Do not call every negative gradient deceleration. Compare the signs of velocity and acceleration.
Do not use the line length as distance. Calculate the area between the graph and the time axis.
Do not add areas below the axis as positive when finding displacement. Use signed areas.

Exam technique

For a comparison of accelerations, compare gradients and state which line is steeper.
For displacement, label the geometric shapes, show every area calculation and combine them with the correct signs.
When a graph crosses the axis, state that the object is momentarily at rest and then changes direction.

Self review

What does the gradient of a velocity–time graph represent?
What does the area between the line and time axis represent?
How is total distance found when part of the graph lies below the axis?
What happens when a velocity–time graph crosses the time axis?

2.2.4 Measuring speed, typical speeds and free-fall acceleration

Measuring speed

Definition

Average speed

Average speed is the total distance travelled divided by the total time taken.

Definition

Resolution

Resolution is the smallest change in a quantity that a measuring instrument can distinguish.

Every speed method measures a distance and the corresponding time, then uses
$speed=distancetime\text{average speed}=\frac{\text{distance}}{\text{time}}$
For a stopwatch method, mark two positions a measured distance apart. Time the object between the markers, repeat the measurement and calculate a mean.
A long measured distance reduces the percentage uncertainty in the distance and makes reaction-time uncertainty a smaller fraction of the measured time.

Using light gates

Attach a card of measured length to the moving object. Connect a light gate to a timer or data logger.
As the card passes through the beam, the logger measures the interruption time. The instantaneous speed at the gate is approximated by
$timev=\frac{\text{card length}}{\text{interruption time}}$
With two light gates, measure their separation and the travel time between them to calculate average speed over that interval.
Light gates remove the delay caused by starting and stopping a handheld stopwatch, so they reduce reaction-time uncertainty.

Other laboratory methods

A ticker timer makes dots on a moving tape at equal time intervals. Measure the distance across several intervals and divide by their total time.
If the mains frequency is $Hz50\ \text{Hz}$ , successive dots are separated by $s\dfrac{1}{50}=0.020\ \text{s}$ . Increasing gaps show increasing speed.
Video analysis uses a known scale in the frame and the recording frame rate. Measure the displacement across a known number of frames and divide by the elapsed time.
Motion sensors can record distance at short regular intervals. The gradient of a distance–time graph then gives speed.

Exam technique

A complete method names the equipment, states what distance and time are measured, gives the speed equation, repeats the readings and calculates a mean.
When explaining an improvement, link it to the relevant uncertainty. Light gates reduce reaction-time uncertainty, while a longer timing distance reduces percentage uncertainty.
Use consistency, reliability, precision or reduced random error when justifying repeats. Do not claim that repeats automatically remove systematic error.

Typical speeds

Typical values are estimates that provide an order-of-magnitude check rather than exact constants.
Walking is about $m/s1.5\ \text{m/s}$ , running about $m/s3\ \text{m/s}$ and cycling about $m/s6\ \text{m/s}$ .
A car in urban traffic is about $m/s13\ \text{m/s}$ , which is roughly $mph30\ \text{mph}$ , while motorway travel is about $m/s30\ \text{m/s}$ .
A train may travel at about $m/s50\ \text{m/s}$ and a cruising passenger aircraft at about $m/s250\ \text{m/s}$ .
A gentle breeze is about $m/s5\ \text{m/s}$ and a gale about $m/s20\ \text{m/s}$ .
Sound in air travels at about $m/s330\ \text{m/s}$ .
Useful conversions are $km/h1\ \text{m/s}=3.6\ \text{km/h}$ and $mph1\ \text{m/s}\approx2.2\ \text{mph}$ .

Example

A person walks for $min20\ \text{min}$ at an estimated $m/s1.5\ \text{m/s}$ .
The time is $s20\times60=1200\ \text{s}$ .
The estimated distance is $ms=vt=(1.5)(1200)=1800\ \text{m}$ , which is about $km2\ \text{km}$ .

Free-fall acceleration

Definition

Free fall

Free fall is motion in which gravity is the only force acting on an object.

Definition

Gravitational acceleration

Gravitational acceleration is the acceleration caused by a gravitational field.

Near the Earth’s surface, the acceleration in free fall is
$m/s2g=10\ \text{m/s}^2$
The acceleration is directed downwards towards the centre of the Earth. An object’s downward velocity therefore increases by about $m/s10\ \text{m/s}$ each second when air resistance is negligible.
All objects have the same free-fall acceleration at the same location, regardless of mass. A greater mass produces a proportionally greater weight, so $a = F / m$ remains the same.
A ball thrown upwards is still accelerating downwards. It slows while rising, has zero instantaneous velocity at its highest point, then speeds up downwards.
The symbol $g$ can also represent gravitational field strength in $N/kg\text{N/kg}$ . The numerical value near Earth is also about $10$ , but acceleration and field strength have different units.

Example

A ball is released from rest and falls freely for $s0.80\ \text{s}$ .
Using $v = u + a t$ with $u = 0$ and $m/s2a=10\ \text{m/s}^2$ gives $v = 0 + (10) (0.80)$ .
The speed after $s0.80\ \text{s}$ is $m/s8.0\ \text{m/s}$ .

Estimating accelerations

Use $a=v−uta=\dfrac{v-u}{t}$ with sensible estimates for the velocity change and time interval.
A train leaving a station may accelerate at about $m/s20.5\ \text{m/s}^2$ , a car pulling away at about $m/s22\ \text{m/s}^2$ and a sprinter at about $m/s25\ \text{m/s}^2$ .
Hard braking can produce a deceleration with magnitude approaching $m/s210\ \text{m/s}^2$ .
Check that the estimate matches the context and quote an appropriate unit and sensible precision.

Common Mistake

Do not state that heavier objects have a larger free-fall acceleration. Differences observed in air result from air resistance.
Do not describe $g$ as a force. Weight is the force, calculated using $W = m g$ .
Do not use a single short timing interval when a longer interval or electronic timer is available.

Self review

What two measurements are required in every speed method?
How does a light gate determine speed from a card?
What are suitable estimates for walking speed and the speed of sound in air?
What is the value and direction of free-fall acceleration near Earth?
Why do different masses have the same free-fall acceleration?

Speed, velocity and acceleration

2.1.1 Scalar and vector quantities

Scalar and vector quantities

Classifying quantities

Direction changes vector results

Speed and velocity

2.2.1 Speed and distance–time graphs

Speed and average speed

Average speed for a whole journey

Distance–time graphs

Calculating a graph gradient

2.2.2 Acceleration and the uniform-acceleration equation

Acceleration

Rearranging the acceleration equation

Uniform acceleration without time

Interpreting acceleration

2.2.3 Velocity–time graphs

Velocity–time graphs

Gradient gives acceleration

Area gives displacement

Describing motion from the graph

2.2.4 Measuring speed, typical speeds and free-fall acceleration

Measuring speed

Using light gates

Other laboratory methods

Typical speeds

Free-fall acceleration

Estimating accelerations

Recap questions

Review 2.2 Speed, velocity and acceleration by teaching Genie

Recap your knowledge with an interactive lesson

Put it into practice with exam-style questions

Remember key concepts with flashcards

2.2 Speed, velocity and acceleration Revision Guide

Speed, velocity and acceleration

2.1.1 Scalar and vector quantities

Scalar and vector quantities

Classifying quantities

Direction changes vector results

Speed and velocity

2.2.1 Speed and distance–time graphs

Speed and average speed

Average speed for a whole journey

Distance–time graphs

Calculating a graph gradient

2.2.2 Acceleration and the uniform-acceleration equation

Acceleration

Rearranging the acceleration equation

Uniform acceleration without time

Interpreting acceleration

2.2.3 Velocity–time graphs

Velocity–time graphs

Gradient gives acceleration

Area gives displacement

Describing motion from the graph

2.2.4 Measuring speed, typical speeds and free-fall acceleration

Measuring speed

Using light gates

Other laboratory methods

Typical speeds

Free-fall acceleration

Estimating accelerations

Recap questions

Review 2.2 Speed, velocity and acceleration by teaching Genie

Recap your knowledge with an interactive lesson

Put it into practice with exam-style questions

Remember key concepts with flashcards

2.2 Speed, velocity and acceleration Revision Guide