Kinematics

What you'll learn

Describe motion using displacement, speed, velocity and acceleration.
Use displacement-time, speed-time and velocity-time graphs quantitatively.
Derive and apply the constant-acceleration equations for straight-line motion.
Explain freefall, terminal velocity, and projectile motion using independent components.

What kinematics is

Kinematics is the description of motion without asking what force caused it. In this topic, you mainly deal with rectilinear motion, which means motion along a straight line, and projectile motion, where an object moves freely under gravity after being launched.

A scalar has size only. A vector has size and direction.

Definition

Core motion quantities

Distance is the total path length travelled. It is a scalar, measured in metres (m).
Displacement, $s$ , is the change in position from start to finish in a stated direction. It is a vector, measured in metres (m).
Mean speed is total distance divided by total time.
Velocity is the rate of change of displacement. It is a vector, measured in metres per second (m s⁻¹).
Mean velocity is displacement divided by time.
Acceleration, $a$ , is the rate of change of velocity, measured in metres per second squared (m s⁻²):

a=ΔvΔta=\frac{\Delta v}{\Delta t}

Instantaneous means “at one instant”. A speedometer gives instantaneous speed; a light gate can measure instantaneous speed by timing how long a card of known length blocks the beam.

Example

Mean speed and mean velocity

A cyclist travels 120 m east, then 40 m west, in a total time of 20.0 s.

The total distance is $m120\,\text{m}+40\,\text{m}=160\,\text{m}$ , so the mean speed is:

s−1\frac{160\,\text{m}}{20.0\,\text{s}}=8.00\,\text{m s}^{-1}

The displacement is $m120\,\text{m}-40\,\text{m}=80\,\text{m}$ east.
The mean velocity is:

east\frac{80\,\text{m}}{20.0\,\text{s}}=4.00\,\text{m s}^{-1}\ \text{east}

Common Mistake

Distance is not displacement

If an object turns around, its distance keeps increasing, but its displacement may decrease or even become zero.

Motion graphs

Graphs let you turn motion into gradients and areas. Always check what is on each axis before choosing a method.

The graphs below show the links between displacement, velocity and acceleration for motion with constant positive acceleration.

Aligned displacement-time, velocity-time and acceleration-time graphs for constant acceleration

Displacement-time graphs

On a displacement-time graph, the gradient gives velocity:

gradient=ΔsΔt=v\text{gradient}=\frac{\Delta s}{\Delta t}=v

A straight line means constant velocity. A curve means changing velocity. For non-uniform motion, draw a tangent to the curve at the instant you care about; the tangent’s gradient is the instantaneous velocity.

Speed-time and velocity-time graphs

On a velocity-time graph, the gradient gives acceleration:

gradient=ΔvΔt=a\text{gradient}=\frac{\Delta v}{\Delta t}=a

The area under a velocity-time graph gives displacement. Area below the time axis counts as negative displacement.

A speed-time graph cannot show direction, so its area gives distance travelled, not displacement. For curved graphs, estimate area using trapezia or count squares carefully.

Key Idea

Gradient and area links

Gradient takes you “forward” through the quantities: displacement-time to velocity, then velocity-time to acceleration. Area under a velocity-time graph takes you back to displacement.

Example

Using a velocity-time graph

A train’s velocity increases uniformly from 4.0 m s⁻¹ to 16.0 m s⁻¹ in 6.0 s. Find its acceleration and displacement.

The acceleration is the gradient of the velocity-time graph:

s−2a=\frac{16.0\,\text{m s}^{-1}-4.0\,\text{m s}^{-1}}{6.0\,\text{s}} =2.0\,\text{m s}^{-2}

The displacement is the trapezium area under the graph:

ms=\frac{(4.0\,\text{m s}^{-1}+16.0\,\text{m s}^{-1})}{2}\times 6.0\,\text{s} =60\,\text{m}

The velocity is always positive, so the displacement is 60 m in the positive direction.

Uniformly accelerated motion

Uniform acceleration means constant acceleration. The velocity-time graph is a straight line, so the standard equations can be derived from gradient and area.

Let:

$u$ be initial velocity
$v$ be final velocity
$a$ be acceleration
$s$ be displacement
$t$ be time

From the velocity-time graph:

Gradient gives $a=v−uta=\frac{v-u}{t}$ , so:

v = u + a t

Area under the graph is a trapezium, so:

s=(u+v)t2s=\frac{(u+v)t}{2}

Combining these gives the other constant-acceleration equations:

s=ut+12at2v2=u2+2as\begin{aligned} s&=ut+\frac{1}{2}at^2\\ v^2&=u^2+2as \end{aligned}

Common Mistake

Only for constant acceleration

These equations apply to straight-line motion with constant acceleration. If acceleration changes, use graph gradients/areas or a smaller time interval model instead.

Example

Stopping under constant deceleration

A car travelling at 27 m s⁻¹ brakes uniformly at $s−2-6.0\,\text{m s}^{-2}$ . Calculate the braking distance.

Time is not needed, so choose:

v^2=u^2+2as

Substitute $s−1v=0\,\text{m s}^{-1}$ , $s−1u=27\,\text{m s}^{-1}$ and $s−2a=-6.0\,\text{m s}^{-2}$ :

s−2)s0=(27\,\text{m s}^{-1})^2+2(-6.0\,\text{m s}^{-2})s

Rearrange for $s$ :

ms=\frac{(27\,\text{m s}^{-1})^2}{12.0\,\text{m s}^{-2}} =60.75\,\text{m}\approx 61\,\text{m}

Tip

Stopping-distance data

For speed-limit or road-safety data, convert speeds to m s⁻¹ before using equations. With the same braking acceleration, braking distance is proportional to $u^2$ , so a small speed increase can produce a much larger stopping distance.

Falling in a gravitational field

Near Earth’s surface, an object in freefall has acceleration $g$ , about $s−29.81\,\text{m s}^{-2}$ downward. If upward is chosen as positive, then a falling object has $a = - g$ .

Without air resistance, all objects fall with the same acceleration, regardless of mass. With air resistance, the motion changes because drag increases as speed increases.

Terminal velocity is the constant velocity reached when the resistive force equals the weight, so the resultant force is zero and acceleration is zero.

Example

Explaining terminal velocity

A skydiver jumps from rest and eventually reaches terminal velocity before opening the parachute.

Just after jumping, speed is small, so air resistance is small. Weight is larger than drag, giving a downward resultant force and downward acceleration.
As speed increases, air resistance increases. The resultant force decreases, so the acceleration decreases.
At terminal velocity, air resistance equals weight. The resultant force is zero, so velocity remains constant.

Specified practical: measuring $g$ by freefall

A common method uses an electromagnet, a steel ball and an electronic timer. Electronic timing is much better than a hand stopwatch because human reaction time would be a large source of uncertainty.

Freefall apparatus for measuring gravitational acceleration with an electromagnet, contact pad and graph of height against time squared

Measure the height $h$ from the bottom of the ball to the contact pad. Release the ball from rest; the timer starts when the electromagnet releases it and stops when the ball hits the pad.

For freefall from rest:

h=12gt2h=\frac{1}{2}gt^2

So a graph of $h$ against $t^2$ should be a straight line through the origin, with:

gradient=g2\text{gradient}=\frac{g}{2}

To improve the investigation, repeat each height, use a small dense ball to reduce air resistance, measure height carefully with a metre rule, and use a best-fit line rather than one pair of raw readings.

Example

Finding g from a freefall graph

A best-fit line on a graph of $h$ against $t^2$ passes through $m)(0.0200\,\text{s}^2,\ 0.098\,\text{m})$ and $m)(0.100\,\text{s}^2,\ 0.490\,\text{m})$ .

Calculate the gradient:

s−2\text{gradient}=\frac{0.490\,\text{m}-0.098\,\text{m}}{0.100\,\text{s}^2-0.0200\,\text{s}^2} =\frac{0.392\,\text{m}}{0.0800\,\text{s}^2} =4.90\,\text{m s}^{-2}

Use $gradient=g2\text{gradient}=\frac{g}{2}$ :

s−2g=2\times 4.90\,\text{m s}^{-2}=9.80\,\text{m s}^{-2}

This is close to $s−29.81\,\text{m s}^{-2}$ , so the result is reasonable; remaining differences could come from air resistance, timing delay or measuring $h$ incorrectly.

Tip

Graph quality check

If the $h$ against $t^2$ graph has a clear non-zero intercept, suspect a systematic error such as a timing offset or measuring height from the wrong point.

Projectile motion

A projectile moving freely under gravity has two independent components of motion:

horizontal motion: constant velocity, if air resistance is neglected
vertical motion: uniform acceleration due to gravity

The same time $t$ applies to both components, but the equations are used separately in each direction.

For horizontal motion:

ax=0,x=uxta_x=0,\qquad x=u_x t

For vertical motion, if upward is positive:

a_y=-g

The path is curved because the object has uniform velocity in one direction and uniform acceleration in the perpendicular direction.

Projectile motion diagram showing constant horizontal velocity and increasing downward vertical velocity

Example

Horizontal launch from a cliff

A ball is launched horizontally at 8.0 m s⁻¹ from a cliff 20 m high. Ignore air resistance. Find the time to hit the ground and the horizontal range.

Use vertical motion to find time. With upward positive, $ms_y=-20\,\text{m}$ , $s−1u_y=0\,\text{m s}^{-1}$ and $s−2a_y=-9.81\,\text{m s}^{-2}$ :

s−2)t2-20\,\text{m}=0+\frac{1}{2}(-9.81\,\text{m s}^{-2})t^2

Solve for $t$ :

s2t^2=\frac{20\,\text{m}}{4.905\,\text{m s}^{-2}}=4.08\,\text{s}^2

st=2.02\,\text{s}

Use horizontal motion to find range:

mx=u_x t=(8.0\,\text{m s}^{-1})(2.02\,\text{s})=16.2\,\text{m}\approx 16\,\text{m}

Common Mistake

Mixing components

Do not put the horizontal velocity into a vertical equation. Resolve first, then use one set of equations horizontally and another vertically.

Exam technique

In the exam

Define your positive direction before substituting values; signs matter for velocity and acceleration.
For graphs, use gradients for rates of change and areas under velocity-time graphs for displacement.
For constant-acceleration questions, list $u$ , $v$ , $a$ , $s$ and $t$ with units, then choose the equation missing the unwanted quantity.
For practical questions, describe the measurement method, repeat readings, uncertainty reduction, and how the graph gradient gives the final value.

Self review

Check yourself

A displacement-time graph becomes steeper with time. What does that tell you about the object’s velocity?
Why does plotting $h$ against $t^2$ help you measure $g$ in the freefall practical?
In projectile motion, which quantity links the horizontal and vertical calculations?

What you'll learn

Describe motion using displacement, speed, velocity and acceleration.
Use displacement-time, speed-time and velocity-time graphs quantitatively.
Derive and apply the constant-acceleration equations for straight-line motion.
Explain freefall, terminal velocity, and projectile motion using independent components.

What kinematics is

A scalar has size only. A vector has size and direction.

Definition

Core motion quantities

Distance is the total path length travelled. It is a scalar, measured in metres (m).
Displacement, $s$ , is the change in position from start to finish in a stated direction. It is a vector, measured in metres (m).
Mean speed is total distance divided by total time.
Velocity is the rate of change of displacement. It is a vector, measured in metres per second (m s⁻¹).
Mean velocity is displacement divided by time.
Acceleration, $a$ , is the rate of change of velocity, measured in metres per second squared (m s⁻²):

a=ΔvΔta=\frac{\Delta v}{\Delta t}

Instantaneous means “at one instant”. A speedometer gives instantaneous speed; a light gate can measure instantaneous speed by timing how long a card of known length blocks the beam.

Example

Mean speed and mean velocity

A cyclist travels 120 m east, then 40 m west, in a total time of 20.0 s.

The total distance is $m120\,\text{m}+40\,\text{m}=160\,\text{m}$ , so the mean speed is:

s−1\frac{160\,\text{m}}{20.0\,\text{s}}=8.00\,\text{m s}^{-1}

The displacement is $m120\,\text{m}-40\,\text{m}=80\,\text{m}$ east.
The mean velocity is:

east\frac{80\,\text{m}}{20.0\,\text{s}}=4.00\,\text{m s}^{-1}\ \text{east}

Common Mistake

Distance is not displacement

If an object turns around, its distance keeps increasing, but its displacement may decrease or even become zero.

Motion graphs

Graphs let you turn motion into gradients and areas. Always check what is on each axis before choosing a method.

The graphs below show the links between displacement, velocity and acceleration for motion with constant positive acceleration.

Aligned displacement-time, velocity-time and acceleration-time graphs for constant acceleration

Displacement-time graphs

On a displacement-time graph, the gradient gives velocity:

gradient=ΔsΔt=v\text{gradient}=\frac{\Delta s}{\Delta t}=v

Speed-time and velocity-time graphs

On a velocity-time graph, the gradient gives acceleration:

gradient=ΔvΔt=a\text{gradient}=\frac{\Delta v}{\Delta t}=a

The area under a velocity-time graph gives displacement. Area below the time axis counts as negative displacement.

A speed-time graph cannot show direction, so its area gives distance travelled, not displacement. For curved graphs, estimate area using trapezia or count squares carefully.

Key Idea

Gradient and area links

Gradient takes you “forward” through the quantities: displacement-time to velocity, then velocity-time to acceleration. Area under a velocity-time graph takes you back to displacement.

Example

Using a velocity-time graph

A train’s velocity increases uniformly from 4.0 m s⁻¹ to 16.0 m s⁻¹ in 6.0 s. Find its acceleration and displacement.

The acceleration is the gradient of the velocity-time graph:

s−2a=\frac{16.0\,\text{m s}^{-1}-4.0\,\text{m s}^{-1}}{6.0\,\text{s}} =2.0\,\text{m s}^{-2}

The displacement is the trapezium area under the graph:

ms=\frac{(4.0\,\text{m s}^{-1}+16.0\,\text{m s}^{-1})}{2}\times 6.0\,\text{s} =60\,\text{m}

The velocity is always positive, so the displacement is 60 m in the positive direction.

Uniformly accelerated motion

Uniform acceleration means constant acceleration. The velocity-time graph is a straight line, so the standard equations can be derived from gradient and area.

Let:

$u$ be initial velocity
$v$ be final velocity
$a$ be acceleration
$s$ be displacement
$t$ be time

From the velocity-time graph:

Gradient gives $a=v−uta=\frac{v-u}{t}$ , so:

v = u + a t

Area under the graph is a trapezium, so:

s=(u+v)t2s=\frac{(u+v)t}{2}

Combining these gives the other constant-acceleration equations:

s=ut+12at2v2=u2+2as\begin{aligned} s&=ut+\frac{1}{2}at^2\\ v^2&=u^2+2as \end{aligned}

Common Mistake

Only for constant acceleration

These equations apply to straight-line motion with constant acceleration. If acceleration changes, use graph gradients/areas or a smaller time interval model instead.

Example

Stopping under constant deceleration

A car travelling at 27 m s⁻¹ brakes uniformly at $s−2-6.0\,\text{m s}^{-2}$ . Calculate the braking distance.

Time is not needed, so choose:

v^2=u^2+2as

Substitute $s−1v=0\,\text{m s}^{-1}$ , $s−1u=27\,\text{m s}^{-1}$ and $s−2a=-6.0\,\text{m s}^{-2}$ :

s−2)s0=(27\,\text{m s}^{-1})^2+2(-6.0\,\text{m s}^{-2})s

Rearrange for $s$ :

ms=\frac{(27\,\text{m s}^{-1})^2}{12.0\,\text{m s}^{-2}} =60.75\,\text{m}\approx 61\,\text{m}

Tip

Stopping-distance data

Falling in a gravitational field

Near Earth’s surface, an object in freefall has acceleration $g$ , about $s−29.81\,\text{m s}^{-2}$ downward. If upward is chosen as positive, then a falling object has $a = - g$ .

Without air resistance, all objects fall with the same acceleration, regardless of mass. With air resistance, the motion changes because drag increases as speed increases.

Terminal velocity is the constant velocity reached when the resistive force equals the weight, so the resultant force is zero and acceleration is zero.

Example

Explaining terminal velocity

A skydiver jumps from rest and eventually reaches terminal velocity before opening the parachute.

Just after jumping, speed is small, so air resistance is small. Weight is larger than drag, giving a downward resultant force and downward acceleration.
As speed increases, air resistance increases. The resultant force decreases, so the acceleration decreases.
At terminal velocity, air resistance equals weight. The resultant force is zero, so velocity remains constant.

Specified practical: measuring $g$ by freefall

A common method uses an electromagnet, a steel ball and an electronic timer. Electronic timing is much better than a hand stopwatch because human reaction time would be a large source of uncertainty.

Freefall apparatus for measuring gravitational acceleration with an electromagnet, contact pad and graph of height against time squared

Measure the height $h$ from the bottom of the ball to the contact pad. Release the ball from rest; the timer starts when the electromagnet releases it and stops when the ball hits the pad.

For freefall from rest:

h=12gt2h=\frac{1}{2}gt^2

So a graph of $h$ against $t^2$ should be a straight line through the origin, with:

gradient=g2\text{gradient}=\frac{g}{2}

Example

Finding g from a freefall graph

A best-fit line on a graph of $h$ against $t^2$ passes through $m)(0.0200\,\text{s}^2,\ 0.098\,\text{m})$ and $m)(0.100\,\text{s}^2,\ 0.490\,\text{m})$ .

Calculate the gradient:

s−2\text{gradient}=\frac{0.490\,\text{m}-0.098\,\text{m}}{0.100\,\text{s}^2-0.0200\,\text{s}^2} =\frac{0.392\,\text{m}}{0.0800\,\text{s}^2} =4.90\,\text{m s}^{-2}

Use $gradient=g2\text{gradient}=\frac{g}{2}$ :

s−2g=2\times 4.90\,\text{m s}^{-2}=9.80\,\text{m s}^{-2}

This is close to $s−29.81\,\text{m s}^{-2}$ , so the result is reasonable; remaining differences could come from air resistance, timing delay or measuring $h$ incorrectly.

Tip

Graph quality check

If the $h$ against $t^2$ graph has a clear non-zero intercept, suspect a systematic error such as a timing offset or measuring height from the wrong point.

Projectile motion

A projectile moving freely under gravity has two independent components of motion:

horizontal motion: constant velocity, if air resistance is neglected
vertical motion: uniform acceleration due to gravity

The same time $t$ applies to both components, but the equations are used separately in each direction.

For horizontal motion:

ax=0,x=uxta_x=0,\qquad x=u_x t

For vertical motion, if upward is positive:

a_y=-g

The path is curved because the object has uniform velocity in one direction and uniform acceleration in the perpendicular direction.

Projectile motion diagram showing constant horizontal velocity and increasing downward vertical velocity

Example

Horizontal launch from a cliff

A ball is launched horizontally at 8.0 m s⁻¹ from a cliff 20 m high. Ignore air resistance. Find the time to hit the ground and the horizontal range.

Use vertical motion to find time. With upward positive, $ms_y=-20\,\text{m}$ , $s−1u_y=0\,\text{m s}^{-1}$ and $s−2a_y=-9.81\,\text{m s}^{-2}$ :

s−2)t2-20\,\text{m}=0+\frac{1}{2}(-9.81\,\text{m s}^{-2})t^2

Solve for $t$ :

s2t^2=\frac{20\,\text{m}}{4.905\,\text{m s}^{-2}}=4.08\,\text{s}^2

st=2.02\,\text{s}

Use horizontal motion to find range:

mx=u_x t=(8.0\,\text{m s}^{-1})(2.02\,\text{s})=16.2\,\text{m}\approx 16\,\text{m}

Common Mistake

Mixing components

Do not put the horizontal velocity into a vertical equation. Resolve first, then use one set of equations horizontally and another vertically.

Exam technique

In the exam

Define your positive direction before substituting values; signs matter for velocity and acceleration.
For graphs, use gradients for rates of change and areas under velocity-time graphs for displacement.
For constant-acceleration questions, list $u$ , $v$ , $a$ , $s$ and $t$ with units, then choose the equation missing the unwanted quantity.
For practical questions, describe the measurement method, repeat readings, uncertainty reduction, and how the graph gradient gives the final value.

Self review

Check yourself

A displacement-time graph becomes steeper with time. What does that tell you about the object’s velocity?
Why does plotting $h$ against $t^2$ help you measure $g$ in the freefall practical?
In projectile motion, which quantity links the horizontal and vertical calculations?

What you'll learn

What kinematics is

Motion graphs

Displacement-time graphs

Speed-time and velocity-time graphs

Uniformly accelerated motion

Falling in a gravitational field

Specified practical: measuring $g$ by freefall

Projectile motion

Review Kinematics by teaching Genie

Remember key concepts with flashcards

Kinematics

What you'll learn

What kinematics is

Motion graphs

Displacement-time graphs

Speed-time and velocity-time graphs

Uniformly accelerated motion

Falling in a gravitational field

Specified practical: measuring $g$ by freefall

Projectile motion

Review Kinematics by teaching Genie

Remember key concepts with flashcards

Kinematics

What you'll learn

What kinematics is

Motion graphs

Displacement-time graphs

Speed-time and velocity-time graphs

Uniformly accelerated motion

Falling in a gravitational field

Specified practical: measuring ggg by freefall

Projectile motion

Review Kinematics by teaching Genie

Remember key concepts with flashcards

Kinematics Revision Guide

Kinematics

What you'll learn

What kinematics is

Motion graphs

Displacement-time graphs

Speed-time and velocity-time graphs

Uniformly accelerated motion

Falling in a gravitational field

Specified practical: measuring ggg by freefall

Projectile motion

Review Kinematics by teaching Genie

Remember key concepts with flashcards

Kinematics Revision Guide

Specified practical: measuring $g$ by freefall

Specified practical: measuring $g$ by freefall