SYSTEM MAP
Complete Kinematics Atlas
Use this page as a complete revision system rather than simply memorising the SUVAT equations. Kinematics questions are primarily about describing motion correctly, choosing a consistent direction, interpreting signs and graphs, and connecting physical language to mathematical equations.
LESSON 01
What Is Kinematics?
Kinematics is the branch of mechanics concerned with describing motion without considering the forces that cause that motion.
Kinematics tells us where a particle is, how quickly its position is changing and how quickly its velocity is changing.
Kinematics
Describes displacement, velocity, acceleration and time. It answers questions such as “How far has it travelled?” or “What is its velocity after 5 seconds?”
Dynamics
Explains motion using forces and Newton's laws. It answers questions such as “What force is required to produce this acceleration?”
The four central variables
| Symbol | Quantity | Typical SI unit | Type |
|---|---|---|---|
| s | displacement | m | vector |
| u | initial velocity | m s⁻¹ | vector |
| v | final velocity | m s⁻¹ | vector |
| a | acceleration | m s⁻² | vector |
| t | time | s | scalar |
LESSON 02
The Particle Model
Mechanics 1 frequently models moving objects as particles. This is a mathematical simplification that allows us to focus on translational motion.
A particle is an object whose size and shape are ignored and whose mass is treated as concentrated at a single point.
Why use a particle model?
- It removes unnecessary complications caused by the object's dimensions.
- It allows us to describe the motion using a single position coordinate.
- It is appropriate when the object's size is negligible compared with the distances involved.
- It is particularly useful for modelling cars, balls, projectiles, trains and other objects moving approximately along a line.
What does the model ignore?
A particle model normally ignores rotation, deformation, size and shape. If an object is rotating or its dimensions are important, the particle model may no longer be appropriate.
If a question says an object is modelled as a particle, do not introduce unnecessary rotational effects or dimensions into the solution.
LESSON 03
Position, Distance and Displacement
To describe one-dimensional motion, we first choose an origin and a positive direction. The particle's position can then be represented by a signed coordinate.
Position
Position tells us where the particle is relative to the chosen origin. It may be positive or negative depending on the chosen direction.
Displacement
It tells us the directed change from the initial position to the final position.
Distance travelled
Distance is the total length of the path travelled. It is always non-negative.
Distance
Scalar. Never negative. Adds the lengths of all parts of the path.
Displacement
Vector. Can be positive, negative or zero depending on the chosen positive direction.
Worked Example — Distance vs Displacement
A particle moves from x = 2 m to x = 11 m and then moves back to x = 7 m.
First part of journey:
11 − 2 = 9 m
Second part:
11 − 7 = 4 m
Therefore distance travelled:
9 + 4 = 13 m
Final displacement:
7 − 2 = 5 m
Do not calculate total distance by simply subtracting the initial position from the final position when the particle changes direction. That gives displacement, not distance.
LESSON 04
Velocity
Velocity measures how rapidly displacement changes with time.
In a straight-line problem, the sign of velocity tells us the direction of motion relative to the chosen positive direction.
Average velocity
Therefore:
Instantaneous velocity
Instantaneous velocity is the velocity at one particular instant. It is represented mathematically by the derivative:
Speed
Speed is the magnitude of velocity.
If velocity = −8 m s⁻¹, then the speed is 8 m s⁻¹. The negative sign belongs to the velocity because it represents direction.
Worked Example — Negative Velocity
A particle has velocity −12 m s⁻¹.
The particle is moving in the negative direction with speed 12 m s⁻¹.
LESSON 05
Acceleration and Deceleration
Acceleration measures the rate at which velocity changes.
Acceleration is a vector quantity. Its sign depends on the chosen positive direction.
Positive acceleration does not always mean speeding up
This is one of the most important conceptual ideas in kinematics. Whether a particle speeds up or slows down depends on the signs of both velocity and acceleration.
| Velocity | Acceleration | Effect |
|---|---|---|
| + | + | speed increases |
| + | − | speed decreases |
| − | + | speed decreases |
| − | − | speed increases |
Deceleration means the speed is decreasing. It does NOT simply mean that acceleration is negative.
Worked Example — Negative Direction
A particle has velocity −6 m s⁻¹ and acceleration −2 m s⁻².
Both velocity and acceleration are negative, so the particle is moving in the negative direction and its speed is increasing.
LESSON 06
Direction and Sign Conventions
Almost every straight-line kinematics question becomes easier if you choose a positive direction before doing any calculations.
Write explicitly: Take the positive direction as ... Then give every velocity, displacement and acceleration a sign consistent with that choice.
Example: upward positive
For vertical motion, if upward is positive:
downward velocity = negative
upward acceleration = positive
downward acceleration = negative
Near the Earth's surface, gravitational acceleration would therefore be approximately:
Example: downward positive
If downward is chosen as positive, gravitational acceleration becomes:
Using +9.8 in one equation and −9.8 in another without changing the direction convention is a common source of incorrect answers.
LESSON 07
Motion Graphs
Graphs are one of the most powerful ways to understand kinematics. You must know what the gradient and area represent.
Displacement–time graph
The gradient of a displacement–time graph gives the velocity.
A straight line on an s–t graph represents constant velocity. A horizontal line represents zero velocity. A curved line indicates changing velocity.
Velocity–time graph
The gradient of a velocity–time graph gives acceleration.
The signed area between the velocity–time graph and the time axis gives displacement.
Acceleration–time graph
The acceleration–time graph gives acceleration directly from its vertical coordinate.
Therefore:
Graph summary
| Graph | Gradient | Area |
|---|---|---|
| s–t | velocity | not usually required |
| v–t | acceleration | displacement |
| a–t | rate of change of acceleration | change in velocity |
Worked Example — v–t Area
A particle moves with constant velocity 8 m s⁻¹ for 5 seconds. Find its displacement.
On a velocity–time graph this is a rectangle.
Area = 8 × 5
Therefore:
s = 40 m
LESSON 08
Constant Acceleration and SUVAT
When acceleration is constant, the five variables s, u, v, a, t are related by the standard SUVAT equations.
What does SUVAT mean?
| Letter | Meaning |
|---|---|
| S | displacement |
| U | initial velocity |
| V | final velocity |
| A | constant acceleration |
| T | time |
The SUVAT equations apply only when the acceleration is constant throughout the interval being considered.
Useful rearrangements
You do not need to memorise every rearrangement separately. Algebraically rearranging the standard equations is often faster and safer.
LESSON 09
Choosing the Correct SUVAT Equation
A strong Mechanics student does not simply choose the equation that looks familiar. Instead, identify the variables you know and the variable you need.
| Equation | Does not contain | Useful when |
|---|---|---|
| v = u + at | s | displacement is irrelevant |
| s = ut + ½at² | v | final velocity is not needed |
| v² = u² + 2as | t | time is unknown or unnecessary |
| s = ½(u+v)t | a | acceleration is unnecessary |
| s = vt − ½at² | u | initial velocity is unnecessary |
Worked Example — No Time Given
A particle has initial velocity 6 m s⁻¹ and accelerates uniformly at 4 m s⁻² until its velocity becomes 14 m s⁻¹. Find its displacement.
The known quantities are u, v and a. Time is not given. Therefore use:
14² = 6² + 2(4)s
196 = 36 + 8s
160 = 8s
s = 20 m
LESSON 10
Motion in Stages
Many challenging kinematics questions involve a journey split into two or more stages. The key is to treat each stage separately while transferring the correct information from one stage to the next.
The boundary principle
The final velocity of one stage is normally the initial velocity of the next stage if the motion is continuous.
Similarly, if the particle's position is tracked:
Recommended procedure
- Divide the journey into stages.
- Write the known values for each stage.
- Choose a direction convention.
- Use the appropriate equation for each stage.
- Carry the final velocity of one stage into the next.
- Check whether the acceleration changes.
- Combine displacements or times only after each stage is solved.
Worked Example — Two Stages
A particle starts from rest and accelerates at 3 m s⁻² for 4 s. It then travels at constant velocity for 6 s. Find the total displacement.
Stage 1:
u = 0, a = 3, t = 4
s = ut + ½at²
s = 0 + ½(3)(16) = 24 m
Its velocity after 4 s is:
v = 0 + 3(4) = 12 m s⁻¹
Stage 2:
Constant velocity = 12 m s⁻¹ for 6 s.
s = vt = 12 × 6 = 72 m
Total displacement:
24 + 72 = 96 m
LESSON 11
Turning Points and Change of Direction
A particle changes direction when its velocity changes sign. At the exact instant of a smooth reversal, its velocity is zero.
However, a velocity of zero does not automatically prove that the particle changes direction. You must examine what happens before and after the instant.
Worked Example — Turning Point
A particle moves with velocity:
Find when it changes direction.
At a turning point:
v = 0
Therefore:
10 − 2t = 0
2t = 10
t = 5 s
Before 5 s, v is positive. After 5 s, v is negative. Therefore the particle changes direction at t = 5 s.
Distance after a turning point
If a particle changes direction, displacement can be found using signed quantities, but total distance requires the individual parts of the journey to be added as positive lengths.
If a particle travels 20 m forward and 8 m backward, its displacement is 12 m forward, but its total distance is 28 m.
LESSON 12
Modelling Assumptions in Kinematics
Mechanics problems simplify real-world situations. Understanding the assumptions tells you what the mathematical model allows you to do.
| Statement | Meaning |
|---|---|
| particle | size and shape are ignored |
| moves in a straight line | one-dimensional motion can be used |
| constant acceleration | SUVAT equations may be used over that interval |
| constant speed | velocity is constant if direction is unchanged |
| smooth surface | friction is ignored |
| rough surface | friction may act |
| light object/string | its mass is neglected |
| inextensible string | its length does not change |
A mathematical model can be extremely useful without being perfectly realistic. The goal is to use the assumptions consistently and state or understand their consequences.
LESSON 13
The IAL Exam Method
For most kinematics problems, use the following sequence.
01 — Identify
Write down the known values of s, u, v, a and t. Do not immediately substitute numbers.
02 — Direction
Choose the positive direction and give all directed quantities consistent signs.
03 — Check acceleration
Ask whether acceleration is constant. Only then use SUVAT.
04 — Choose equation
Pick an equation containing the variables you know and the variable you need, while avoiding unnecessary unknowns.
05 — Solve
Substitute carefully, show the equation and solve algebraically.
06 — Interpret
Check the sign, units and physical meaning of your final answer.
LESSON 14
Common Kinematics Errors
1. Confusing distance and displacement
Distance is total path length. Displacement is final position minus initial position.
2. Treating speed as velocity
Velocity has direction. Speed does not.
3. Ignoring negative signs
Negative velocity or acceleration is not “wrong”. It usually indicates direction relative to the chosen positive direction.
4. Using SUVAT when acceleration is not constant
SUVAT is derived under constant acceleration.
5. Using the wrong time interval
In multi-stage questions, do not accidentally use the total journey time in an equation describing only one stage.
6. Forgetting that velocity can be zero
At a turning point, instantaneous velocity is zero even though acceleration may be non-zero.
7. Assuming negative acceleration means slowing down
Deceleration depends on the relationship between velocity and acceleration, not simply on the sign of acceleration.
8. Misreading graph areas
On a velocity–time graph, signed area gives displacement, not necessarily total distance.
If a velocity–time graph goes below the time axis, the area there represents negative displacement. For total distance, use the magnitude of each area and add the distances travelled.
FORMULA ATLAS
Complete Kinematics Formula Sheet
Gradient of s–t = velocity.
Gradient of v–t = acceleration.
Area under v–t = displacement.
Area under a–t = change in velocity.
Displacement: m
Velocity: m s⁻¹
Acceleration: m s⁻²
Time: s
MASTERCLASS 01
12 High-Value Worked Examples
Example 1 — Find acceleration
A particle changes velocity from 5 m s⁻¹ to 17 m s⁻¹ in 4 seconds.
a = (v − u)/t
a = (17 − 5)/4
a = 3 m s⁻²
Example 2 — Find displacement
A particle starts at 4 m s⁻¹ and accelerates uniformly at 3 m s⁻² for 5 seconds.
s = ut + ½at²
s = 4(5) + ½(3)(25)
s = 20 + 37.5
s = 57.5 m
Example 3 — Find final velocity
u = 12 m s⁻¹, a = −2 m s⁻², t = 4 s.
v = u + at
v = 12 − 8
v = 4 m s⁻¹
Example 4 — No time given
u = 10 m s⁻¹, v = 22 m s⁻¹, a = 4 m s⁻². Find displacement.
v² = u² + 2as
484 = 100 + 8s
384 = 8s
s = 48 m
Example 5 — Starting from rest
A particle accelerates from rest at 2.5 m s⁻² for 8 seconds.
u = 0
v = u + at
v = 2.5 × 8
v = 20 m s⁻¹
s = ½at²
s = ½(2.5)(64)
s = 80 m
Example 6 — Uniform deceleration
A car travels at 25 m s⁻¹ and comes to rest with acceleration −5 m s⁻². Find the stopping time.
v = u + at
0 = 25 − 5t
5t = 25
t = 5 s
Example 7 — Stopping distance
Using the previous situation, find the stopping distance.
v² = u² + 2as
0 = 625 − 10s
s = 62.5 m
Example 8 — Two-stage motion
A particle accelerates from rest at 4 m s⁻² for 3 seconds and then travels at constant velocity for 7 seconds.
Stage 1:
v = 4(3) = 12 m s⁻¹
s₁ = ½(4)(9) = 18 m
Stage 2:
s₂ = 12(7) = 84 m
Total displacement:
102 m
Example 9 — Vertical motion
A ball is thrown vertically upwards at 19.6 m s⁻¹. Take upward as positive and g = 9.8 m s⁻². Find the time to reach maximum height.
At maximum height:
v = 0
0 = 19.6 − 9.8t
t = 2 s
Example 10 — Maximum height
Using the previous example, find the maximum height.
v² = u² + 2as
0 = 19.6² − 19.6s
Therefore:
s = 19.6 m
Example 11 — Turning point
A particle has velocity v = 18 − 3t.
At a turning point, v = 0.
18 − 3t = 0
t = 6 s
Example 12 — v–t graph
A particle accelerates uniformly from 0 to 20 m s⁻¹ in 5 seconds. Find the displacement.
The v–t graph forms a triangle.
Area = ½ × base × height
= ½ × 5 × 20
s = 50 m
FINAL ASSESSMENT
100-Question Kinematics Examination
A comprehensive Mechanics 1 assessment covering definitions, vectors, displacement, velocity, acceleration, SUVAT, graphs, turning points, vertical motion, multi-stage motion, modelling and exam technique.