Kinematics: Complete O Level Physics Cheatsheet
O Level Physics, Chapter 2 · Read time: ~9 minutes
Key Formulas
Distance vs displacement, speed vs velocity
Distance is the total length of path travelled, a scalar (no direction). Displacement is the straight-line distance from the starting point to the finishing point, in a stated direction, a vector.
Speed is distance ÷ time, a scalar. Velocity is displacement ÷ time, a vector, it includes direction.
If an object travels away from and then back to its starting point, its total distance travelled is not zero, but its displacement is zero, since it ends up back where it started.
Trap: A common exam trap: 'the car travelled 500 m north then 200 m south, find its average velocity' is NOT solved using the total distance. Distance travelled = 700 m, but displacement = 500 − 200 = 300 m north, use displacement (not distance) whenever the question asks for velocity rather than speed.
Acceleration and deceleration
Acceleration is the rate of change of velocity: a = (v − u) / t.
If an object slows down, its acceleration is negative (in the direction of motion), this is sometimes called deceleration or retardation.
Uniform acceleration means the velocity changes by the same amount every second, this is the only case where the equations of motion (v = u + at, etc.) apply directly.
Trick: Always define a positive direction first (e.g. 'taking forward/upward as positive'), then keep every velocity and acceleration consistent with that sign convention throughout the whole question. This is what prevents sign errors in multi-step problems.
Distance-time and velocity-time graphs
Distance (or displacement)-time graph: the gradient at any point gives the speed (or velocity) at that instant. A horizontal line means the object is stationary. A straight sloped line means constant speed. A curve means the speed is changing.
Velocity-time graph: the gradient gives acceleration. The area under the graph (between the line and the time-axis) gives the distance (or displacement) travelled.
For a velocity-time graph with a non-uniform shape (e.g. a trapezium), split the area into simple shapes (rectangles, triangles) and add them up separately, rather than trying to use one formula for the whole thing.
Trap: On a velocity-time graph, if the line dips below the time-axis (negative velocity), the 'area' there still represents distance travelled, but in the opposite direction. Don't just take the magnitude and add it to the rest without thinking about direction, if the question asks for displacement rather than total distance.
Worked Example
A car starts from rest and accelerates uniformly at 2.0 m/s² for 5.0 s. It then travels at a constant velocity for a further 10 s, before decelerating uniformly to rest in 4.0 s.
(a) Calculate the velocity of the car at the end of the first 5.0 s. [2]
v = u + at
= 0 + (2.0 × 5.0)
= 10 m/s
(b) Calculate the total distance travelled by the car during the entire journey. [4]
Stage 1 (accelerating): s1 = ((u + v)/2) t = ((0 + 10)/2) × 5.0 = 25 m
Stage 2 (constant velocity): s2 = v × t = 10 × 10 = 100 m
Stage 3 (decelerating to rest): s3 = ((u + v)/2) t = ((10 + 0)/2) × 4.0 = 20 m
Total distance = 25 + 100 + 20 = 145 m
(c) Calculate the average speed for the whole 19 s journey. [2]
Average speed = total distance ÷ total time
= 145 ÷ 19
= 7.6 m/s (2 s.f.)
Why this question is a good test of the topic: it forces you to treat each stage of motion separately (different equations apply to constant-acceleration stages vs the constant-velocity stage), then combine the results, rather than trying to apply one single equation across the whole journey.
Frequently Asked Questions
What's the difference between distance and displacement?
Distance is the total path length travelled (scalar). Displacement is the straight-line distance from start to end point, in a stated direction (vector). They're only equal if the motion is in a straight line with no reversal.
When can I use the equations of motion (v = u + at, etc.)?
Only when acceleration is constant (uniform). If acceleration changes, you need to break the motion into stages of constant acceleration, or use a velocity-time graph instead.
What does the area under a velocity-time graph represent?
The distance (or displacement) travelled during that time interval.
What does the gradient of a velocity-time graph represent?
Acceleration. A steeper gradient means a bigger acceleration.
Is this topic tested in Paper 1 (MCQ) or Paper 2 (structured)?
Both. MCQs often test graph interpretation, Paper 2 usually has a multi-stage motion calculation similar to the worked example above.
Struggling with multi-stage motion questions?
Small group O Level Physics classes at TGC Academy's Bishan, Bukit Timah and Potong Pasir centres, taught by Andrew Seah, MOE Award-Winning Teacher and Marshall Cavendish textbook author.