Before any calculus happens, you need to be fluent in five words: displacement, distance, velocity, speed, and acceleration. In everyday English these words get used loosely — "speed" and "velocity" basically mean the same thing to most people. In maths, they don't. Getting this distinction wrong is the single most common way students lose marks in this topic, so let's build it properly.
aDisplacement vs. Distance — the "can it go negative?" test
Picture a number line. You're standing at 0, and you can only move left or right along it (that's the "straight line" restriction — kinematics never deals with curved paths in this course).
Displacement is your position relative to that starting point, right now, in a straight line. If you're 5 m to the right, your displacement is +5. If you're 5 m to the left, it's −5. Crucially, displacement doesn't care about the wiggly path you took to get there — only where you ended up relative to the fixed point.
Distance is the total length of the path you actually walked, and it only ever adds up — it can never be negative and it can never "cancel out." Walk 3 m right then 3 m left, and your displacement is back to 0, but you've covered a distance of 6 m.
Analogy
Think of a bus route that starts and ends at the same depot. When the bus gets back, its
displacement is zero — it's exactly where it started. But the odometer doesn't reset — the
distance travelled is the full length of the route. Same journey, two totally different numbers, because they're answering two different questions ("where are you?" vs. "how far did you go?").
Practice Question 1
A particle starts at the origin, moves 8 m to the right, then moves 3 m back to the left. State (a) its final displacement, and (b) the total distance it has travelled.
bVelocity vs. Speed — the same relationship, one level up
Exactly the same logic applies one level up the calculus chain. Velocity is the rate of change of displacement — it tells you not just how fast something is moving, but which way. A velocity of −6 m/s means the object is moving at 6 metres per second, but in the negative direction. Speed is the magnitude of velocity — written |v| — and it strips the direction away, leaving only "how fast," which is always positive (or zero).
So if v = 4, speed is |v| = 4. If v = −6, speed is |v| = 6. Same idea as distance vs. displacement, just one derivative up.
A velocity of exactly zero has a special name worth knowing: the particle is said to be at rest or stationary.
cAcceleration — the trickiest one, because sign alone isn't enough
Acceleration is the rate of change of velocity. Here's the part students often get backwards: negative acceleration does not automatically mean "slowing down." You have to compare the sign of acceleration to the sign of velocity together:
| Velocity sign | Acceleration sign | What's happening |
|---|
| Same sign as each other | e.g. v > 0, a > 0 | Accelerating (speeding up) |
| Opposite signs | e.g. v > 0, a < 0 | Decelerating (slowing down) |
| Acceleration is zero | a = 0 | Moving at constant velocity |
Why does this trip people up? Because if an object is moving in the negative direction (v < 0) and has negative acceleration (a < 0), it is actually speeding up — it's getting "more negative," i.e. going faster in the negative direction. Both signs match, so it's accelerating, even though "negative" sounds like it should mean slowing down. Always check both signs, never judge acceleration on its own.
Useful phrases to decode instantly
- "At rest" → v = 0
- Moving "due east" / "to the right" → positive direction, so v > 0
- "Dropped from a cliff" / moving "down" → negative vertical direction, so v < 0
Practice Question 2
At a certain instant, a particle has velocity v = −3 m/s and acceleration a = −2 m/s². Is the particle speeding up or slowing down? Explain.
dReading a velocity-time graph like a pro
A velocity-time graph packs in three pieces of information at once, and knowing where to look for each one is a skill in itself:
The area interpretation has a subtlety worth sitting with. The area between the graph and the x-axis, over some interval, gives the change in displacement for that interval — but you have to treat areas below the axis as negative contributions if you want net displacement (they represent backward movement, cancelling out forward movement). If instead you want total distance travelled, you add up all areas as positive, regardless of whether they're above or below the axis — because distance never "cancels."
Area above the axis = forward displacement. Area below = backward displacement. Net displacement subtracts the two; total distance adds them.
Practice Question 3
A velocity-time graph shows a particle with an area of 12 m above the x-axis between t = 0 and t = 5, and an area of 4 m below the x-axis between t = 5 and t = 8. Find (a) the displacement and (b) the distance travelled over the full 8 seconds.