Kinematics models motion in a straight line using three time-dependent quantities: displacement, velocity, and acceleration. The everyday words have precise technical meanings here — and crucially they carry a sign that tells you direction. Get the definitions and the sign rules straight and the calculus that follows falls into place.
📘 What you need to know
Displacement s (m): position relative to a fixed point — signed; zero when back at the start.
Velocity v (m s−1): rate of change of displacement — signed; zero means at rest.
Acceleration a (m s−2): rate of change of velocity — signed.
Speed = |v|: velocity with the direction stripped off — always ≥ 0.
Distance: total length travelled — always ≥ 0; differs from displacement when the object turns back.
Sign = direction: positive is right/up, negative is left/down.
The three quantities
All three are functions of time t (in seconds), with t = 0 the start. Displacement is measured from a fixed point — usually, but not always, the particle’s initial position.
🤔 Why do we keep the sign instead of just using size?
Motion in a line has two directions. A single signed number captures both how much and which way: v = +5 and v = −5 are the same speed but opposite directions. Throwing the sign away (taking the modulus) gives speed or distance — useful, but it loses the direction the calculus needs to track turning points and returns.
🧠 “Signed twins, unsigned shadows”
Velocity and displacement are signed (direction matters). Their unsigned shadows are speed = |v| and distance. If a quantity can’t be negative, it’s a shadow.
Displacement vs distance
DisplacementsignedPosition relative to a fixed point. A bus returning to its depot has displacement 0.
Distance≥ 0Total length travelled. The same bus has travelled the full length of its route.
Quantity
Symbol
Units
Can be negative?
Displacement
s
m
Yes
Velocity
v
m s−1
Yes
Acceleration
a
m s−2
Yes
Speed
|v|
m s−1
No
Distance
d
m
No
Speeding up or slowing down?
Acceleration alone doesn’t tell you whether a particle is speeding up — you must compare its sign with the velocity’s.
🧭 Reading the motion from signs
Same signs (v and a): the particle is accelerating — speeding up.
Different signs: the particle is decelerating — slowing down.
a = 0: constant velocity.
v = 0: instantaneously at rest.
Direction of travel is always set by the sign of v, never a.
The velocity–time graph
A particle thrown up, returning after 8 s
Gradient = acceleration. Area above the axis = forward displacement; area below = backward. Here both areas are 8, so total displacement = 0 (it returns) but distance = 16 m.
Graph facts: the gradient is the acceleration; the area between the line and the time axis is the change in displacement. Add areas with sign for displacement; add their magnitudes for distance.
Worked examples
A particle is projected vertically upwards from ground level, returning after 8 seconds. Its motion is the velocity–time graph above (a straight line from (0, 4) to (8, −4)).
WE 1
How long does the particle take to reach maximum height? Give a reason.
At maximum height the particle is instantaneously at rest, so v = 0.
max height ⟺ v = 0from the graph, v = 0 at t = 44 seconds (because v = 0 there)
WE 2
Is the particle accelerating or decelerating at t = 3? Give a reason.
Compare the signs of velocity and acceleration at t = 3.
at t = 3: v > 0 (above axis)a = gradient < 0 (line slopes down)different signsdecelerating
WE 3
Use areas to find the particle’s displacement and the distance travelled over the 8 seconds.
Two triangles, each base 4 and height 4, so each area is 12(4)(4) = 8.
above axis = +8, below axis = −8displacement = 8 + (−8) = 0distance = |8| + |−8|displacement 0 m, distance 16 mdisplacement 0 confirms it returns to the ground
WE 4
A particle has velocity v = −6 m s−1. State its speed and direction of travel.
Speed is the magnitude; direction comes from the sign.
speed = |−6|speed = 6 m s−1, moving in the negative direction
WE 5
A particle’s velocity is v(t) = t2 − 5t + 6. Find the times when it is at rest.
“At rest” means v = 0 — solve the quadratic.
t² − 5t + 6 = 0(t − 2)(t − 3) = 0at rest at t = 2 s and t = 3 s
💡 Top tips
Watch the sign — it encodes direction for s, v, and a.
Speed and distance are never negative — take the modulus.
“At rest” means v = 0, not a = 0.
Compare signs of v and a to decide speeding up vs slowing down.
Sketch the velocity–time graph — gradient gives a, area gives displacement.
Decode the words: “initially” → t = 0; “due east/right” → v > 0; “dropped/down” → v < 0.
⚠ Common mistakes
Confusing displacement with distance — they differ once the object turns back.
Thinking negative acceleration always means slowing down — it depends on the velocity’s sign.
Reporting a negative speed or distance.
Using a = 0 for “at rest” instead of v = 0.
Adding signed areas for distance — distance needs magnitudes.
Next up — Calculus for Kinematics. You now have the definitions and sign rules for displacement, velocity, and acceleration, and can read them off a velocity–time graph. The next topic makes the links exact with calculus: differentiate to go s → v → a, and integrate to go back the other way — turning these ideas into equations you can solve.
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