Describing motion
There and back
- A runner does one lap of a $400\ \text{m}$ track and stops where they started.
- Distance travelled: $400\ \text{m}$. Displacement: $0$.
- Describing motion well starts with getting words like these exactly right.
Five key words
- distance (scalar) — total path length; displacement (vector) — straight line, start to end.
- speed (scalar) — rate of change of distance; velocity (vector) — rate of change of displacement.
- acceleration (vector) — rate of change of velocity. Units: $\dfrac{\text{m}}{\text{s}}$ and $\dfrac{\text{m}}{\text{s}^2}$.
The rate of change of velocity with time is called ____.
Acceleration is how fast the velocity changes — a vector, in $\dfrac{\text{m}}{\text{s}^2}$.
Which quantity is the rate of change of displacement?
Velocity is the rate of change of displacement (a vector). Speed is the rate of change of distance (a scalar).
Deceleration is not a new quantity — it is just acceleration pointing opposite to the velocity (a negative acceleration).
Deceleration is a completely separate quantity from acceleration.
No — deceleration is just acceleration in the opposite direction to the motion (a negative acceleration).
Two graphs do the work
- Most motion questions use a displacement–time or a velocity–time graph.
- The whole skill is knowing what the gradient and the area mean on each.
Displacement–time graph
- The gradient (steepness) = the velocity.
- Flat → at rest. Straight → constant velocity. Curved → take a tangent.
On a displacement–time graph, a flat (horizontal) line means the object is at rest.
A flat line has zero gradient, and the gradient is the velocity — so the velocity is zero (at rest).
Velocity–time graph
- The gradient = the acceleration.
- The area under the line = the displacement.
Match each graph feature to what it tells you.
Gradients give rates of change; the area under a velocity–time graph builds up the displacement.
Displacement from the area
- Split the area into triangles and rectangles, then add them up.
- Triangle $= \tfrac{1}{2} \times \text{base} \times \text{height}$; rectangle $= \text{base} \times \text{height}$.
A car speeds up from rest to $20\ \dfrac{\text{m}}{\text{s}}$ in $10\ \text{s}$ (a straight velocity–time line). How far does it travel?
Displacement = area under the line = $\tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 10 \times 20 = 100\ \text{m}$.
You've got it
- distance & speed are scalars; displacement, velocity & acceleration are vectors
- displacement–time gradient = velocity; velocity–time gradient = acceleration
- area under a velocity–time graph = the displacement