Kinematics and Motion Graphs: Revision Notes

Condensed for the final weeks. For the full explanation, use the Kinematics and Motion Graphs study guide.

Definitions — scalar or vector

Quantity	Type	Definition
Distance	Scalar	Total path length travelled
Displacement	Vector	Straight-line distance in a stated direction
Speed	Scalar	Distance per unit time
Velocity	Vector	Displacement per unit time (speed in a given direction)
Acceleration	Vector	Rate of change of velocity

Acceleration occurs whenever speed changes, direction changes, or both — a car turning a corner at a constant speedometer reading is still accelerating, because its direction is changing.

Worked example. A runner completes one full lap of a 400 m track. Distance travelled = 400 m (the total path length). Displacement = 0 m (they end up back at the start, so the straight-line distance from start to finish is zero) — a clean illustration of why distance and displacement are not the same quantity, even for real motion.

Equations

speed         v = d / t
acceleration  a = (v - u) / t

u = initial velocity   v = final velocity
a in m/s2              deceleration is NEGATIVE acceleration

Worked example. A cyclist accelerates from 2 m/s to 8 m/s in 4 s.

a = (8 - 2) / 4 = 1.5 m/s^2

Worked example. A trolley decelerates from 12 m/s to 4 m/s in 2 s.

a = (4 - 12) / 2 = -4 m/s^2

The negative sign itself shows this is a deceleration — there’s no separate “deceleration formula” to remember.

Reading the two graphs — learn this table

	Distance–time	Speed–time
Gradient gives	Speed	Acceleration
Horizontal line	At rest	Constant speed
Straight slope	Constant speed	Constant acceleration
Curve upward	Accelerating	Increasing acceleration
Area under graph	No meaning	Distance travelled

The single most useful fact: area under a speed–time graph = distance. Split awkward shapes into triangles and rectangles.

Worked example. An object accelerates uniformly from rest to 10 m/s in 5 s, then travels at 10 m/s for a further 8 s. Find the total distance travelled.

Stage 1 (triangle):   1/2 x 5 x 10  = 25 m
Stage 2 (rectangle):  8 x 10        = 80 m
                                      -----
Total distance                       = 105 m

Free fall

• Acceleration of free fall g = 9.8 m/s².
• Without air resistance, all objects fall with the same acceleration regardless of mass — a feather and a hammer dropped in a vacuum land together.
• With air resistance: drag rises with speed until drag = weight, resultant force = 0, and the object falls at constant terminal velocity.
• On a speed–time graph, terminal velocity is where the curve flattens — the object does not slow down.
• A skydiver’s speed–time graph before and after the parachute opens shows: a rising curve that flattens as the object approaches its (higher) terminal velocity in free fall, then a sharp drop in speed the instant the parachute opens (drag suddenly increases), followed by the curve flattening again at a much lower terminal velocity for the rest of the descent.

Exam traps

• Deceleration is negative acceleration, not a separate quantity.
• Gradient of a distance–time graph is speed, never acceleration.
• Area under a distance–time graph means nothing — don’t calculate it.
• An object moving at constant speed in a circle is accelerating, because direction changes.
• Use u for initial and v for final consistently; swapping them flips the sign.
• Check units: convert km/h to m/s by dividing by 3.6.
• Confusing distance with displacement, or speed with velocity — the first of each pair is a scalar, the second a vector.
• Mixing up which axis a gradient is taken from before deciding what it represents.

Self-test

1. A car goes from 5 m/s to 25 m/s in 8 s. Find the acceleration.
2. What does the area under a speed–time graph represent?
3. Sketch the speed–time graph of a skydiver before and after the parachute opens.
4. Why is an object in circular motion at constant speed accelerating?
5. A graph of distance against time is a horizontal line. What is happening?
6. A runner completes one lap of a 400 m track. State the distance travelled and the displacement.
7. An object accelerates uniformly from rest to 10 m/s in 5 s, then travels at 10 m/s for 8 s. Find the total distance.

Answers: 1. a = (25 − 5)/8 = 2.5 m/s². 2. Distance travelled. 3. Rising curve flattening to terminal velocity, sharp drop when the parachute opens, then flattening to a lower terminal velocity. 4. Velocity is a vector; direction is changing continuously, so velocity changes and there is acceleration. 5. The object is stationary — distance is not changing. 6. Distance = 400 m; displacement = 0 m, since the runner returns to the starting point. 7. 25 m (triangle) + 80 m (rectangle) = 105 m.

For the full worked explanations behind every example above, see the Kinematics and Motion Graphs study guide; for exam-style questions with full mark schemes, see the Kinematics practice questions.