Motion (P2.1)

Motion questions appear in every OCR Gateway A Physics paper. Examiners want you to interpret graphs accurately, perform calculations, and use the correct equations with the right units.

Key definitions

Quantity	Definition	Type	Unit
Distance	Total path length travelled	Scalar	m
Displacement	Straight-line distance in a given direction from start to finish	Vector	m
Speed	Rate of change of distance	Scalar	m/s
Velocity	Rate of change of displacement (speed in a given direction)	Vector	m/s
Acceleration	Rate of change of velocity	Vector	m/s²

Scalars have magnitude only. Vectors have magnitude AND direction.

Speed and velocity

speed (or velocity) = distance (or displacement) ÷ time
v = s / t

Typical speeds to know:

• Walking: ~1.5 m/s
• Running: ~3 m/s
• Cycling: ~6 m/s
• Car on motorway: ~30 m/s
• Speed of sound in air: ~340 m/s
• Speed of light: ~3 × 10⁸ m/s

Acceleration

acceleration = change in velocity ÷ time taken
a = (v − u) / t

Where u = initial velocity, v = final velocity, t = time.

Units: m/s² (metres per second per second).

A negative acceleration (deceleration) means slowing down.

Distance–time (s–t) graphs

Section	Meaning
Straight upward slope	Constant speed
Steeper slope	Greater speed
Horizontal (flat) line	Stationary (speed = 0)
Curved, getting steeper	Accelerating
Curved, getting less steep	Decelerating

The gradient of a distance–time graph = speed.

• Gradient = Δs / Δt

Velocity–time (v–t) graphs

Section	Meaning
Straight upward slope	Constant acceleration
Horizontal (flat) line	Constant velocity (zero acceleration)
Straight downward slope	Constant deceleration
Curve	Non-uniform acceleration or deceleration

The gradient of a velocity–time graph = acceleration.

• Gradient = Δv / Δt

The area under a velocity–time graph = distance travelled.

• Area = ½ × base × height (triangle) or length × height (rectangle).

Equations of motion (suvat)

For uniform (constant) acceleration:

v = u + at                     ... (1)
s = ½(u + v)t                  ... (2)
s = ut + ½at²                  ... (3)
v² = u² + 2as                  ... (4)

Where: s = displacement (m), u = initial velocity (m/s), v = final velocity (m/s), a = acceleration (m/s²), t = time (s).

At GCSE you need (1) and (3). Higher Tier may need (4).

Worked example: A car accelerates from rest (u = 0) at 3 m/s² for 8 s. Find the distance travelled.

• s = ut + ½at² = 0 × 8 + ½ × 3 × 8² = 0 + ½ × 3 × 64 = 96 m

Terminal velocity

An object falling through a fluid experiences increasing drag as speed increases.

1. Initially, weight > drag → object accelerates (net downward force).
2. As speed increases, drag increases.
3. Eventually, drag = weight → net force = 0 → constant velocity = terminal velocity.

On a velocity–time graph: starts with steep upward gradient (accelerating), gradient decreases, curve levels off at terminal velocity (horizontal).

Common Gateway-paper mistakes

1. Mixing up distance–time and velocity–time graphs (gradient of d–t = speed; area under v–t = distance).
2. Forgetting that constant speed ≠ constant velocity if direction changes (e.g. circular motion).
3. Using total distance instead of displacement for velocity calculations.
4. Confusing u (initial) and v (final) in the equations.
5. Not converting units — km/h to m/s (divide by 3.6) or cm to m (divide by 100).