OCR A Level Physics A: Kinematics, equations of motion and projectiles (H556) – Revision Notes

These notes condense section 3.1 (Motion, outcomes 3.1.1 to 3.1.3) of OCR A Level Physics A (H556), Version 3.0 (March 2026). Full explanations and worked examples are in the study guide; test yourself with the practice questions. Hub: OCR A Level Physics. Checklist: printable checklist. Not sure where to start? Take a free diagnostic.

3.1.1 Definitions

Quantity	Definition	Type	Unit
Distance	Length of path travelled	Scalar	m
Displacement	Straight-line distance from start in a stated direction	Vector	m
Average speed	Total distance / total time	Scalar	m s⁻¹
Instantaneous speed	Speed at one instant	Scalar	m s⁻¹
Velocity	Rate of change of displacement	Vector	m s⁻¹
Acceleration	Rate of change of velocity, a = Δv/Δt	Vector	m s⁻²

Acceleration is in the direction of the change in velocity. A decelerating body has acceleration opposite to its velocity.

3.1.1(b)-(d) Graph rules

Graph	Gradient	Area under line
Displacement-time	Velocity	None used
Velocity-time	Acceleration	Displacement
Acceleration-time	None used	Change in velocity

• Straight line on s-t: constant velocity. Curve getting steeper: accelerating.
• Straight line on v-t: constant acceleration. Horizontal line: constant velocity. Line crossing the axis: the direction reverses.
• Area below the time axis counts as negative displacement.
• Curved graph, instantaneous gradient: draw a tangent and use a large triangle.
• Curved graph, area: split into equal strips and use trapezia (area of strip = width × mean of the two heights), or count squares.

Method in steps: estimating an area

1. Read the heights at equal time steps.
2. Find the mean of each pair of heights.
3. Multiply by the strip width and add.
4. Comment on whether the estimate is high or low: a concave-down curve gives an underestimate, a concave-up curve an overestimate.

Worked reminder: speeds 0, 4, 6, 7 m s⁻¹ at 0, 3, 6, 9 s give 3 × (2 + 5 + 6.5) = 40.5 m.

3.1.2(a) Equations of motion

Constant acceleration, straight line:

Equation	Missing variable
v = u + at	s
s = ½(u + v)t	a
s = ut + ½at²	v
v² = u² + 2as	t

• Choose a positive direction first. Signs follow from it: upward positive makes a = −g for a projectile.
• At the highest point of vertical motion, v = 0.
• Falling from rest: u = 0 and s = ½gt².
• Use these equations only when a is constant. They apply to free fall without air resistance with a = g.

Method in steps

1. Sketch and mark the positive direction.
2. List the five variables with signs.
3. Choose the equation with the unknown and three knowns.
4. Rearrange first, substitute second.

Worked reminder: a train slows from 30 m s⁻¹ to 12 m s⁻¹ in 360 m. a = (12² − 30²)/(2 × 360) = −756/720 = −1.05 m s⁻².

Investigating motion and collisions (3.1.2(a)(ii))

Apparatus	What it gives
Ticker timer	Dots at fixed time intervals; spacing shows velocity
Light gates and timer	Average speed over a card, d/Δt; two gates give acceleration
Data-logger and motion sensor	Direct s-t or v-t graphs
Video	Position against time from frame rate, with a scale in view
Air track	Almost frictionless motion for constant acceleration and collisions

3.1.2(b) Free fall and measuring g

• g is the acceleration of free fall, about 9.81 m s⁻² near the Earth’s surface, downwards.
• Trapdoor and electromagnet: drop height h, time t, u = 0. h = ½gt², so plot h against t². Gradient = g/2, so g = 2 × gradient.
• Light gates: v = d/Δt for card width d. Two gates: g = (v₂ − v₁)/t. One gate after a drop of h: v² = 2gh, so a graph of v² against h has gradient 2g.
• Improve results by repeating and averaging, using a wide range of heights, and measuring h with a set square or plumb line.
• Possible systematic errors: late release from residual magnetism, and a card that is not vertical.

Worked reminder: gradient of h against t² is 4.95 m s⁻², so g = 2 × 4.95 = 9.90 m s⁻².

3.1.2(c) Stopping distance

Term	Meaning	Depends on
Thinking distance	Speed × reaction time	Speed (proportional), reaction time
Braking distance	v²/(2a) for constant deceleration	Speed squared, road, tyres, brakes, load
Stopping distance	Thinking + braking	All of the above

Reaction time is lengthened by tiredness, alcohol, drugs and distraction.

Worked reminder: 15 m s⁻¹, reaction time 0.80 s, deceleration 5.0 m s⁻². Thinking = 12 m, braking = 225/10 = 22.5 m, stopping = 34.5 m.

3.1.3 Projectiles

• Horizontal velocity is constant: x = vₓt.
• Vertical motion has a = g downwards, using the equations of motion.
• Only t is shared between the two directions.
• Launch at speed v and angle θ: vₓ = v cos θ, vᵧ = v sin θ.
• On level ground: time of flight T = 2vᵧ/g, range = vₓT, maximum height = vᵧ²/(2g).
• At the top, vᵧ = 0 but vₓ is unchanged, so the speed is vₓ.
• Speed on landing = √(vₓ² + vᵧ²), at tan⁻¹(vᵧ/vₓ) to the horizontal.
• Launch from a height: solve s = vᵧt − ½gt² with s equal to minus the height. Take the positive root.

Worked reminder: ball thrown horizontally at 6.0 m s⁻¹ from 2.0 m. t = √(2 × 2.0/9.81) = 0.64 s, so x = 3.8 m.

Worked reminder (angled launch): 20 m s⁻¹ at 30° from level ground. vₓ = 17.3 m s⁻¹, vᵧ = 10.0 m s⁻¹. T = 2 × 10.0/9.81 = 2.04 s, range = 17.3 × 2.04 = 35 m, maximum height = 100/19.62 = 5.1 m.

Worked reminder (light gates): a 0.040 m card takes 0.020 s to pass gate 1 and 0.008 s to pass gate 2, with 0.30 s between the gates. v₁ = 2.0 m s⁻¹, v₂ = 5.0 m s⁻¹, so a = 3.0/0.30 = 10 m s⁻².

Explaining in words

• Independence: the horizontal velocity is constant because no horizontal force acts, while gravity changes only the vertical velocity. Two balls released from the same height, one dropped and one launched sideways, land together.
• Why braking distance grows faster than thinking distance: kinetic energy rises with v², so the work done by the brakes, and the distance needed, rises with v².
• Why a longer fall improves a g measurement: the timing uncertainty is a smaller fraction of a longer time.

Must-know distinctions

• Speed is a scalar; velocity is a vector. Average speed and average velocity differ when direction changes.
• Gradient of s-t is velocity; gradient of v-t is acceleration.
• Thinking distance is proportional to speed; braking distance to speed squared.
• Horizontal: constant velocity. Vertical: constant acceleration.

Quick self-test

1. A cyclist rides once round a 300 m circular track in 40 s. State the average speed and the average velocity.
2. A car accelerates from rest at 2.5 m s⁻² for 8.0 s. Find the distance travelled.
3. A ball is dropped from rest from 20 m. Find its speed on landing (g = 9.81 m s⁻²).
4. State what the area under a velocity-time graph represents.
5. A body’s velocity rises steadily from 5.0 m s⁻¹ at 0 s to 20 m s⁻¹ at 5.0 s. Find the acceleration and the displacement.
6. In a trapdoor experiment h = 0.50 m and t = 0.32 s. Calculate g.
7. State why a narrower card improves a light-gate speed measurement.
8. A car at 24 m s⁻¹ has a reaction time of 0.50 s. Find the thinking distance.
9. A stone leaves a horizontal surface at 3.0 m s⁻¹. How long does it take to fall 0.80 m?
10. State two factors that increase braking distance but not thinking distance.

Answers

1. Average speed = 300/40 = 7.5 m s⁻¹; displacement is zero, so average velocity = 0.
2. s = ½at² = 0.5 × 2.5 × 64 = 80 m.
3. v² = 2gs = 2 × 9.81 × 20 = 392.4, so v = 19.8 m s⁻¹.
4. Displacement (distance on a speed-time graph).
5. a = 15/5.0 = 3.0 m s⁻²; s = ½ × (5.0 + 20) × 5.0 = 62.5 m.
6. g = 2h/t² = 1.0/0.1024 = 9.8 m s⁻² (9.77 before rounding).
7. The card’s average speed is then closer to the speed at the gate.
8. 24 × 0.50 = 12 m.
9. t = √(2 × 0.80/9.81) = 0.40 s (0.404 s); the horizontal speed does not matter.
10. Any two of: higher mass or load, worn tyres, worn brakes, wet or icy road.

Where marks are usually lost

• Writing a distance when a displacement with direction is asked for.
• Not stating a sign convention, then mixing up the signs of g and u.
• Reading the gradient from a short chord instead of a long tangent.
• Counting the area below the axis as positive.
• Using v = u + at for the whole of a problem in which the acceleration changes.
• Plotting h against t when the straight-line graph is h against t².
• Giving thinking distance as proportional to speed squared.
• Using the vertical speed alone as the landing speed.
• Missing units, or giving answers to too many significant figures.

Official syllabus

OCR, A Level GCE Physics A H556 Specification, Version 3.0 (March 2026), first assessment 2017, section 3.1 Motion (3.1.1 to 3.1.3) in Module 3: Forces and motion.