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

This guide covers section 3.1 (Motion) of Module 3, Forces and motion, in the OCR A Level Physics A (H556) specification, Version 3.0 (March 2026). It follows the learning outcomes 3.1.1 (kinematics), 3.1.2 (linear motion) and 3.1.3 (projectile motion) in order. Use it with the revision notes and the practice questions. The course hub is OCR A Level Physics, the tick-list is the printable checklist, and the free diagnostics show where you stand.

What this unit covers

Spec reference	What you must be able to do
3.1.1(a)	Use displacement, instantaneous speed, average speed, velocity and acceleration
3.1.1(b)	Interpret graphs of displacement, speed, velocity and acceleration
3.1.1(c)	Displacement-time graphs: velocity is the gradient
3.1.1(d)	Velocity-time graphs: acceleration is the gradient, displacement is the area, including estimating areas under curves
3.1.2(a)(i)	Apply the equations of motion for constant acceleration in a straight line, including free fall without air resistance
3.1.2(a)(ii)	Describe techniques for investigating motion and collisions
3.1.2(b)(i)-(ii)	Know g and how to determine it (trapdoor and electromagnet; light gates and timer)
3.1.2(c)	Reaction time, thinking distance, braking distance and stopping distance
3.1.3(a)-(b)	Independence of horizontal and vertical motion; two-dimensional projectiles

3.1.1(a) Describing motion

Displacement is the straight-line distance from the start, in a stated direction. It is a vector. Distance is the length of path travelled. It is a scalar.

Velocity is the rate of change of displacement. Speed is the rate of change of distance. Average speed is total distance divided by total time. Instantaneous speed is the speed at one moment, found from the gradient of a distance-time graph at that point, or from a very short time interval. Acceleration is the rate of change of velocity, a = Δv/Δt, in m s⁻².

Velocity and acceleration are vectors, so direction matters. A ball thrown upwards has acceleration downwards even at the top, where its velocity is momentarily zero.

Worked example 1

A courier walks 3.0 km east in 10 minutes, then 4.0 km north in 15 minutes.

• Distance = 7000 m, time = 1500 s. Average speed = 7000 / 1500 = 4.7 m s⁻¹.
• Displacement magnitude = √(3.0² + 4.0²) = 5.0 km. Direction: tan θ = 4.0/3.0, so θ = 53° north of east.
• Average velocity = 5000 / 1500 = 3.3 m s⁻¹ at 53° north of east.

Average velocity is smaller than average speed because the path was not straight.

3.1.1(b)-(d) Motion graphs

Three rules cover almost every graph question.

• Displacement-time: gradient = velocity. A straight line means constant velocity. A curve that steepens means increasing speed. A horizontal line means at rest.
• Velocity-time: gradient = acceleration. Area between the line and the time axis = displacement. Area below the axis is displacement in the negative direction.
• Acceleration-time: the area gives the change in velocity.

On a speed-time graph the area gives distance.

For a curved graph, draw a tangent to find the instantaneous gradient. Choose a long tangent, read two points far apart from the axes, and show the triangle you used.

Estimating area under a non-linear graph

Split the time axis into equal strips and treat each strip as a trapezium. For strips of width h with readings y₀, y₁, y₂, … the area is the sum of h(yᵢ + yᵢ₊₁)/2. Alternatively, count squares on the grid, counting a square if more than half is under the line, and convert using the area of one square.

Worked example 2

A cyclist’s speed is read every 2.0 s.

t / s	0	2	4	6	8	10
v / m s⁻¹	0	3.1	5.2	6.4	7.0	7.2

Area = 2 × [(0 + 3.1)/2 + (3.1 + 5.2)/2 + (5.2 + 6.4)/2 + (6.4 + 7.0)/2 + (7.0 + 7.2)/2] = 2 × (1.55 + 4.15 + 5.8 + 6.7 + 7.1) = 2 × 25.3 = 50.6 m.

The graph is concave (the gradient falls), so each chord lies below the curve and the estimate is a slight underestimate.

3.1.2(a)(i) The equations of motion

For constant acceleration in a straight line, with initial velocity u, final velocity v, acceleration a, displacement s and time t, the specification lists:

• v = u + at
• s = ½(u + v)t
• s = ut + ½at²
• v² = u² + 2as

Each equation omits one variable. This also covers a body falling in a uniform gravitational field without air resistance, where a = g.

Method. Choose a positive direction. List s, u, v, a, t with signs and mark the unknown. Pick the equation with the unknown and three known quantities. Rearrange before substituting, then give a unit and sensible significant figures.

Worked example 3

A motorbike speeds up from 6.0 m s⁻¹ to 22 m s⁻¹ over 120 m on a straight road. Find the acceleration and the time taken.

• Use v² = u² + 2as. Then a = (v² − u²)/(2s) = (22² − 6.0²)/(2 × 120) = (484 − 36)/240 = 1.9 m s⁻² (1.87 before rounding).
• Use s = ½(u + v)t. Then t = 2s/(u + v) = 240/28 = 8.6 s.

Worked example 4: vertical motion

A stone is thrown straight up at 17 m s⁻¹ from the ground. Take up as positive and g = 9.81 m s⁻² (a = −9.81).

• At the top v = 0. Height: s = (v² − u²)/(2a) = (0 − 289)/(−19.62) = 14.7 m.
• Time to the top: t = (v − u)/a = (0 − 17)/(−9.81) = 1.73 s. Time in the air = 2 × 1.73 = 3.47 s.
• Speed 6.0 m above the ground: v² = 17² − 2 × 9.81 × 6.0 = 171.3, so v = 13 m s⁻¹. It passes this height going up (after 0.40 s) and again going down (after 3.07 s).

3.1.2(a)(ii) Investigating motion and collisions

The specification names trolleys, air-track gliders, ticker timers, light gates, data-loggers and video techniques. Know what each measures.

• Ticker timer and tape: dots at equal time intervals. Evenly spaced dots mean constant velocity. Spacing that grows steadily means constant acceleration. Velocity is the distance between dots divided by the time between them.
• Light gates and timer: a card of known width d blocks the beam for time Δt, so average speed over the card is d/Δt. A narrower card gives a better estimate of the speed at the gate. Two gates and a timer give acceleration from a = (v₂ − v₁)/t.
• Data-logger with motion sensor: gives velocity-time or displacement-time graphs directly. The gradient and area then give acceleration and displacement.
• Video analysis: frames at a known rate give position against time. A scale object in the picture fixes distances. Keep the camera square on to the motion.
• Air track: removes most friction, so acceleration is nearly constant on a slope and momentum can be compared before and after a collision.

Repeat timings and average them to reduce uncertainty.

3.1.2(b) Acceleration of free fall

g is the acceleration of a freely falling object, close to 9.81 m s⁻² near the Earth’s surface. It acts downwards on every object whatever its mass when air resistance is negligible.

Trapdoor and electromagnet

An electromagnet holds a steel ball. Switching it off starts a timer and releases the ball. The ball hits a trapdoor (a pad that opens a contact) and stops the timer. Measure the drop height h with a metre rule or tape, from the underside of the ball to the pad.

Since u = 0, h = ½gt², so plotting h on the vertical axis against t² on the horizontal axis gives a line passing through the origin, and its gradient = g/2.

Worked example 5

h / m	0.40	0.60	0.80	1.00	1.20
t / s	0.286	0.350	0.404	0.452	0.495
t² / s²	0.0818	0.1225	0.1632	0.2043	0.2450

The best-fit line through the plotted points has gradient 4.90 m s⁻². So g = 2 × 4.90 = 9.80 m s⁻².

Improve accuracy by repeating each time, using a large height range, allowing for a late release caused by residual magnetism, and measuring h with a set square or plumb line.

Light gates and timer

Drop a card of width 0.050 m through two gates a known time apart (or one gate a measured height below the release point). Then v₁ = 0.050/0.0250 = 2.00 m s⁻¹ and v₂ = 0.050/0.0100 = 5.00 m s⁻¹, with 0.306 s between the gates. Then g = (v₂ − v₁)/t = 3.00/0.306 = 9.80 m s⁻².

With one gate, v² = 2gh, so a graph of v² against h has gradient 2g.

3.1.2(c) Reaction time and stopping distance

• Thinking distance = speed × reaction time, travelled before the brakes are applied.
• Braking distance is travelled while the vehicle slows. For constant deceleration a, braking distance = v²/(2a).
• Stopping distance = thinking distance + braking distance.

Thinking distance is proportional to speed. Braking distance is proportional to speed squared. Reaction time rises with tiredness, alcohol, drugs and distraction. Braking distance rises with higher speed, wet or icy roads, worn tyres or brakes, and a heavier load.

Worked example 6

A car travels at 20 m s⁻¹. The driver’s reaction time is 0.65 s and the deceleration is a steady 7.0 m s⁻².

• Thinking distance = 20 × 0.65 = 13 m.
• Braking distance = 20²/(2 × 7.0) = 28.6 m.
• Stopping distance = 42 m (41.6 m before rounding).

At 30 m s⁻¹ the thinking distance is 19.5 m and the braking distance is 64.3 m, a total of 84 m. A 50% rise in speed has raised the stopping distance by about 100%.

3.1.3 Projectile motion

A projectile moves under gravity alone. The horizontal and vertical motions are independent: horizontal velocity stays constant, vertical acceleration is g downwards, and they share only the time.

A ball rolled off a table and a ball dropped from the table edge land together.

Method. Resolve the launch velocity: vₓ = v cos θ and vᵧ = v sin θ. Use x = vₓt horizontally. Use the equations of motion with a = −g vertically. The same t links them.

Worked example 7: horizontal launch

A ball leaves the edge of a 45 m cliff horizontally at 14 m s⁻¹. Take g = 9.81 m s⁻².

• Vertical: 45 = ½ × 9.81 × t², so t = √(90/9.81) = 3.03 s.
• Horizontal distance = 14 × 3.03 = 42 m.
• Landing: vᵧ = 9.81 × 3.03 = 29.7 m s⁻¹. Speed = √(14² + 29.7²) = 33 m s⁻¹, at tan⁻¹(29.7/14) = 65° below the horizontal.

Worked example 8: launch at an angle

A ball is struck from level ground at 28 m s⁻¹, 35° above the horizontal.

• vₓ = 28 cos 35° = 22.9 m s⁻¹. vᵧ = 28 sin 35° = 16.1 m s⁻¹.
• Time of flight: T = 2vᵧ/g = 3.27 s.
• Range = vₓT = 75 m.
• Maximum height = vᵧ²/(2g) = 13.1 m, reached after 1.64 s, when the ball still has vₓ = 22.9 m s⁻¹.

For a launch from height, put s = −(height) in s = vᵧt − ½gt² and solve the quadratic, taking the positive root.

Common errors

• Using suvat when acceleration is not constant.
• Forgetting that the area under a v-t graph below the axis is negative displacement.
• Adding horizontal and vertical velocities as plain numbers instead of using Pythagoras.
• Stating thinking distance “increases with speed squared”. It is proportional to speed.

Work through the practice questions, then use the revision notes for the final weeks.

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.