Linear motion, motion graphs and SUVAT: worked solutions

Answer: Distance = 165 m; displacement = +75 m (east); average speed = 2.2 m/s; average velocity = +1.0 m/s (east).

1. Distance: 120 + 45 = 165 m. Distance is a scalar and does not subtract travel in the opposite direction.
2. Displacement: +120 − 45 = +75 m. The positive sign means east of the starting point.
3. Average speed = total distance / total time = 165 / 75 = 2.2 m s⁻¹.Average velocity = displacement / total time = 75 / 75 = +1.0 m s⁻¹.
4. Reversing direction adds to distance but reduces the net displacement. The total distance is therefore greater than the magnitude of displacement, while both averages use the same time.

Check: The final position is still east of the start. Neither average tells you the speed during the pause or at a particular instant.

Mark checklist · 5 marks

• 1 mark — Total distance 165 m.
• 1 mark — Displacement +75 m or 75 m east.
• 1 mark — Average speed 2.2 m/s using the whole 75 s.
• 1 mark — Average velocity +1.0 m/s or 1.0 m/s east.
• 1 mark — Explains that reversal makes distance greater than displacement magnitude, with the same total time in both averages.

Common mistake: Using 120 − 45 for distance, or finding the average of two unknown walking speeds. Whole-journey averages use total distance or displacement divided by total elapsed time.

Exam technique: PhysicsUK guidance: give a direction or a signed convention for velocity. A numerical magnitude alone does not distinguish it from speed.

Answer: Acceleration = +2.0 m/s²; displacement = +25 m (right); distance = 35 m.

1. a = Δv/Δt = (6 − 0)/(3 − 0) = +2.0 m s⁻².The units are velocity divided by time.
2. Between 5 and 9 s, a = (−2 − 6)/(9 − 5) = −2.0 m s⁻². The time to fall from +6 to 0 m s⁻¹ is 6/2 = 3 s, so the reversal is at t = 8 s.
3. Positive areas: 0–3 s triangle = ½ × 3 × 6 = 9 m; 3–5 s rectangle = 2 × 6 = 12 m; 5–8 s triangle = ½ × 3 × 6 = 9 m. Total positive area = 30 m.
4. Negative areas: 8–9 s triangle = −½ × 1 × 2 = −1 m; 9–11 s rectangle = −2 × 2 = −4 m. Total negative area = −5 m.
5. Displacement: 30 − 5 = +25 m. Distance: 30 + 5 = 35 m. You may use a signed trapezium for displacement, but you must split at the zero crossing to obtain distance.

Check: 35 m ≥ |25 m|, as required. The carriage reverses when velocity crosses zero at 8 s, rather than when acceleration first becomes negative at 5 s.

Mark checklist · 6 marks

• 1 mark — Shows a correct gradient calculation (6 − 0)/(3 − 0).
• 1 mark — Acceleration +2.0 m/s² with units.
• 1 mark — Uses signed areas for displacement, including the sections below v = 0.
• 1 mark — Displacement +25 m or 25 m to the right.
• 1 mark — For distance, separates the positive and negative portions at 8 s and adds their magnitudes.
• 1 mark — Total distance 35 m.

Common mistake: Treating every area as positive for displacement, or calling the whole descending line ‘slowing down’. From 8 to 9 s both velocity and acceleration are negative, so the carriage speeds up to the left.

Exam technique: The 2023 report supports showing the coordinates used for a gradient. PhysicsUK extension: annotate positive and negative areas to make your distance/displacement distinction visible.

Related examiner evidence: OCR H156/02, June 2023 — Q1(a)(i), Q1(a)(iii)–(b); report pp. 6–8

Answer: Average velocity = +4.0 m/s; instantaneous velocity at 2.0 s = +6.0 m/s.

1. Average velocity = Δs/Δt = (8.0 − 0)/(2.0 − 0) = +4.0 m s⁻¹.This is the gradient of the chord joining the two positions on the curve.
2. Instantaneous velocity at 2.0 s = tangent gradient = (20.0 − 2.0)/(4.0 − 1.0) = +6.0 m s⁻¹.The tangent touches the motion curve at 2.0 s even though A and B have different times.
3. The increasing gradient shows that the cart’s velocity increases. Its velocity at the end of the first 2 s exceeds the average over those 2 s.

Check: The tangent is steeper than the 0–2 s chord. Dividing the single displacement 8 m by 2 s gives the average, so it cannot also supply the instantaneous velocity here.

Mark checklist · 4 marks

• 1 mark — Uses change in displacement / elapsed time between the curve points at 0 and 2 s.
• 1 mark — Average velocity +4.0 m/s.
• 1 mark — Uses the tangent points A and B: (20.0 − 2.0)/(4.0 − 1.0).
• 1 mark — Instantaneous velocity +6.0 m/s.

Common mistake: Using two points on the curve for instantaneous velocity, or assuming the tangent’s marked points must themselves lie on the curve. They are construction points on a straight line.

Exam technique: PhysicsUK guidance: draw a long tangent and use well-separated points on that tangent. Label both coordinates and subtract them; s/t is only valid for a line through the origin.

Answer: Trapezium estimate = 20.75 m ≈ 21 m. The whole-interval SUVAT method is invalid because acceleration changes.

1. s ≈ ½(0 + 4.0) × 1.0 + ½(4.0 + 6.0) × 1.0 + ½(6.0 + 7.0) × 1.0 + ½(7.0 + 7.5) × 1.0.
2. The strip areas are 2.00, 5.00, 6.50 and 7.25 m. Their sum is 20.75 m, or about 21 m to two significant figures. It is an estimate because each strip replaces a curve with a straight line.
3. The velocity gains are 4.0, 2.0, 1.0 and 0.5 m s⁻¹ over equal 1 s intervals. The gradient, and therefore the acceleration, cannot be constant. The expression s = ½(u + v)t assumes a straight velocity–time line across the whole interval. Using just the endpoints would give ½(0 + 7.5) × 4 = 15 m and would miss much of the area.

Check: The estimate is less than the rectangle 7.5 × 4 = 30 m and greater than the 15 m triangle obtained from the endpoints.

Mark checklist · 4 marks

• 1 mark — Uses the trapezium area rule with 1.0 s widths.
• 1 mark — Correctly combines all four strip areas, without omitting an internal reading.
• 1 mark — Estimate 20.75 m, accept 20.8 m or about 21 m with units.
• 1 mark — Explains that the curved graph has a changing gradient/acceleration, violating the constant-acceleration assumption.

Common mistake: Assuming any known u, v and t can be substituted into SUVAT. Knowing the variables does not establish constant acceleration.

Exam technique: PhysicsUK guidance: write ‘estimate’ when approximating a curved area, show the strips used, and state the assumption before using constant-acceleration equations.

Answer: Total time = 35.0 s: 6.00 s accelerating, 25.0 s cruising and 4.00 s braking.

1. Accelerate:v = u + at = 0 + 2.00 × 6.00 = 12.0 m s⁻¹.s₁ = ut + ½at² = ½ × 2.00 × 6.00² = 36.0 m.
2. Brake: use u = 12.0 m s⁻¹, v = 0 and a = −3.00 m s⁻².s₃ = (v² − u²)/(2a) = (0 − 12.0²)/(2 × −3.00) = 24.0 m.t₃ = (v − u)/a = (0 − 12.0)/(−3.00) = 4.00 s.
3. Cruise: s₂ = 360 − 36.0 − 24.0 = 300 m.t₂ = s₂/v = 300/12.0 = 25.0 s.
4. Total: t = 6.00 + 25.0 + 4.00 = 35.0 s. Each stage has constant acceleration, but the whole trip has three different accelerations.

Check: The velocity–time area is ½ × 6 × 12 + 25 × 12 + ½ × 4 × 12 = 360 m. The whole-trip average speed 360/35 ≈ 10.3 m/s is below the maximum 12.0 m/s.

Mark checklist · 5 marks

• 1 mark — Peak velocity 12.0 m/s and accelerating distance 36.0 m, with an appropriate SUVAT method.
• 1 mark — Braking distance 24.0 m from signed SUVAT, or an equivalent graph method.
• 1 mark — Braking time 4.00 s.
• 1 mark — Remaining cruise distance 300 m and cruise time 25.0 s.
• 1 mark — Adds all three times to obtain 35.0 s. Allow consistent follow-through after one earlier arithmetic error.

Common mistake: Using 360 = ½(0 + 0)t for the entire journey. The endpoint velocities are zero, but acceleration is not constant across the whole trip.

Exam technique: PhysicsUK guidance: make a small table of u, v, a, s and t for each stage. State which values carry into the next stage, and which ones change.

Answer: Thinking distance = 6.0 m; braking distance = 16 m; stopping distance = 22 m, so the cyclist cannot stop before the barrier. In reality, gradual brake application can make the initial braking gradient change.

1. s(thinking) = v × t(reaction) = 8.0 × 0.75 = 6.0 m.
2. Once braking begins, u = 8.0 m s⁻¹, v = 0 and a = −2.0 m s⁻².s(braking) = (v² − u²)/(2a) = (0 − 8.0²)/(2 × −2.0) = 16 m.
3. s(stopping) = s(thinking) + s(braking) = 6.0 + 16 = 22 m.The predicted stopping point without an obstruction is 2.0 m beyond the barrier, so the cyclist reaches it before stopping.
4. For example, the braking force builds up while the brake is squeezed. Acceleration therefore changes during that part of the motion. Instead of an immediate corner followed by one straight descending line, the initial braking section would be curved, with its negative gradient increasing in magnitude as the brakes take effect. A changing resistance force with speed is another acceptable developed explanation.

Check: Braking time = 8.0/2.0 = 4.0 s. The braking graph area is ½ × 4.0 × 8.0 = 16 m; the reaction rectangle adds 6.0 m.

Mark checklist · 7 marks

• 1 mark — Thinking distance 6.0 m.
• 1 mark — Appropriate braking equation with v = 0 and a = −2.0 m/s², or an equivalent area calculation.
• 1 mark — Braking distance 16 m.
• 1 mark — Adds thinking and braking distances to obtain 22 m.
• 1 mark — Compares 22 m with 20 m and concludes the cyclist does not stop before the barrier.
• 1 mark — A physical reason for non-constant braking acceleration, such as brake force building up or resistance varying with speed.
• 1 mark — Connects that reason to a changing gradient/curvature of the braking section of the velocity–time graph.

Common mistake: Comparing only 16 m with the available 20 m gives the wrong safety conclusion. Saying ‘acceleration is not constant’ alone also fails to explain the physical cause.

Exam technique: The 2023 report distinguishes braking from stopping distance. The 2025 report asks for a physical explanation of the simplified motion, rather than just repeating that acceleration changes. Apply both lessons to this new context.

Related examiner evidence: OCR H156/02, June 2023 — Q1(a)(i), Q1(a)(iii)–(b); report pp. 6–8 · OCR H156/02, June 2025 — Q1(b)–(c); report p. 6

Answer: Maximum height = 6.26 m above ground; time to ground = 1.95 s; impact velocity = −11.1 m/s (downwards); acceleration at the highest point = −9.81 m/s².

1. Throughout the flight a = −9.81 m s⁻². At the highest point v = 0.s(rise) = (0 − 8.00²)/(2 × −9.81) = 3.26198… m.Height above the ground = 3.00 + 3.26198… = 6.26 m.
2. At the ground s = −3.00 m.−3.00 = 8.00t − 4.905t².4.905t² − 8.00t − 3.00 = 0.
3. t = [8.00 ± √(8.00² + 4 × 4.905 × 3.00)]/(2 × 4.905).The roots are 1.94538… s and −0.314396… s. The flight after launch uses the positive root: 1.95 s. The negative root is a mathematical extension of the model to before launch.
4. v = u + at = 8.00 − 9.81 × 1.94538… = −11.0842… m s⁻¹ ≈ −11.1 m s⁻¹.The sign specifies downwards; the speed is 11.1 m s⁻¹.
5. At the highest point, the velocity is momentarily zero, but gravity still acts. The acceleration remains −9.81 m s⁻².

Check: Independently, v² = 8.00² + 2(−9.81)(−3.00) = 122.86, so the impact speed is √122.86 = 11.1 m/s. It exceeds the launch speed because the ground is lower than the launch point.

Mark checklist · 7 marks

• 1 mark — Uses v = 0 and a = −g to find a rise of 3.26 m.
• 1 mark — Adds the platform height and gives 6.26 m above ground.
• 1 mark — Sets s = −3.00 m in s = ut + ½at² with consistent signs.
• 1 mark — Solves the resulting quadratic, using an appropriate method.
• 1 mark — Selects the positive post-launch root, 1.95 s.
• 1 mark — Impact velocity −11.1 m/s, or 11.1 m/s downwards, from v = u + at or signed v².
• 1 mark — Acceleration at the highest point −9.81 m/s² or 9.81 m/s² downwards.

Common mistake: Setting s = +3.00 while keeping upwards positive, treating the maximum height as just the rise, or setting a = 0 when v = 0. A turning point is not a period of equilibrium.

Exam technique: PhysicsUK guidance: write the positive direction and the displacement origin before substituting. Keep the full calculator value through linked calculations, and explain which quadratic root belongs to the actual motion.

Answer: Best-fit gradient = 4.80 m/s²; g = 9.60 m/s²; worst-line g = 10.0 m/s²; percentage uncertainty ≈ 4.2% (absolute uncertainty ≈ 0.40 m/s²).

1. Measure h vertically from the bottom of the held ball to the trapdoor contact surface, using a metre rule and a set square to locate the two levels. A centre-to-trapdoor distance would include a radius the ball’s centre does not actually travel before contact. Secure the apparatus and use a tray to catch the ball.
2. Electronic timing links the start to release and the stop to contact. A hand-operated stopwatch adds the operator’s reaction-time variation at both events; this can be a large fraction of a short fall time. Electronic timing reduces that problem, although release delay and trigger offsets must still be checked.
3. From h = ut + ½gt² with u = 0, the ideal relationship is h = ½gt².Gradient m = Δh/Δ(t²) = (0.970 − 0.202)/(0.200 − 0.0400) = 4.80 m s⁻².Use construction points on the fitted line, rather than two arbitrary measured points. The small fitted intercept is a reason to investigate an offset; do not force this best-fit line through zero.
4. For these axes m = g/2, so g = 2m = 9.60 m s⁻². Equivalently, compare h = ½gt² + c with y = mx + c if an additive height offset is included. This does not mean every timing error produces a constant intercept.
5. Worst-line g = 2 × 5.00 = 10.0 m s⁻². The estimated absolute uncertainty is |10.0 − 9.60| = 0.40 m s⁻².Percentage uncertainty = (0.40/9.60) × 100 = 4.17…% ≈ 4.2%.You obtain the same result from (5.00 − 4.80)/4.80 × 100. Multiplying by the exact constant 2 doubles the absolute uncertainty, but leaves the percentage uncertainty unchanged.

Check: The units of gradient are m/s². The result 9.60 ± 0.40 m/s² includes 9.81 m/s². A constant multiplier changes absolute uncertainty, not fractional uncertainty.

Mark checklist · 6 marks

• 1 mark — Measures vertical fall distance with a metre rule, identifying the bottom of the held ball and trapdoor contact surface (or an equivalent correct diameter correction).
• 1 mark — Explains that electronic release/contact timing reduces operator reaction-time error compared with a hand-operated stopwatch.
• 1 mark — Correct fitted gradient (0.970 − 0.202)/(0.200 − 0.0400) = 4.80 m/s².
• 1 mark — Derives m = g/2 from the given axes and obtains g = 9.60 m/s².
• 1 mark — Uses the greatest acceptable gradient departure: Δm = 0.20 m/s² or Δg = 0.40 m/s².
• 1 mark — Percentage uncertainty about 4.2%, relative to the best-fit value. Accept 4.17% with clear working.

Common mistake: Using g = 2/m² from a different graph, timing by hand without considering reaction time, measuring from the ball’s centre without correction, or doubling percentage uncertainty merely because g = 2m.

Exam technique: The 2024 report supports a developed measurement method, fitted-line coordinates and a worst acceptable gradient. Our graph deliberately uses different axes from that paper: derive g = 2m here, rather than importing its inverse-square gradient relationship. This is a structured point-marked task, not the OCR six-mark levels-of-response question.

Related examiner evidence: OCR H156/02, June 2024 — Q3; report pp. 12–13