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Specification: AQA A-level Physics (7408) 3.4.1.3  |  Paper: 1 (and 2 as assumed knowledge)  |  Data and formulae booklet: given in the exam

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The equation

displacement = ½ × (initial velocity + final velocity) × time    s = ½(u + v)t    (booklet form: s = ((u + v) ÷ 2) t)    No a: this equation does not contain acceleration, so choose it when a is neither given nor asked for. ½(u + v) is the average velocity, which is only true when the acceleration is uniform.

Quantity	Symbol	Unit
displacement	s	metres (m)
initial velocity	u	metres per second (m s⁻¹)
final velocity	v	metres per second (m s⁻¹)
time	t	seconds (s)

Rearranged: t = 2s ÷ (u + v)  |  v = (2s ÷ t) − u  |  u = (2s ÷ t) − v  |  Signs: choose a positive direction (e.g. take up as positive) and give every vector (s, u, v, a) a sign; a deceleration is a negative a, and a ball thrown upwards has a = −9.81 m s⁻².

Use FIFA for every answer

• F – Formula: write the equation as it appears on the sheet.
• I – Insert: put in the numbers (convert to standard units first).
• F – Fix: rearrange to make the unknown the subject.
• A – Answer: calculate and always give the unit.

Core: Q1–4.   Exam standard: Q5–8.   Stretch: Q9–10.

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Questions

Q1 (Core) A car accelerates uniformly from 10 m s⁻¹ to 20 m s⁻¹ in 8.0 s. Calculate the distance it travels in this time.

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F: s = ½(u + v)t
I: s = 0.5 × (10 + 20) × 8.0
F: s is already the subject
A: 15 × 8.0 = 120 m

Q2 (Core) A sprinter starts from rest and reaches 9.0 m s⁻¹ in 2.0 s with uniform acceleration. Calculate the distance she covers.

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F: s = ½(u + v)t
I: s = 0.5 × (0 + 9.0) × 2.0
F: s is already the subject
A: 9.0 m

Q3 (Core) A train slows uniformly from 108 km h⁻¹ to rest over a distance of 1.5 km. Calculate the time taken.

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Convert first: 108 km h⁻¹ = 108 000 ÷ 3600 = 30 m s⁻¹; 1.5 km = 1500 m
F: s = ½(u + v)t
I: 1500 = 0.5 × (30 + 0) × t
F: t = 2 × 1500 ÷ 30
A: 100 s

Q4 (Core) A cyclist accelerating uniformly travels 0.45 km in 30 s, reaching a velocity of 18 m s⁻¹. Calculate her initial velocity.

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Convert first: 0.45 km = 450 m
F: s = ½(u + v)t
I: 450 = 0.5 × (u + 18) × 30
F: u + 18 = 2 × 450 ÷ 30 = 30, so u = 30 − 18
A: 12 m s⁻¹
Watch out: 2s ÷ t = 30 m s⁻¹ is (u + v), not u. Remember to subtract v.

Q5 (Exam) A car travelling at 31 m s⁻¹ decelerates uniformly to rest in 4.2 s. Calculate its braking distance, to 2 significant figures.

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F: s = ½(u + v)t
I: s = 0.5 × (31 + 0) × 4.2
F: s is already the subject
A: 15.5 × 4.2 = 65.1 m, so 65 m (2 s.f.)
Watch out: the deceleration is not given and not needed – that is exactly why this is the equation to choose.

Q6 (Exam) A ball is thrown vertically upwards at 12.0 m s⁻¹. 2.00 s later it is moving downwards at 7.62 m s⁻¹. Calculate its displacement from the point of release at this time, to 3 significant figures. (Ignore air resistance)

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Signs: take up as positive, so u = +12.0 m s⁻¹ and v = −7.62 m s⁻¹
F: s = ½(u + v)t
I: s = 0.5 × (12.0 + (−7.62)) × 2.00
F: s = 0.5 × 4.38 × 2.00
A: +4.38 m, i.e. 4.38 m above the point of release (3 s.f.)
Watch out: using v = +7.62 m s⁻¹ gives 19.6 m, which is above the ball’s maximum height. Displacement is not the same as distance travelled.

Q7 (Exam) A proton in a linear accelerator is accelerated uniformly from 1.2 × 10⁶ m s⁻¹ to 4.8 × 10⁶ m s⁻¹ over a distance of 25 cm. Calculate the time taken, to 2 significant figures.

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Convert first: 25 cm = 0.25 m
F: s = ½(u + v)t
I: 0.25 = 0.5 × (1.2 × 10⁶ + 4.8 × 10⁶) × t
F: t = 2 × 0.25 ÷ 6.0 × 10⁶ = 8.333… × 10⁻⁸ s
A: 8.3 × 10⁻⁸ s (2 s.f.)

Q8 (Exam) A velocity–time graph for a car is a straight line from 24 m s⁻¹ at t = 0 to 6.0 m s⁻¹ at t = 9.0 s. Calculate (a) the displacement in the 9.0 s and (b) the acceleration.

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(a) Area under the graph (trapezium)
F: s = ½(u + v)t
I: s = 0.5 × (24 + 6.0) × 9.0
F: s is already the subject
A: 15 × 9.0 = 135 m
(b) Gradient of the graph: a = (v − u) ÷ t = (6.0 − 24) ÷ 9.0 = −2.0 m s⁻² (a deceleration of 2.0 m s⁻²)
Watch out: the area gives displacement and the gradient gives acceleration – do not mix them up.

Q9 (Stretch) A lift starts from rest and accelerates uniformly to 2.4 m s⁻¹ in 2.0 s. It then travels at constant velocity for 9.0 s and decelerates uniformly to rest in 3.0 s. Calculate the total distance travelled and the average speed for the whole journey. Give both answers to 2 significant figures.

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Step 1 – accelerating: s = ½(u + v)t = 0.5 × (0 + 2.4) × 2.0 = 2.4 m
Step 2 – constant velocity: s = vt = 2.4 × 9.0 = 21.6 m
Step 3 – decelerating: s = ½(u + v)t = 0.5 × (2.4 + 0) × 3.0 = 3.6 m
Step 4 – total: 2.4 + 21.6 + 3.6 = 27.6 m, so 28 m (2 s.f.)
Step 5 – average speed: total time = 2.0 + 9.0 + 3.0 = 14 s; average speed = 27.6 ÷ 14 = 1.971… m s⁻¹, so 2.0 m s⁻¹ (2 s.f.)
Watch out: average speed is total distance ÷ total time, not (0 + 2.4) ÷ 2. The ½(u + v) shortcut only works for a single stage of uniform acceleration.

Q10 (Stretch) A driver travelling at 27 m s⁻¹ sees a hazard. Her reaction time is 0.70 s, after which she brakes with a uniform deceleration of 6.5 m s⁻². Calculate the total stopping distance, to 2 significant figures.

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Step 1 – thinking distance: constant speed, so s = vt = 27 × 0.70 = 18.9 m
Step 2 – braking time: v = u + at, so 0 = 27 + (−6.5 × t), t = 27 ÷ 6.5 = 4.1538… s
Step 3 – braking distance
F: s = ½(u + v)t
I: s = 0.5 × (27 + 0) × 4.1538
F: s = 56.076… m
A: total = 18.9 + 56.076 = 74.976… m, so 75 m (2 s.f.)
Watch out: keep t unrounded – rounding it to 4.2 s first gives 75.6 m and the wrong 2 s.f. answer.

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