The Physics Question Bank

Course: Combined Science + Separate Physics  |  Tier: Foundation + Higher  |  Equation sheet: given in the exam

The equation

(final velocity)² − (initial velocity)² = 2 × acceleration × distance    v² − u² = 2 a s

Quantity	Symbol	Unit
final velocity	v	metres per second (m/s)
initial velocity	u	metres per second (m/s)
acceleration	a	metres per second squared (m/s²)
distance	s	metres (m)

Rearranged: s = (v² − u²) ÷ 2a  |  a = (v² − u²) ÷ 2s  |  v = √(u² + 2as)  |  “From rest” means u = 0

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.

Foundation: aim for Q1–8.   Higher: aim for Q4–10. Every question needs rearranging.

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Questions

Q1 (F) A cyclist accelerates from rest to 6 m/s at 1.5 m/s². How far does she travel?

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F: v² − u² = 2 a s
I: 6² − 0² = 2 × 1.5 × s, so 36 = 3 × s
F: s = 36 ÷ 3
A: 12 m

Q2 (F) A sprinter starts from rest and reaches 8 m/s after 16 m. Calculate her acceleration.

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F: v² − u² = 2 a s
I: 8² − 0² = 2 × a × 16, so 64 = 32 × a
F: a = 64 ÷ 32
A: 2 m/s²

Q3 (F) A car starts from rest and accelerates at 4 m/s² over 50 m. Calculate its final speed.

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F: v² − u² = 2 a s
I: v² − 0² = 2 × 4 × 50
F: v² = 400, so v = √400
A: 20 m/s

Q4 (F) A car speeds up from 10 m/s to 20 m/s over a 75 m slip road. Calculate its acceleration.

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F: v² − u² = 2 a s
I: 20² − 10² = 2 × a × 75, so 300 = 150 × a
F: a = 300 ÷ 150
A: 2 m/s²

Q5 (F) A car travelling at 72 km/h brakes to a stop with a deceleration of 8 m/s². Calculate its braking distance.

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Convert first: 72 km/h = 72 000 ÷ 3600 = 20 m/s; deceleration means a = −8 m/s²
F: v² − u² = 2 a s
I: 0² − 20² = 2 × −8 × s, so −400 = −16 × s
F: s = −400 ÷ −16
A: 25 m

Q6 (F/H) A plane starts from rest and reaches its take-off speed of 80 m/s over 1600 m of runway. Calculate its acceleration.

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F: v² − u² = 2 a s
I: 80² − 0² = 2 × a × 1600, so 6400 = 3200 × a
F: a = 6400 ÷ 3200
A: 2 m/s²

Q7 (F/H) A car brakes with a deceleration of 6.0 m/s² and stops in 48 m. How fast was it going before braking?

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F: v² − u² = 2 a s
I: 0² − u² = 2 × −6.0 × 48, so −u² = −576
F: u² = 576, so u = √576
A: 24 m/s

Q8 (F/H) An electron starts from rest and accelerates at 4.5 × 10¹⁴ m/s² over a distance of 0.010 m. Calculate its final speed.

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F: v² − u² = 2 a s
I: v² − 0² = 2 × 4.5 × 10¹⁴ × 0.010
F: v² = 9.0 × 10¹², so v = √(9.0 × 10¹²)
A: 3.0 × 10⁶ m/s

Q9 (H) A 1200 kg car travelling at 30 m/s brakes with a braking force of 7200 N. Calculate its braking distance.

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Step 1 – deceleration: a = F ÷ m = 7200 ÷ 1200 = 6 m/s², so a = −6 m/s²
Step 2 – distance
F: v² − u² = 2 a s
I: 0² − 30² = 2 × −6 × s, so −900 = −12 × s
F: s = −900 ÷ −12
A: 75 m

Q10 (H) A ball is thrown straight up at 15 m/s. Calculate the maximum height it reaches, to 2 significant figures. (Acceleration due to gravity = 9.8 m/s²; ignore air resistance.)

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Think: at the top, v = 0; gravity slows it, so a = −9.8 m/s²
F: v² − u² = 2 a s
I: 0² − 15² = 2 × −9.8 × s, so −225 = −19.6 × s
F: s = −225 ÷ −19.6 = 11.47… m
A: 11 m (2 s.f.)