Physics Speed Tricks

Check the unit before computing, push a variable to its limit, read the area under a graph, and keep everything as ratios for as long as possible.

11 min readLevel ★★Four techniques

1. Dimensional analysis: units first, numbers second

Write the units next to every symbol while you build the expression. If the units do not reduce to the required quantity, stop — there is no point computing a number. This rejects most wrong options in a multiple-choice paper without any arithmetic.

2. Special values & limits: push a variable to the boundary

Example: acceleration on a smooth incline

a = g sin θ. At θ = 90° this must give a = g (free fall) ✓, and at θ = 0° it must give a = 0 ✓. Substitute θ = 30° to get a = 0.5 g for a quick estimate.

Example: total resistance in parallel

Set R₁ = R₂ = R: the answer must be R/2. Candidates R₁ + R₂ and R₁R₂/(R₁+R₂) — only the second gives R/2. Then the limit check: as R₂ → ∞ the total must approach R₁, which the product-over-sum form also satisfies.

3. Graphs & geometry: the area is the answer

Uniform acceleration from 0 to 20 m/s in 10 s: the v–t graph is a triangle, so the displacement is ½ × 10 × 20 = 100 m. Gradients give rates, areas give accumulated quantities — no formula needed.

Equal time intervals under uniform acceleration: displacement ratio 1 : 3 : 5 : 7
Equal displacements under uniform acceleration: time ratio 1 : (√2−1) : (√3−√2)

4. Ratios & common constants

Free fall: h = ½gt², with g = 10 this is h = 5t²
Projectile: t = √(2h/g), horizontal range x = v₀t
g ≈ 10 N/kg for estimating
sin 37° = 0.6, cos 37° = 0.8 (the 3-4-5 triangle)
√2 ≈ 1.41, √3 ≈ 1.73, √5 ≈ 2.24
Water density 1.0 × 10³ kg/m³, 1 atm ≈ 1.0 × 10⁵ Pa
c = 3 × 10⁸ m/s, k = 9 × 10⁹ N·m²/C²
Avogadro 6.02 × 10²³ /mol, molar volume at STP 22.4 L/mol

5. Practice

A car goes from 0 to 20 m/s in 10 s: acceleration and distance covered
Free fall for 3 s, g = 10: how far?
A 1 kg mass on a scale in a lift accelerating upward at 2 m/s²: what does the scale read?
R₁ = 2 Ω and R₂ = 3 Ω in parallel: total resistance, and the R₂ → ∞ limit
Uniform acceleration, first interval covers 5 m: displacement during the fourth interval, and the total after four intervals

Answers

1) a = 2 m/s²; s = ½ × 20 × 10 = 100 m. 2) 5 × 3² = 45 m. 3) m(g + a) = 1 × 12 = 12 N (overweight). 4) 2 × 3 / (2 + 3) = 1.2 Ω; as R₂ → ∞ the total → 2 Ω = R₁ ✓. 5) ratio 1:3:5:7 → fourth interval 35 m, total 5 + 15 + 25 + 35 = 80 m.

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