Equations of motion (suvat) and free fall

1. (a) Up = +: u = 5.6, v = 0, a = −9.81, s = ? No t, so \(v^2 = u^2 + 2as\).
2. \(0 = 5.6^2 + 2(-9.81)s\) → \(s = \dfrac{5.6^2}{2 \times 9.81} = 1.6\) m C1
3. Height above the ground = 3.1 + 1.6 = 4.7 m A1 The answer must include the 3.1 m: the 1.6 m is measured from the hand.
4. (b) From the top: u = 0 (not 5.6), s = 4.7 m down, a = 9.81 down. \(s = \tfrac12at^2\): \(4.7 = \tfrac12(9.81)t^2\) C1
5. \(t = \sqrt{\dfrac{2 \times 4.7}{9.81}} =\) 0.98 s A1

The June 2022 report: some used u = 5.6 m s⁻¹ at the top, some gave t² as the answer (0.96), and g = 10 gave an inaccurate answer.

1. (i) A: u = 0, s = 14 m down, a = 9.81. \(s = \tfrac12at^2\) → \(t = \sqrt{\dfrac{2 \times 14}{9.81}}\) C1 = 1.7 s A1
2. (ii) B takes the same 1.7 s. Down = +: s = +3.6 m, a = +9.81, and B starts moving up, so its velocity is −u.
3. \(3.6 = -u(1.7) + \tfrac12(9.81)(1.7)^2\) C1
4. \(u = \dfrac{14.2 - 3.6}{1.7} =\) 6.2 m s⁻¹ A1 The scheme writes (−)6.2: the size is what's asked.

Other way (also in the scheme): find B's speed at the ground first, \(v = (3.6 + \tfrac12 \times 9.81 \times 1.7^2)/1.7 = 10\) m s⁻¹, then \(10 = -u + 9.81 \times 1.7\) → u = 6.2 m s⁻¹.

1. (a)(i) Gradient: \(a = \dfrac{16.5 - 0}{0.30 - 0} = 55\) m s⁻² A1
2. (b)(i) Stage 1 (rocket): u = 0, a = 55, t = 0.30. \(s = \tfrac12at^2 = \tfrac12 \times 55 \times 0.30^2\) C1 = 2.5 m A1 Or the area under the graph: ½ × 0.30 × 16.5.
3. (b)(ii) Stage 2 (metal): it leaves the rocket moving up at 16.5 m s⁻¹, so u = 16.5 m s⁻¹ upwards, not 0 C1. Now only gravity acts: a = 9.81 down.
4. Down = +: u = −16.5, s = +2.5, a = +9.81. \(v^2 = u^2 + 2as = 16.5^2 + 2 \times 9.81 \times 2.5\) C1
5. v = 18 m s⁻¹ A1

It rises, stops, and falls back past the start at 16.5 m s⁻¹, then falls another 2.5 m, so it must hit the ground a bit faster than 16.5 m s⁻¹: 18 m s⁻¹ fits.

1. Goods train (constant speed): \(s = 10T\).
2. Express (from rest): \(s = \tfrac12 \times 0.5 \times T^2\).
3. Same s at the overtaking time: \(10T = 0.25T^2\) → \(T = 40\) s → D 1

C (20 s) is the trap: that is when the two trains have the same speed (0.5 × 20 = 10 m s⁻¹). At that moment the express is still behind; it only catches up when the areas under the two v–t graphs are equal.

1. SI units: v = 85 ÷ 3.6 = 23.6 m s⁻¹; s = 1200 m; u = 0.
2. No t: \(v^2 = u^2 + 2as\) → \(a = \dfrac{v^2}{2s} = \dfrac{23.6^2}{2 \times 1200}\)
3. a = 0.23 m s⁻² → A 1

The March 2025 report: the most common error was using 85 without converting (giving C); a minority dropped the 2 in 2as (giving B).

1. Stopping: v = 0, so \(0 = u^2 - 2as\) → \(s = \dfrac{u^2}{2a}\).
2. a is fixed, so \(s \propto u^2\).
3. u × 1.20 → s × 1.20² = 1.44x → C 1

The November 2016 report: weaker candidates chose A, assuming s ∝ u.