Standard difficulty · 20 questions · 40 marks · Time: 36 minutes
Show your working for Sections B and C. Answers are on the last page — try every question first!
Understanding Motion through Experience (Advanced) · AW-9S-AMO-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.
| Q | Answer | Working |
|---|---|---|
| 1 | (d) 60 m | sn = u + a(2n − 1)/2 = 3 + 6 × 19/2 |
| 2 | (c) 100 m | the two walks are at right angles: √(602 + 802) = 100 |
| 3 | (d) speed | speed has magnitude only |
| 4 | (a) 20 km/h | same direction: 90 − 70 |
| 5 | (d) an inertial frame | Newton's laws hold in it without change |
| 6 | (a) 30 m west | opposite directions: 38 − 8, towards the larger one |
| 7 | (d) 5 : 9 | from rest sn = a(2n − 1)/2, so the ratio is 5 : 9 |
| 8 | (a) Both A and R are true, and R is the correct explanation of A. | A is true: sn = 5 + 8 × 15/2 = 65. R is true and explains A. |
| 9 | 99.5 m | sn = 14 + 9(2 × 10 − 1)/2 |
| 10 | 2 m/s2 | 23 = 6 + a × 17/2, so a = 2 × 17 ÷ 17 |
| 11 | 261 km | √(1892 + 1802) = 261 |
| 12 | (a) 1147 m; (b) 925 m | (a) 259 + 888; (b) √(2592 + 8882) |
| 13 | 272 m | 0 = u2 − 2as gives s ∝ u2: 17 × (4)2 |
| 14 | (a) 2 s; (b) 24 m | (a) t = (17 − 7) ÷ 5; (b) s = (172 − 72) ÷ (2 × 5) |
| 15 | 8 s | relative speed 21 − 17 = 4 m/s; 32 ÷ 4 |
| 16 | 114 m | from rest the distances in successive seconds are in the ratio 1 : 3 : 5 …; 6 × 19 |
| 17 | (a) 4 km; (b) 13 km; (c) 13 km | (a) AB = 4; (b) BC + CD = 12 + 1; (c) AD = √(52 + 122) = 13 |
| 18 | (a) 6 m; (b) 2 m/s2; (c) 1 m/s; (d) 9 m/s | (a) 12 − 6; (b) each second's distance grows by a = 2; (c) s1 = u + a/2 = 2; (d) v = u + 4a |
| 19 | (a) 88 m; (b) 520 m; (c) 92 m/s | (a) u + a(2n − 1)/2; (b) ut + ½at2 = 120 + 400; (c) u + at |
| 20 | (a) 97 km/h; (b) 83 km/h; (c) 7 km/h; (d) 57 km/h | (a) 90 + 7; (b) 90 − 7; (c) the seated passenger shares the train's motion; (d) 97 − 40 |