Mathematics Advanced • Year 12 • Module 6 • Lesson 14
Motion Applications
Practise HSC-style writing on motion problems, including a multi-part distance-vs-displacement extended response.
1. Short-answer questions
1.1 A particle has a(t) = 6 − 2t m/s². At t = 0, v = 1 m/s and x = 2 m. Find v(t), x(t), and evaluate x(3). Show all working. 3 marks Band 4
1.2 A ball is thrown vertically upward from ground level with initial velocity 25 m/s. Taking g = 9.8 m/s² (downward) and up positive, find the maximum height (to 1 d.p.) and the time at which the ball returns to ground level. 4 marks Band 4-5
1.3 Explain in 2-3 sentences the difference between displacement and distance travelled. Give a one-sentence numerical example using a particle that moves 4 m forward then 3 m back. 2 marks Band 3
Stuck on 1.2? Use v = u + at and the symmetry of projectile motion (up-time = down-time when starting and ending at the same height).2. Extended response
2.1 A particle moves along a straight line with velocity v(t) = t² − 5t + 6 m/s, for t ≥ 0.
(a) Find the times at which the particle is instantaneously at rest.
(b) Find the displacement of the particle from t = 0 to t = 4.
(c) Find the total distance travelled by the particle from t = 0 to t = 4.
(d) Sketch a clearly-labelled graph of v(t) over [0, 4], indicating where v(t) ≥ 0 and v(t) ≤ 0, and explain in one sentence how the sketch supports your answer to (c). 8 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — factorises to (t − 2)(t − 3) and states t = 2, t = 3.
Part (b) — 2 marks
• 1 mark — sets up the integral ∫₀⁴ v(t) dt.
• 1 mark — evaluates correctly to 16/3 m.
Part (c) — 3 marks
• 1 mark — identifies the sign pattern: v ≥ 0 on [0, 2], v ≤ 0 on [2, 3], v ≥ 0 on [3, 4].
• 1 mark — sets up the three sub-integrals correctly (signs handled).
• 1 mark — evaluates to 17/3 m.
Part (d) — 2 marks
• 1 mark — sketches the parabola opening upward with roots at t = 2, t = 3, labelled axes.
• 1 mark — one-sentence explanation linking the sign-change at t = 2 (and t = 3) to the need to split the integral when computing distance.
Your response:
Stuck on (c)? After (a), the zeros split [0, 4] into three intervals; integrate |v(t)| piecewise.How did this worksheet feel?
What I'll revisit before next class:
1.1 — a = 6 − 2t, v(0) = 1, x(0) = 2; find x(3) (3 marks)
Sample response. v(t) = ∫(6 − 2t) dt = 6t − t² + C₁; v(0) = 1 ⇒ C₁ = 1 ⇒ v(t) = 6t − t² + 1.
x(t) = ∫ v dt = 3t² − t³/3 + t + C₂; x(0) = 2 ⇒ C₂ = 2 ⇒ x(t) = 3t² − t³/3 + t + 2.
x(3) = 27 − 9 + 3 + 2 = 23 m.
Marking notes. 1 mark — correct v(t) with C₁ applied. 1 mark — correct x(t) with C₂ applied. 1 mark — correct numerical x(3). Students who forget either constant lose marks; rounding/arithmetic errors in the final evaluation cost the third mark.
1.2 — Ball thrown upward at 25 m/s, g = 9.8 (4 marks)
Sample response. a(t) = −9.8; v(t) = −9.8 t + 25; v = 0 at t = 25/9.8 ≈ 2.551 s. x(t) = −4.9 t² + 25 t; max height x(2.551) = −4.9(6.508) + 25(2.551) ≈ −31.89 + 63.78 ≈ 31.9 m (1 d.p.).
Return to ground: 0 = −4.9 t² + 25 t = t(−4.9 t + 25) ⇒ t = 0 (launch) or t = 25/4.9 ≈ 5.10 s.
Marking notes. 1 mark — correct v(t) and time at max height. 1 mark — correct x(t) and max height. 1 mark — correct setup for return to ground. 1 mark — correct positive root for return time. Note return-time is double the time-to-peak by symmetry — a useful sanity check.
1.3 — Displacement vs distance (2 marks)
Sample response. Displacement is the net change in position, x(t2) − x(t1); it can be positive, negative or zero. Distance travelled is the total path length, ∫ |v(t)| dt; it is always non-negative and equals or exceeds |displacement|. Example: 4 m forward then 3 m back gives displacement = 1 m, distance = 7 m.
Marking notes. 1 mark — correct definitions distinguishing signed vs unsigned. 1 mark — correct numerical example. Students who confuse the two (e.g. say distance = 1 m) score 0.
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) v(t) = t² − 5t + 6 = (t − 2)(t − 3). v = 0 when t = 2 s and t = 3 s. [1 mark — factorisation and times.]
(b) Displacement = ∫₀⁴ (t² − 5t + 6) dt = [t³/3 − 5t²/2 + 6t]₀⁴ = (64/3 − 40 + 24) − 0 = 64/3 − 16 = 16/3 m. [1 mark — integral set up; 1 mark — correct evaluation.]
(c) Sign check: v(0) = 6 > 0; v(2.5) = 6.25 − 12.5 + 6 = −0.25 < 0; v(3.5) = 12.25 − 17.5 + 6 = 0.75 > 0.
So v ≥ 0 on [0, 2] and [3, 4], and v ≤ 0 on [2, 3]. [1 mark — sign pattern identified.]
Distance = ∫₀² v dt − ∫₂³ v dt + ∫₃⁴ v dt.
Let F(t) = t³/3 − 5t²/2 + 6t. Then F(0) = 0, F(2) = 8/3 − 10 + 12 = 14/3, F(3) = 9 − 22.5 + 18 = 9/2, F(4) = 64/3 − 40 + 24 = 16/3.
Distance = (F(2) − F(0)) − (F(3) − F(2)) + (F(4) − F(3))
= 14/3 − (9/2 − 14/3) + (16/3 − 9/2)
= 14/3 − (27/6 − 28/6) + (32/6 − 27/6)
= 14/3 − (−1/6) + (5/6)
= 14/3 + 1/6 + 5/6 = 28/6 + 1/6 + 5/6 = 34/6 = 17/3 m. [1 mark — sub-integrals; 1 mark — correct evaluation.]
(d) Sketch (described): standard upward parabola y = t² − 5t + 6 with roots at t = 2 and t = 3, vertex (2.5, −0.25). Axes labelled t (horizontal) and v (vertical). The portion of the curve below the t-axis lies between t = 2 and t = 3. [1 mark — labelled parabola.]
Because v changes sign at t = 2 and t = 3, the integrand for distance, |v(t)|, has a "V"-shaped reflection there: the area below the t-axis is added rather than subtracted, which is exactly why distance exceeds displacement by 1/3 m here (17/3 − 16/3 = 1/3 m, accounting for the small backward excursion on [2, 3]). [1 mark — links sketch to need for split integral.] ▮
Total: 8/8.
Band descriptors for marker.
Band 3: Solves (a) only; treats distance and displacement the same in (b)/(c); no sketch. ≈ 2-3 marks.
Band 4: Correct (a) and (b); (c) attempted but sign-splitting omitted (e.g. answers 16/3 again). Sketch missing key labels. ≈ 4-5 marks.
Band 5: Full (a)-(c) with correct arithmetic; (d) sketch correct but explanation only partially connects sign-change to split integral. ≈ 6-7 marks.
Band 6: All parts complete, sign analysis explicit, exact fractions throughout, sketch labelled, explanation in (d) explicitly invokes the reflection of v below the t-axis. 8/8.