Mathematics Advanced • Year 12 • Module 6 • Lesson 14
Motion Applications
Apply integration of motion to braking distance, projectile, rocket-landing, train signalling and GPS contexts.
Problem 1 — Braking a car (deceleration)
A car travelling at 25 m/s applies the brakes, giving a constant deceleration of 5 m/s². Take the moment braking begins as t = 0.
Set up: What are we solving for?
(i) Write a(t), find v(t) and the time tstop at which the car comes to rest. 2 marks
(ii) Find the braking distance — the distance travelled between brake application and full stop. 2 marks
(iii) If the deceleration drops to 4 m/s² in wet conditions, by what percentage does the braking distance increase? 3 marks
Stuck? Use the relation distance = v² / (2 |a|) for uniform deceleration as a sanity check.Problem 2 — Diving board (vertical projectile)
A diver leaves a 10 m diving board with an initial upward velocity of 4 m/s. Take upward positive, gravity a(t) = −9.8 m/s², t = 0 at take-off, x(0) = 10 m.
Set up: What are we solving for?
(i) Find v(t) and x(t). 2 marks
(ii) Find the maximum height above the water. 2 marks
(iii) Find the time at which the diver hits the water (x = 0), to 2 d.p. 3 marks
Problem 3 — Rocket landing burn (Falcon-9 style)
During a landing burn, a Falcon-9 first stage has acceleration a(t) = 0.5t − 6 m/s² in the upward direction (t in seconds since the burn starts). Initial velocity v(0) = −100 m/s (downward), initial height x(0) = 1000 m.
Set up: What are we solving for?
(i) Find v(t) and x(t). 3 marks
(ii) Find the time t0 at which the rocket is momentarily at rest (v = 0). Solve algebraically. 2 marks
(iii) For a successful landing, the rocket must reach v = 0 at exactly x = 0. Using your formulas, is the landing successful? Justify with a one-sentence numerical check. 3 marks
Problem 4 — Sydney train approaching a station
A train approaches a station with velocity v(t) = 20 − 2t m/s (t in seconds, train starts braking at t = 0 from 20 m/s).
Set up: What are we solving for?
(i) Find the stopping time and the displacement up to the moment it stops. 2 marks
(ii) The signalling system requires the train to stop at least 5 m before the end of the platform. If the train starts braking 110 m from the end of the platform, will it stop in time? Justify with a numerical comparison. 2 marks
(iii) A late-running scenario: the same train brakes from 25 m/s instead of 20 m/s with the same deceleration (2 m/s²). Find the new stopping distance and decide whether it still satisfies the 5 m margin. 3 marks
Problem 5 — Dead-reckoning a drone (accelerometer integration)
A drone's accelerometer reports a(t) = 0.4 m/s² (a small, constant forward acceleration) for 30 seconds, starting from rest at the origin. GPS is lost during this period and the flight computer must integrate to estimate position.
Set up: What are we solving for?
(i) Find v(t) and x(t), and evaluate x(30). 2 marks
(ii) The accelerometer has a small bias error: it reports a' = 0.42 instead of 0.40 m/s². Find the wrongly-integrated x'(30) and the position error Δx at t = 30 s. 3 marks
(iii) Explain in 1-2 sentences (a) why a small acceleration error grows into a much larger position error after two integrations, and (b) one way GPS data is used to correct this. 2 marks
Stuck? Look at how Δa scales after each ∫dt: linear in t, then quadratic in t.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Braking
Set up. Constant deceleration: integrate to v(t), find rest time, integrate v over [0, tstop] for distance.
(i) a(t) = −5; v(t) = −5t + 25; v = 0 ⇒ tstop = 5 s.
(ii) Distance = ∫₀⁵ (−5t + 25) dt = [−2.5t² + 25t]₀⁵ = −62.5 + 125 = 62.5 m. (Sanity check: v²/(2|a|) = 625/10 = 62.5 ✓.)
(iii) With |a| = 4: distance = 25²/(2 × 4) = 625/8 = 78.125 m. Increase = (78.125 − 62.5)/62.5 = 0.25 = 25%.
Problem 2 — Diver
Set up. Vertical motion under gravity from a 10 m board with upward initial velocity.
(i) v(t) = −9.8 t + 4. x(t) = −4.9 t² + 4t + 10.
(ii) v = 0 ⇒ t = 4/9.8 ≈ 0.408 s. x(0.408) ≈ −4.9(0.1665) + 4(0.408) + 10 ≈ −0.816 + 1.633 + 10 ≈ 10.82 m above the water (≈ 0.82 m above the board).
(iii) 0 = −4.9 t² + 4t + 10 ⇒ 4.9 t² − 4t − 10 = 0. Quadratic formula: t = (4 ± √(16 + 196))/9.8 = (4 ± √212)/9.8 = (4 ± 14.56)/9.8. Take positive root: t ≈ 18.56/9.8 ≈ 1.89 s.
Problem 3 — Falcon-9 landing burn
Set up. Time-varying acceleration. Integrate to v(t), then x(t), apply initial conditions, then check landing.
(i) v(t) = ∫(0.5t − 6) dt = 0.25 t² − 6t + C₁; v(0) = −100 ⇒ C₁ = −100. So v(t) = 0.25 t² − 6t − 100.
x(t) = ∫ v dt = (1/12) t³ − 3t² − 100 t + C₂; x(0) = 1000 ⇒ C₂ = 1000. So x(t) = t³/12 − 3t² − 100 t + 1000.
(ii) v = 0: 0.25 t² − 6t − 100 = 0 ⇒ t² − 24 t − 400 = 0 ⇒ t = (24 ± √(576 + 1600))/2 = (24 ± √2176)/2 ≈ (24 ± 46.65)/2. Positive root: t0 ≈ 35.32 s.
(iii) At t0 ≈ 35.32: x(35.32) ≈ (35.32)³/12 − 3(35.32)² − 100(35.32) + 1000 ≈ 3676 − 3742 − 3532 + 1000 ≈ −2598 m. No — the rocket has already crashed (x became 0 long before v reached 0, so the burn profile is insufficient). A successful landing would require x(t0) = 0 to coincide with v(t0) = 0.
Problem 4 — Train signalling
Set up. Standard deceleration problem; check stopping distance against platform.
(i) 20 − 2t = 0 ⇒ tstop = 10 s. Distance = ∫₀¹⁰ (20 − 2t) dt = [20t − t²]₀¹⁰ = 200 − 100 = 100 m.
(ii) Stopping distance 100 m, available 110 m → ends 10 m from end. 10 m > 5 m required margin, so safe.
(iii) New v(0) = 25, a = −2 ⇒ v(t) = 25 − 2t; rest at t = 12.5 s. Distance = 25²/(2 × 2) = 156.25 m. Since 156.25 > 110, train overshoots by 46.25 m — well beyond the 5 m margin. Not safe.
Problem 5 — Drone dead-reckoning
Set up. Twice-integrate constant acceleration; compare nominal vs biased reading.
(i) v(t) = 0.4 t; x(t) = 0.2 t². x(30) = 0.2 × 900 = 180 m.
(ii) x'(t) = 0.21 t². x'(30) = 0.21 × 900 = 189 m. Position error Δx = 189 − 180 = 9 m at t = 30 (acceleration error of just 0.02 m/s² → 9 m position error in 30 s).
(iii) (a) After two integrations, an acceleration error Δa grows into a position error (Δa/2) t² — quadratic in t, so it explodes over time. (b) GPS provides an independent absolute position measurement; the flight computer uses it to reset the integrated estimate (Kalman filter or equivalent), preventing the drift from growing without bound.