Mathematics Advanced • Year 12 • Module 6 • Lesson 14

Motion Applications

Build fluency in moving up and down the chain x ↔ v ↔ a, handling initial conditions and distinguishing displacement from distance.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the chain (differentiation moves you which way? integration the other):

x(t) →(d/dt)→ ________ →(d/dt)→ ________; a(t) →(∫dt)→ ________ →(∫dt)→ ________.

Q1.2 Distinguish:

Displacement = ∫_t₁^t₂ __________ dt Distance travelled = ∫_t₁^t₂ __________ dt

Q1.3 Two integrations introduce ________ arbitrary constants, so a second-order motion problem needs ________ initial conditions.

Stuck? Revisit lesson § Kinematics.

2. Worked example — constant acceleration a(t) = 2 m/s²

Problem. A particle has a(t) = 2 m/s². At t = 0: v = 5 m/s, x = 10 m. Find x(t) and evaluate x(3).

Step 1 — Integrate a(t) for v(t), apply v(0) = 5.

v(t) = ∫ 2 dt = 2t + C₁; v(0) = 5 ⇒ C₁ = 5

∴ v(t) = 2t + 5

Step 2 — Integrate v(t) for x(t), apply x(0) = 10.

x(t) = ∫ (2t + 5) dt = t² + 5t + C₂; x(0) = 10 ⇒ C₂ = 10

∴ x(t) = t² + 5t + 10

Step 3 — Evaluate.

x(3) = 9 + 15 + 10 = 34 m

Check. dx/dt = 2t + 5 = v(t) ✓; dv/dt = 2 = a(t) ✓; x(0) = 10, v(0) = 5 ✓.

3. Faded example — fill in the missing steps

A particle has a(t) = 6t − 4 m/s², initially at rest at the origin. Find x(2). 4 marks

Step 1 — Find v(t).

v(t) = ∫ (6t − 4) dt = ____________________ + C₁

v(0) = 0 ⇒ C₁ = ________ ⇒ v(t) = ____________________

Step 2 — Find x(t).

x(t) = ∫ v(t) dt = ____________________ + C₂

x(0) = 0 ⇒ C₂ = ________ ⇒ x(t) = ____________________

Step 3 — Evaluate x(2) = ____________ m.

Stuck? Revisit lesson § Examples — Example 1.

4. Graduated practice

Foundation — single integration (4 questions)

Find the requested function or value. Show one line of working.

Q	Given	Find
4.1 1	a(t) = 4 m/s², v(0) = 0	v(t)
4.2 1	v(t) = 3t² m/s, x(0) = 0	x(t)
4.3 1	v(t) = 2t − 5	Time when at rest
4.4 1	v(t) = t² − 9	Zeros of v(t)

Standard — two integrations or displacement (6 questions)

4.5 a(t) = 4t − 2 m/s², v(0) = 3 m/s, x(0) = 0. Find x(2). 2 marks

4.6 A car decelerates uniformly: v(t) = 12 − 2t m/s. Find the distance travelled before it stops. 2 marks

4.7 A ball is thrown upward with a(t) = −10 m/s², v(0) = 20 m/s, x(0) = 0. Find the maximum height. 3 marks

4.8 v(t) = t² − 4t + 3 m/s. Find the displacement from t = 0 to t = 3. 2 marks

4.9 a(t) = 6 − 2t m/s², v(0) = 1, x(0) = 2. Find x(3). 3 marks

4.10 v(t) = t² − 5t + 6 m/s. Find when the particle is at rest. 2 marks

Extension — distance vs displacement (2 questions)

4.11 v(t) = t² − 4t + 3 m/s. Find the total distance travelled from t = 0 to t = 3. Compare with the displacement (4.8). 3 marks

4.12 v(t) = 3t² − 12t + 9 m/s. Find the total distance travelled from t = 0 to t = 4. 3 marks

Stuck? Factor v(t), find where it changes sign, then split the integral.

5. Self-check the easy 3

Tick once verified.

For 4.1 I integrated to v(t) = 4t + C, then used v(0) = 0 to get v(t) = 4t.

For 4.3 I set v(t) = 0 to find the rest time t = 5/2.

For 4.6 I found that v = 0 at t = 6, then integrated v over [0, 6] to get distance 36 m.

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What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Chain

x →(d/dt)→ v(t) →(d/dt)→ a(t); a →(∫)→ v(t) →(∫)→ x(t).

Q1.2 — Displacement vs distance

Displacement = ∫ v(t) dt (signed). Distance = ∫ |v(t)| dt (always non-negative).

Q1.3 — Constants

Two integrations introduce two arbitrary constants; a second-order motion problem needs two initial conditions (typically x(0) and v(0)).

Q3 — Faded example a(t) = 6t − 4, v(0) = 0, x(0) = 0

Step 1: v(t) = 3t² − 4t + C₁; v(0) = 0 ⇒ C₁ = 0 ⇒ v(t) = 3t² − 4t.
Step 2: x(t) = t³ − 2t² + C₂; x(0) = 0 ⇒ C₂ = 0 ⇒ x(t) = t³ − 2t².
Step 3: x(2) = 8 − 8 = 0 m.

Q4.1 — 4.4 Foundation

4.1 v(t) = 4t. 4.2 x(t) = t³. 4.3 t = 5/2 s. 4.4 t = 3 and t = −3 (in motion typically t ≥ 0, so t = 3).

Q4.5 — a = 4t − 2, v(0) = 3, x(0) = 0

v(t) = 2t² − 2t + 3; x(t) = (2/3)t³ − t² + 3t. x(2) = 16/3 − 4 + 6 = 22/3 m ≈ 7.33 m.

Q4.6 — Car decelerating: v(t) = 12 − 2t

Stops when 12 − 2t = 0 ⇒ t = 6 s. Distance (v ≥ 0 on [0, 6]) = ∫₀⁶ (12 − 2t) dt = [12t − t²]₀⁶ = 72 − 36 = 36 m.

Q4.7 — Ball thrown up, a = −10, v(0) = 20

v(t) = −10t + 20. Max height when v = 0: t = 2 s. x(t) = −5t² + 20t. x(2) = −20 + 40 = 20 m.

Q4.8 — v(t) = t² − 4t + 3, displacement on [0, 3]

∫₀³ (t² − 4t + 3) dt = [t³/3 − 2t² + 3t]₀³ = 9 − 18 + 9 = 0 m (the particle ends where it started in net terms).

Q4.9 — a = 6 − 2t, v(0) = 1, x(0) = 2

v(t) = 6t − t² + 1; x(t) = 3t² − t³/3 + t + 2. x(3) = 27 − 9 + 3 + 2 = 23 m.

Q4.10 — v(t) = t² − 5t + 6, at rest

v(t) = (t − 2)(t − 3) = 0 ⇒ t = 2 s and t = 3 s.

Q4.11 — Total distance for v = t² − 4t + 3, [0, 3]

v(t) = (t − 1)(t − 3); v = 0 at t = 1, 3. On [0, 1]: v ≥ 0 (e.g. v(0) = 3). On [1, 3]: v ≤ 0 (e.g. v(2) = −1).
Distance = ∫₀¹ v dt − ∫₁³ v dt = [t³/3 − 2t² + 3t]₀¹ − [t³/3 − 2t² + 3t]₁³ = (4/3) − (0 − 4/3) = 8/3 m.
Displacement (from 4.8) was 0 m, but the particle still travelled 8/3 m — it went forward then back.

Q4.12 — v(t) = 3t² − 12t + 9, distance on [0, 4]

v(t) = 3(t − 1)(t − 3); zeros at t = 1, 3. Sign: + on [0, 1], − on [1, 3], + on [3, 4].
Position x(t) = t³ − 6t² + 9t. x(0) = 0, x(1) = 4, x(3) = 0, x(4) = 4.
Distance = |x(1) − x(0)| + |x(3) − x(1)| + |x(4) − x(3)| = 4 + 4 + 4 = 12 m.