Loav · VCE Specialist Mathematics · Units 3 & 4
S34-K01 · Recommended sequence: Unit 4
1. Position is signed; distance is accumulated
Rectilinear motion takes place along a straight line. Choose an origin and a positive direction, then let x(t) be the particle's signed position. Its displacement from time a to time b is x(b) − x(a), which may be positive, negative or zero. Total distance travelled counts the length of every part of the journey and cannot be negative.
Velocity is v = dx/dt and acceleration is a = dv/dt = d²x/dt². Positive velocity means motion in the chosen positive direction; negative velocity means motion in the opposite direction. Speed is |v|, the magnitude of velocity. A particle can have negative position and positive velocity if it lies on the negative side of the origin while moving toward increasing x.
Acceleration describes change in velocity, not automatically increase in speed. When v and a have the same nonzero sign, speed increases. When their signs are opposite, speed decreases. At a turning instant where v = 0, inspect nearby signs rather than applying a same-sign shortcut mechanically. A zero acceleration can occur while the particle is moving at a nonzero speed.
Use consistent units: position in metres and time in seconds give velocity in m/s and acceleration in m/s². Label the time domain, often t ≥ 0, and retain any physical restriction such as the interval before a model changes. The origin and positive direction are choices, but every signed quantity must use those choices consistently.
2. Differentiate to describe motion; integrate to recover it
Given x(t), differentiate once for velocity and again for acceleration. Given a(t), integrate to obtain v(t) and use an initial velocity to determine the constant. Integrate velocity to obtain position and use an initial position for the second constant. An initial position cannot replace an initial velocity: they determine different information.
For displacement over an interval, integrate velocity: x(b) − x(a) = ∫ from a to b of v(t) dt. For distance, integrate speed: distance = ∫ from a to b of |v(t)| dt. Find all direction changes in the interval and split the integral there, or calculate successive absolute changes in position. Integrating acceleration gives change in velocity, not distance.
Rest occurs when v = 0, but a zero alone does not prove a reversal. For v = (t − 2)², the particle momentarily stops at t = 2 and continues in the same direction. For v = t − 2, velocity changes from negative to positive, so direction reverses. Check signs on either side of each zero.
Worked example 1: Account for two changes of direction
A particle has x(t) = t³ − 6t² + 9t + 2 metres for 0 ≤ t ≤ 4 seconds. Then v = 3t² − 12t + 9 = 3(t − 1)(t − 3), and a = 6t − 12. Velocity is positive on (0, 1), negative on (1, 3) and positive on (3, 4), so the particle reverses direction at both t = 1 and t = 3.
The key positions are x(0) = 2, x(1) = 6, x(3) = 2 and x(4) = 6. Displacement is 6 − 2 = 4 m. Distance is |6 − 2| + |2 − 6| + |6 − 2| = 12 m. Returning to an earlier position does not erase the distance already travelled.
Acceleration changes sign at t = 2. Comparing signs shows decreasing speed on (0, 1) and (2, 3), and increasing speed on (1, 2) and (3, 4). The particle is accelerating in the negative direction throughout (0, 2), but it slows during one part and speeds up during the other.
3. Read motion from a velocity-time graph
The height of a velocity-time graph is velocity, its slope is acceleration, and its signed area is displacement. The area between the graph and the time axis, counted positively on every part, is distance. These are three distinct geometric readings of the same graph. A line below the axis represents negative velocity, not negative speed.
A horizontal segment means constant velocity and zero acceleration. A segment on the time axis means the particle is at rest over that interval. A straight sloping segment means constant acceleration. A curved velocity graph means changing acceleration. If the graph is piecewise, examine the stated behaviour at joining times before claiming that acceleration is defined there.
For a position-time graph, the slope is velocity and the curvature reflects acceleration. A local extremum of position often corresponds to a direction reversal, while a stationary point without a sign change of velocity does not. Never transfer an area interpretation from velocity-time to position-time without checking its units and meaning.
4. Constant acceleration is a special case
If acceleration a is constant and v(0) = u, integration gives v = u + at. A second integration gives x = x₀ + ut + at²/2. Writing s = x − x₀ gives s = ut + at²/2. Eliminating t yields v² = u² + 2as, and averaging the endpoint velocities gives s = (u + v)t/2.
These formulas assume constant acceleration over the interval being used. They are not general replacements for calculus when acceleration depends on time, position or velocity. Signed values of u, v, a and s must follow the chosen direction. A negative root for velocity may be physically appropriate after a reversal, even if the initial velocity was positive.
When a particle comes to rest, solve u + at = 0 and retain a time in the stated domain. Decide whether the model continues after that instant. Constant negative acceleration would make the velocity negative afterwards; it does not force the particle to remain at rest unless the model explicitly changes.
Worked example 2: Rest is an instant, not automatically the end of the model
A particle starts at x = 5 m with velocity 12 m/s and constant acceleration −3 m/s². Its velocity is v = 12 − 3t and position is x = 5 + 12t − 3t²/2. It first comes to rest at t = 4 s, at position x = 29 m.
If the same motion model continues to t = 6 s, then v(6) = −6 m/s and x(6) = 23 m. The displacement from t = 0 is 23 − 5 = 18 m. The distance is (29 − 5) + (29 − 23) = 30 m, since the particle reverses at t = 4.
The average velocity over six seconds is 18/6 = 3 m/s, while the average speed is 30/6 = 5 m/s. Using absolute final displacement for distance would miss the six metres travelled back after the turn. The constant-acceleration formulas calculate positions and velocities correctly, but distance still requires direction awareness.
5. Choose an acceleration form suited to the given variable
Acceleration is always dv/dt when velocity is regarded as a function of time. On a suitable motion branch where velocity can be expressed differentiably in terms of position, the chain rule gives dv/dt = (dv/dx)(dx/dt) = v dv/dx. Since d(v²/2)/dx = v dv/dx, three useful representations are:
If a is supplied as a function of t, integrate with respect to time. If a is supplied as a function of x, integrate v dv = a(x) dx to obtain a relationship between velocity and position. If a is supplied as a function of v, separation in dv/dt = a(v) may recover velocity as a function of time. Choose the form that matches the information rather than introducing an unnecessary variable.
The position form needs care at a turning point: x may cease to be a one-to-one local parameter for the motion, and dv/dx may become unbounded even when dv/dt is finite. Work on monotonic position branches and interpret limiting values. Do not divide by v at v = 0 and silently discard a rest or turning event.
After integrating a position-dependent acceleration, v² is determined first. Taking a square root gives two signs. The initial velocity and the direction on the branch select the sign. A negative value inside the proposed square root is not a real speed; it signals an inaccessible position under that branch's relationship or a calculation error.
Worked example 3: Integrate acceleration as a function of position
A motion model has a = −2x, with x in metres and time in seconds; the coefficient has units s⁻². Initially x = 0 and v = 6 m/s. On the initial rightward branch, v dv/dx = −2x. Integrating gives v²/2 = −x² + C. The initial data make C = 18, so v² = 36 − 2x².
The positive velocity branch is v = √(36 − 2x²). It first reaches zero at x = 3√2 m. Negative x = −3√2 is another algebraic zero, but it is not the first turning position reached by a particle initially moving right from zero.
Before the first turn, separation and the initial condition give t = (1/√2)arcsin[x/(3√2)]. Rearranging gives x = 3√2 sin(√2t), so v = 6cos(√2t). The first zero of this velocity is t = π/(2√2) s, confirming the turning position. After the turn the velocity branch is negative; retaining the positive square root would give the wrong direction.
6. Verify a motion model from several viewpoints
A symbolic check differentiates position to recover velocity and acceleration, then substitutes the initial data. A graphical check compares the signs of v and a with the described direction and speed changes. A units check ensures that every term in a position formula has units of length. These checks complement each other: one cannot detect every error alone.
A non-routine comparison asks whether two journeys can have equal displacement but different distances. They can: add a return segment to one journey without changing the endpoint. Conversely, equal total distance does not determine the final position because the directions may differ. This explains why a velocity-time graph needs both signs and area, rather than only its total shaded size.
With technology, solve v(t) = 0 on the requested interval and verify every returned root by substitution and sign checks. Use exact roots to split distance integrals before rounding the final answer. A numerical integrator applied directly to |v| may give a useful check, but a written split integral or position-change calculation makes the physical reasoning visible.
The subject here is motion described by calculus. No force law is needed to use a supplied acceleration equation. State what the equation predicts, preserve its time and position restrictions, and avoid importing assumptions about collisions, friction or mechanisms that were never included in the model.
Practise, then check
- A particle has x = 2t² − 8t + 5 metres, with t in seconds and 0 ≤ t ≤ 3. Find rest time, displacement and distance.
- For v = t² − 4 m/s, with t in seconds and 0 ≤ t ≤ 3, find displacement and distance.
- A particle has v = −5 m/s and a = −2 m/s² at an instant. Is it speeding up or slowing down?
- With a = 4 m/s², v(0) = −6 m/s and x(0) = 2 m, find the rest time and distance travelled in the first 3 seconds.
- A motion has a = 3x and v = 2 at x = 0. Find v² as a function of x and the initial positive-velocity branch.
- Solve dv/dt = −0.5v with v(0) = 8 and x(0) = 0, where x is in metres and t in seconds. Find the limiting position.
- Does v = (t − 1)² cause a direction reversal at t = 1? Explain using signs.
Fully explained answers
- Velocity is 4t − 8, so rest occurs at t = 2 and direction changes there. Positions are x(0) = 5, x(2) = −3 and x(3) = −1. Displacement is −1 − 5 = −6 m; distance is 8 + 2 = 10 m, using the stated metre and second units.
- Velocity changes sign at t = 2. The signed integral is [t³/3 − 4t] from 0 to 3 = −3. The distance is ∫₀²(4 − t²)dt + ∫₂³(t² − 4)dt = 16/3 + 7/3 = 23/3. Both displacement and distance are measured in metres.
- It is speeding up because both velocity and acceleration are negative. The velocity is becoming more negative, so its magnitude increases. Negative acceleration does not universally mean slowing down; direction must be included.
- Here v = −6 + 4t, so rest occurs at t = 3/2 s. Position is x = 2 − 6t + 2t², with x(3/2) = −5/2 and x(3) = 2. Distance is |−5/2 − 2| + |2 + 5/2| = 9 m. Its final displacement is zero despite nonzero distance.
- Use v dv/dx = 3x, giving v²/2 = 3x²/2 + C. At x = 0, C = 2, so v² = 3x² + 4. The branch matching positive initial velocity is v = √(3x² + 4). Its radicand is positive for all real x.
- Separation gives v = 8e−0.5t. Integrating with the position condition gives x = 16(1 − e−0.5t). For t ≥ 0 the position increases toward 16 m and the speed approaches zero. It does not reach the limiting position at a finite time in this model.
- No. Velocity is positive on both sides of t = 1 and equals zero only at that instant. The particle momentarily rests and then continues in the same positive direction. A reversal requires a sign change, not merely a zero.
Curriculum and sources
VCE Specialist Mathematics Units 3 and 4, Areas of Study 4 as applicable, with Outcomes 1–3. Scope follows the VCAA Mathematics Study Design 2023–2027 and the current Specialist Mathematics FAQ. Curriculum checked 15 October 2026. Teaching reference: Cambridge Senior Mathematics VCE Specialist Mathematics 3&4, second edition (2023), sections 12A–12E (pp538–573). All teaching prose, examples, practice questions and diagrams are original. Independent teaching material; not a VCAA publication.
