Loav · VCE Specialist Mathematics · Units 3 & 4
S34-V06 · Vectors · Motion
1. Differentiate position component by component
A position vector r(t) records where an object is at time t relative to a fixed origin. Its derivative is velocity, v(t) = r′(t), and the derivative of velocity is acceleration, a(t) = r″(t). Differentiate each component using the familiar scalar rules. In this guide the coordinate unit vectors are fixed, so their derivatives are zero.
Velocity is a vector. It gives instantaneous direction of travel and rate of displacement. At a time when velocity is nonzero, it points along the tangent to the path. Acceleration describes how the velocity vector changes, including changes of direction. An object can have constant speed but nonzero acceleration because its direction is changing.
v(t) = (x′(t), y′(t), z′(t))
a(t) = (x″(t), y″(t), z″(t))
Speed = |v(t)| = √[x′(t)2 + y′(t)2 + z′(t)2]
If position is measured in metres and time in seconds, velocity has units metres per second and acceleration metres per second squared. The parameter domain is part of the model. A formula may make algebraic sense at negative time while the described journey begins at zero. Solve the mathematics, then enforce the stated physical interval.
Being momentarily stationary means every velocity component is zero at the same time. A zero horizontal component alone means no instantaneous horizontal motion; vertical motion may continue. At a point with zero velocity, its direction is undefined, so a claimed angle between velocities needs special care.
2. Integrate with vector constants and initial conditions
To recover velocity from acceleration, integrate each component and include a constant vector. In three dimensions it has three independent scalar components. An initial velocity determines those constants. Integrating once more gives position and another independent constant vector, determined by a position condition.
Do not use the same constant in every component unless the data require it. Nor should the position constant be confused with the velocity constant. A position measurement alone does not usually determine the initial velocity. State the supplied conditions explicitly before using them, and substitute them into the appropriate stage of the calculation.
For constant acceleration a, integration gives v(t) = u + ta and r(t) = r₀ + tu + (t2/2)a, when the initial conditions are at t = 0. These vector equations apply componentwise. If acceleration varies with time, integrate its actual expression instead of importing a constant-acceleration shortcut.
Worked example 1: Recover a three-dimensional motion
An object has acceleration a(t) = (6t, −4, 2), initial velocity (1, 3, −1), and initial position (2, −1, 0), for t ≥ 0. Use metres and seconds. Find its position function and speed at t = 1.
Integration gives v(t) = (3t2 + C₁, −4t + C₂, 2t + C₃). The initial velocity sets the constants to 1, 3 and −1. Thus v(t) = (3t2 + 1, 3 − 4t, 2t − 1).
Integrating again and imposing the initial position gives r(t) = (t3 + t + 2, 3t − 2t2 − 1, t2 − t). At t = 1 the position is (4, 0, 0), while velocity is (4, −1, 1). The speed is √(16 + 1 + 1) = 3√2 metres per second.
Differentiate the recovered position twice to check the acceleration, then separately substitute t = 0 into position and velocity. Derivative checks alone would not detect wrong constants. Here the object can never stop because its first velocity component 3t2 + 1 is always positive.
3. Speed, displacement and distance are different outputs
The displacement over an interval is r(b) − r(a), a vector joining the initial and final positions. Its magnitude is the straight-line separation of those positions. Distance travelled follows the actual path and equals the integral of speed, ∫ab|v(t)| dt, for a ≤ b. This can exceed the displacement magnitude, particularly when the path curves or retraces itself.
Average velocity is displacement divided by elapsed time. Average speed is distance divided by elapsed time. Even if an object returns to its starting point, it can have travelled a positive distance: average velocity is then zero while average speed is positive. Taking a magnitude after integrating velocity does not generally recover total distance.
Differentiate |v|2 = v · v to get d(|v|2)/dt = 2v · a. Where speed is positive, its rate of change is (v · a)/|v|. A positive dot product means increasing speed and a negative one decreasing speed. Perpendicular velocity and acceleration give zero instantaneous rate of speed change, even though direction may change.
At a zero-velocity time, do not divide by speed. Investigate the speed function directly. Also, a zero dot product at one instant does not prove that speed is constant over an interval. Constant-speed motion requires the corresponding condition throughout that interval.
4. Projectile motion is a vector model
Choose horizontal x and upward vertical y. A simple projectile model uses constant acceleration (0, −g), with g > 0 supplied or chosen as stated in the problem. It assumes the horizontal acceleration is zero and the downward acceleration remains constant. These are modelling assumptions, so do not silently add resistance or other effects.
If launch velocity has magnitude U at angle θ above the positive horizontal, its components are (U cos θ, U sin θ). From launch position (x₀, y₀), the position is (x₀ + U cos θ t, y₀ + U sin θ t − gt2/2). A height equation determines landing time, and that time must then be used in the horizontal coordinate.
The highest point occurs when vertical velocity is zero, provided that time belongs to the flight interval. The total velocity usually remains nonzero there. If landing height differs from launch height, the familiar symmetric-flight shortcut does not apply. Solve the actual height equation and select the physically valid root.
Worked example 2: A launch from above ground
A projectile starts at (0, 5) metres with velocity (12, 16) metres per second. Use acceleration (0, −10) metres per second squared. Find its maximum height, ground-impact time and horizontal range.
The position is r(t) = (12t, 5 + 16t − 5t2) and velocity is (12, 16 − 10t). At the highest point 16 − 10t = 0, so t = 8/5 seconds. The height is 5 + 128/5 − 64/5 = 89/5 metres, or 17.8 metres. The speed there is 12 metres per second, not zero.
Ground impact requires 5 + 16t − 5t2 = 0. The roots are (8 ± √89)/5. Since √89 > 8, the minus root is negative and lies before launch. The impact time is T = (8 + √89)/5 seconds, approximately 3.49 seconds.
The range is 12T = (96 + 12√89)/5 metres, approximately 41.8 metres. The physical flight domain is 0 ≤ t ≤ T. The maximum-height time lies inside it. As a check, substituting T makes the height zero and gives downward vertical velocity −2√89 metres per second.
5. Collisions combine position and timing
For two objects sharing a common time variable, a collision requires equality of every position component at the same permitted time. Solve the simplest component first, then test the others. If the problem contains an unknown parameter, use the remaining coordinate equations or stated velocity conditions to determine it.
Path crossings alone use independent parameters and can occur at different times. A velocity condition cannot replace a position condition: two objects may have perpendicular velocities while far apart. The required event must satisfy all conditions simultaneously. If the question asks for perpendicular velocities, verify that both velocities are nonzero at the event.
Relative position rA − rB is a convenient way to organise the equations. A collision makes it zero. Its derivative is relative velocity. For a closest-approach investigation, minimise the squared magnitude of relative position over the allowed time interval, including endpoints. A minimum separation need not be zero.
Worked example 3: Collision with perpendicular velocities
For t ≥ 0, let rA(t) = (t, t2, 0) and rB(t) = (2 − t, 1 + k(t − 1), 0). Find the collision time and the value of k that makes the velocities perpendicular there.
Equality of x-coordinates gives t = 2 − t, hence t = 1. At that time both y-coordinates equal one, regardless of k, and both z-coordinates are zero. The objects therefore collide at (1, 1, 0) at time one.
The velocities are vA(t) = (1, 2t, 0) and vB(t) = (−1, k, 0). At collision their dot product is −1 + 2k. Setting it to zero gives k = 1/2. Both velocities are nonzero, so the right-angle interpretation is valid. The position equations establish the event; the dot-product equation selects its velocity geometry.
6. Exact algebra, technology and model checks
Use symbolic differentiation to verify a position function and symbolic substitution to test proposed collision times. For numerical roots, inspect the permitted interval and check every component residual. A graph showing two paths crossing does not show a collision unless it also establishes matching times.
When a speed integral is not elementary, state the exact definite integral before giving a numerical approximation. Ensure the numerical routine sees the non-negative speed, not a single signed velocity component. Increasing numerical precision can test stability of an approximation, but it does not repair an incorrect integrand or domain.
A useful algorithm scans a permitted time interval for candidate roots of a simple relative-position component, refines each candidate and checks the other components. It must also inspect interval endpoints and special cases where the chosen component vanishes identically. For this reason, a single-component sign-change search is a diagnostic aid rather than a universal collision proof.
Practise, then check
- For r(t) = (3 cos t, 3 sin t, 2t), find velocity, acceleration and speed.
- Given v(t) = (2t, 3, −1) and r(0) = (1, −2, 4), find position.
- An object has velocity (t − 2, t2 − 4). When is it stationary for t ≥ 0?
- For r(t) = (4 cos t, 4 sin t), explain why nonzero acceleration is consistent with constant speed.
- A projectile is launched horizontally from (0, 20) metres at 10 metres per second, with acceleration (0, −10). Find impact time, range and impact velocity.
- For t ≥ 0, A is at (t, 2t) and B at (6 − 2t, 1 + kt). Find k for a collision and give its time.
- For r(t) = ((t − 1)2, 0), 0 ≤ t ≤ 2, find displacement and distance travelled.
Fully reasoned answers
- Velocity is (−3 sin t, 3 cos t, 2), and acceleration is (−3 cos t, −3 sin t, 0). Speed is √(9 sin2t + 9 cos2t + 4) = √13. The trigonometric identity shows it is constant for every t.
- Integration gives (t2 + C₁, 3t + C₂, −t + C₃). The initial position yields r(t) = (t2 + 1, 3t − 2, 4 − t). Differentiating and substituting t = 0 verify both requirements.
- The first component is zero only at t = 2; the second is then also zero. Thus the object is stationary at t = 2. Solving only the second component would also give −2, but that fails the first component and the time domain.
- Velocity is (−4 sin t, 4 cos t), with constant magnitude four. Acceleration is (−4 cos t, −4 sin t), also of magnitude four, and is perpendicular to velocity. It changes the velocity's direction while its magnitude stays constant.
- Position is (10t, 20 − 5t2). Ground contact gives t = 2 seconds after rejecting −2. The range is 20 metres, and velocity at impact is (10, −20) metres per second. The downward component is negative because y was chosen upward.
- Equality of x gives t = 6 − 2t, hence t = 2. Equality of y then requires 4 = 1 + 2k, so k = 3/2. Both position vectors become (2, 4), confirming the collision.
- The initial and final positions are both (1, 0), so displacement is (0, 0). Speed is 2|t − 1|. Its integral from zero to one is one and from one to two is one, giving total distance two. The object travels to the origin and back.
Review path domains and timing or perpendicularity and projections when setting up a motion constraint.
Curriculum and sources
VCE Specialist Mathematics Units 3 and 4. Scope follows the VCAA Mathematics Study Design 2023–2027, Specialist Mathematics Area of Study 5: Space and measurement, with Area of Study 1 proof connections as applicable, and the current Specialist Mathematics FAQ. Unit 2 vector and conic foundations are retained where required. Curriculum checked 15 October 2026. Teaching reference: Cambridge Senior Mathematics VCE Specialist Mathematics 3&4, second edition (2023), sections 13C–13D, pp. 594–607. All explanations, examples, practice questions and diagrams are original. Independent teaching material; not a VCAA publication.
