eduKateSG · Kinematics · Worked learning guide
Find the motion question you need
Start with the observer, build the relative vector, then decide whether two moving objects actually meet. Worked examples distinguish catching up from merely passing nearby.
1. Name the observer before choosing an equation
3. Solve a one-dimensional catch-up without losing signs
5. Find closest approach with the relative position vector
7. Separate point coincidence, clearance and uncertainty
9. Practise, check and transfer the method
Full chapter index · Practice and checking · Kinematics subject guide
A velocity becomes meaningful when we say what it is measured relative to. A passenger can walk forwards inside a vehicle while another observer sees a very different speed. The gold standard of kinematics is being able to connect those descriptions without changing the event being described.
This guide develops relative velocity and closest approach from coordinates. It is for students who can already work with displacement, time and simple vectors, and want to understand why a closing speed sometimes predicts a meeting and sometimes does not. The numerical cases are original teaching models.
Begin with The Gold Standard of Kinematics for distance, displacement and motion graphs. Here the new task is to track two objects within a shared clock and coordinate system. The model is classical, with speeds far below the speed of light; it does not cover rotating coordinate systems or navigation under real operating conditions.
Chapter index
Chapters 1–4
Chapters 5–8
Chapters 9–10
Imagine a small inspection trolley travelling along a straight gallery. Relative to the floor, the trolley moves east at 3 m/s. A marker on the trolley remains at a fixed position relative to the trolley. Someone who says the marker is stationary and someone who says it moves at 3 m/s can both be correct. Each statement answers a different measurement question.
A reference frame supplies coordinates and a clock for describing events. For these exercises, the frames use parallel, nonrotating axes and a shared time. We choose east as positive x and north as positive y. This choice is a convention, so a negative component means west or south; it does not mean the calculation has failed.
Write velocity labels in words before shortening them. “A relative to ground” and “A relative to B” are different quantities. Let v_AG mean A’s velocity relative to ground, and v_BG mean B’s velocity relative to ground. The relative velocity is v_AB = v_AG − v_BG. The second subscript identifies the observer.
The notation prevents a common error: adding two speeds merely because two moving things appear in a question. Subtraction follows from subtracting their positions. If r_AB = r_AG − r_BG, differentiating with respect to the same time gives the velocity relationship. There is a reason behind the rule.
A check before calculation
State the object, observer, axis direction and time interval in one sentence. For example: “Find the marker’s eastward velocity relative to the floor during the trolley’s steady motion.” If the question instead concerns the trolley driver, change the observer label before doing arithmetic. This small habit protects the meaning of the answer.
Let the moving frame’s origin have ground position X(t). An object at ground position x(t) has moving-frame coordinate x′(t) = x(t) − X(t), provided the axes remain parallel and use the same length scale. Differentiation gives v′ = v − V, where V is the moving origin’s ground velocity.
If V is constant, a′ = a. The observers disagree about velocity but agree about acceleration within this classical model. This is why an object can have a turning point in one inertial frame and no turning point in another. A turning point concerns velocity relative to that frame.
If V changes, a′ = a − A, where A is the frame origin’s acceleration. The coordinate relation still describes motion, but the accelerating frame is not inertial. Applying Newton’s laws there requires additional care. Kinematics can calculate the relative acceleration without pretending that the force description is unchanged.
Consider x = 2t² metres and X = 5t metres, for t in seconds. Then x′ = 2t² − 5t, v′ = 4t − 5 m/s and a′ = 4 m/s². In the ground frame the object moves in the positive direction for t > 0. In the moving frame it initially falls behind, becomes instantaneously level in velocity at t = 1.25 s, and then gains.
At that instant, being equal in velocity does not mean being at the same position. The relative position is −3.125 m. Position equality and velocity equality are separate conditions. Keeping them separate is central to the closest-approach method later.
A coordinate transformation is also not permission to mix data recorded at different times. Two location measurements must refer to the same instant before their difference can be interpreted as a separation vector. A delayed sensor report needs time alignment first.
Two carts travel along a straight teaching track. Cart A starts at x = 0 m and moves at 1.8 m/s. Cart B starts at x = 12 m and moves in the same direction at 1.2 m/s. These are ideal constant velocities, and neither cart changes direction during the interval.
Using A relative to B gives x_AB(0) = −12 m and v_AB = 1.8 − 1.2 = 0.6 m/s. Thus x_AB(t) = −12 + 0.6t. A meeting requires x_AB = 0, so t = 20 s. Ground positions verify the result: x_A = 36 m, and x_B = 12 + 24 = 36 m.
A speed-only shortcut gives 12 divided by 0.6, but the signed equation tells us why this is the appropriate closing speed. It also handles an impossible catch. If A instead moves at 1.0 m/s, x_AB = −12 − 0.2t, so the separation grows for future time. The algebraic zero occurs at t = −60 s, outside the stated interval.
A negative meeting time is not a request to report its magnitude. It says that the extrapolated straight lines meet in the past. Whether that past event was real depends on whether the motion model applied before t = 0. An equation can be extended farther than its evidence supports.
Now let B travel towards A at −1.2 m/s from x = 12 m. The relative velocity becomes 3.0 m/s and the meeting time becomes 4 s. Subtracting a negative component creates the familiar addition of approaching speeds. The vector rule has not changed; only B’s signed velocity has changed.
For every catch-up answer, substitute the time into both original position equations. This independent positional check is stronger than repeating the closing-speed arithmetic.
A model boat has speed 5 m/s relative to still water. A uniform current flows east at 3 m/s relative to the bank. We neglect changing current, wind and steering transients. The boat’s bank velocity equals its water-relative velocity plus the water’s bank velocity.
Suppose the boat points due north across a 40 m wide channel. Its bank velocity is (3, 5) m/s. The northward component, rather than the total speed, determines crossing time: 40/5 = 8 s. In those 8 s the boat drifts 24 m east. Its bank speed is √34 m/s, approximately 5.83 m/s.
Dividing 40 m by 5.83 m/s would be wrong because the 40 m width is northward displacement, while 5.83 m/s measures speed along the diagonal path. Components must match the displacement being calculated.
To arrive directly opposite the starting point, the boat needs a westward water-relative component of −3 m/s. With total water-relative speed 5 m/s, its northward component becomes √(25 − 9) = 4 m/s. Crossing then takes 10 s. The upstream heading removes drift while reducing progress across the channel.
Feasibility matters before a final angle
For an exactly opposite arrival with positive crossing speed, the boat’s available water-relative speed must exceed the current speed in this ideal uniform-current setup. At equality, all of the boat’s available velocity is spent opposing the current, leaving no northward progress. If the current is faster, no heading can cancel the eastward component.
This is an educational vector example, not a boating recommendation. Actual waterways require information that this two-dimensional constant-current model does not contain. The transferable skill is deciding which component controls the target condition, rather than treating the magnitude of a vector as the answer to every question.
A meeting in two dimensions requires both relative components to become zero at the same time. Having a positive closing component along one axis is insufficient. The clean method uses the squared separation, which avoids differentiating a square root.
Let r(t) = r₀ + ut, where r₀ is the initial relative position and u is the constant relative velocity. The squared distance is D²(t) = r₀·r₀ + 2(r₀·u)t + (u·u)t². If u is nonzero, its unconstrained minimum occurs at t* = −(r₀·u)/(u·u).
Consider r₀ = (30, 10) m and u = (−4, 1) m/s. The dot product r₀·u is −110 m²/s, and u·u is 17 m²/s². Therefore t* = 110/17 s, approximately 6.47 s.
At this time r = (70/17, 280/17) m. The minimum distance is √(4900/17) m, approximately 16.98 m. The objects approach, then recede, but never meet. Setting only the x component to zero would produce 7.5 s, when the y separation is 17.5 m. That instant is not the true closest approach.
Why the dot product becomes zero
At the minimum, r(t*)·u = 0. The remaining separation is perpendicular to the relative velocity. Continuing a little farther no longer reduces distance because the motion has passed the nearest point on its straight relative track.
This perpendicularity supplies a geometric check on the calculus. Substituting the calculated vector gives (70/17)(−4) + (280/17)(1) = 0. The numerical result and geometric interpretation agree. A student who can explain this condition has gained more than a formula for a single problem.
The expression for t* gives a minimum over the whole infinite mathematical line. A real question often asks about a finite future interval, such as 0 ≤ t ≤ T. If t* lies inside the interval, evaluate it. If it lies outside, compare the allowed endpoints. For this convex quadratic, that is equivalent to clamping t* to the interval.
In the preceding example, suppose the velocities apply only for the next 4 s. The unconstrained closest time is 6.47 s, outside the model window. At t = 4 s, r = (14, 14) m and D = √392 m, approximately 19.80 m. That is the smallest separation during those four seconds. Reporting 16.98 m would use motion after the evidence ends.
If r₀·u is positive, separation is already increasing at t = 0. The unconstrained t* is negative, so the future minimum occurs at the initial instant. If the dot product is zero, the current separation is already perpendicular to the relative track.
If u = 0, the division in the formula is undefined. Physically, the two objects have the same velocity, so the relative position remains constant. Their separation is the same at every time. If r₀ also equals zero, the point models coincide throughout the interval; otherwise they never meet.
An observation-window check is particularly useful when data come in stages. A trolley may turn at 5 s, or a person may change walking speed. Each stage needs its own initial relative position and relative velocity. Solve within that stage, keep only allowed times, and compare all candidate distances and stage endpoints.
The procedure remains simple because its scope remains visible. A constant-velocity equation should never silently continue through a change in direction or speed.
A zero separation concerns mathematical points. Physical objects occupy space. If two ideal circular footprints have radii R_A and R_B, the centre separation at possible contact is R_A + R_B. In that simplified geometry, comparing the minimum centre separation with the radius sum distinguishes a clear pass from overlap.
For example, a calculated minimum centre separation of 1.4 m is positive. That alone does not establish clearance if the two effective radii total 1.8 m. Conversely, describing the objects only as points discards the dimensions needed for that question.
Real shapes may rotate, deform or have appendages, so replacing every object with a circle is itself a model choice. Its conservatism depends on how the radius was selected. An enclosing circle and a roughly equal-area circle do not support the same clearance claim.
Measurements introduce another distinction. A location reported to the nearest metre is not an exact coordinate. A velocity estimated from a short interval is not a promise about the next minute. If the predicted minimum separation is comparable to the measurement uncertainty, a crisp numerical answer can overstate what is known.
For a learning exercise, perturb one initial component or velocity component and recalculate. This does not produce a formal confidence interval, but it reveals sensitivity. If a tiny change switches the conclusion from a clear pass to overlap, report that the conclusion depends strongly on the assumed inputs.
This article’s examples are not collision-avoidance systems. Their purpose is to develop disciplined reasoning: name the point or shape being represented, state the interval, retain units, and distinguish a calculated trajectory from verified future behaviour. Those habits transfer to robotics and tracking without treating a classroom model as an operational guarantee.
| Wrong move | Why it fails | Repair |
|---|---|---|
| Add every pair of speeds | Directions and observers may differ | Write both signed velocities in one frame |
| Use the diagonal speed for a channel width | Magnitude and component answer different questions | Match each displacement component to its velocity component |
| Set one relative coordinate to zero | A two-dimensional meeting requires both coordinates | Substitute the same time into x and y |
| Accept a negative meeting time | The event may lie outside the model interval | Check the allowed time domain |
| Treat minimum distance as clearance | Objects have dimensions and uncertain trajectories | State the geometry and uncertainty |
A useful correction starts at the first invalid decision. If a student has chosen the wrong observer, improving arithmetic later will not repair the model. Ask them to rewrite each velocity label in full. If the observer is consistent but the vector signs are wrong, redraw the axes before recalculating.
If the algebra is correct but the time is outside the allowed interval, preserve the valid derivation and repair the interpretation. This helps the student recognise that mathematics can be performed correctly on the wrong domain.
A short explanation test
Ask: “Two objects have equal speeds. Is their relative velocity zero?” The answer is only yes when their velocity vectors are equal in the same reference frame. Objects with equal speeds in opposite directions can have a large relative velocity.
Then ask: “At closest approach, must both objects be instantaneously at rest?” No. It is the component of relative motion along the line of separation that is zero at the minimum, for a smooth interior minimum. Each object may continue moving quickly.
A clear explanation should identify the quantity being set to zero. Vague statements about motion stopping often conceal confusion between individual speed, relative speed and rate of change of separation.
Exercise 1: catch or fall behind?
A starts at 2 m and moves at 1.5 m/s. B starts at 14 m and moves at 1.1 m/s on the same axis. Find a future meeting time and verify both positions.
The relative equation is −12 + 0.4t = 0, so t = 30 s. Both positions are 47 m. If A’s speed changes to 0.9 m/s, the relative position is −12 − 0.2t, so there is no future meeting under the model.
Exercise 2: a crossing with drift
A model craft moves north at 4 m/s relative to water while the current moves east at 1.5 m/s. The width is 24 m. Find crossing time, eastward drift and bank speed.
The answers are 6 s, 9 m and √18.25 m/s, approximately 4.27 m/s. The crossing time uses the northward component. To eliminate drift at the same water-relative speed, use westward component −1.5 m/s and northward component √13.75 m/s.
Exercise 3: closest approach inside an interval
Let r₀ = (8, 6) m and u = (−2, 0) m/s for 0 ≤ t ≤ 3 s. The unconstrained minimum is t* = 4 s, so it lies outside the interval. At t = 3 s, r = (2, 6) m and D = √40 m, approximately 6.32 m.
Transfer check: change the observer
Add the same constant velocity vector to both objects’ ground velocities. Their relative velocity is unchanged because the added vector cancels. Their closest-approach result is therefore unchanged within the same classical model and shared time interval.
The most valuable final line is not a bare number. Write: “The point models are closest after ___ seconds, within the stated interval, at centre separation ___.” If a question concerns meeting, state whether both coordinates agree at that time.
Frequently asked questions
Can two observers disagree about velocity without either making a mistake? Yes. A velocity is relative to a frame. Check the frame labels before deciding that values conflict.
Does relative speed always equal the sum of the two speeds? No. In one dimension it is the magnitude of the signed velocity difference. In more dimensions, subtract the vectors before taking the magnitude.
Does closest approach mean collision? No. It identifies the smallest separation in the selected motion model and interval. Contact needs dimensions; real predictions need further evidence.
Why minimise distance squared? Distance is nonnegative, so distance and its square have the same minima. The squared expression is a quadratic for constant relative velocity, making the stationary point straightforward.
Connected learning directory
Return to Kinematics for the underlying quantities. Use Linear Algebra to strengthen vector components and dot products, and Calculus for differentiation and minimisation.
The wider purpose is described in The Gold Standard of Education | Learning That Transfers into Life: an idea becomes useful when its conditions and checks travel with it. Continue through the subject root to the related measurement, curvature and changing-acceleration guides.
Sources and scope
The reference-frame and classical velocity-composition foundations were checked against OpenStax University Physics: relative motion. MIT OpenCourseWare’s kinematics sequence supplies a further-study route. Checked 22 October 2026. The exercises and closest-approach derivations here are original explanatory work; the examples are idealised and do not establish navigation or safety specifications.
Continue with connected motion guides
Sampled positions and hidden reversals