1.1.13 · D5Measurement, Vectors & Kinematics
Question bank — Position vector, displacement, distance
Vocabulary you must already own from the parent note: position vector (arrow from origin O to the point), displacement (arrow from start position to end position ), distance (length of the actual path walked).
One picture to hold in your head before you start. A curved/looping path and its straight shortcut, side by side:
The green feet-path (distance) is always at least as long as the blue bird-arrow (displacement). Every trap below is a variation on which one the question is secretly asking for.
True or false — justify
A closed loop always has zero displacement.
True — the end position equals the start position, i.e. , so , regardless of how long or twisty the loop was.
Zero displacement forces zero distance.
False — a round trip has but a large distance; only the endpoints matched, the feet still walked.
If distance equals the magnitude of displacement, the motion was a straight line without reversing.
True — the straight segment is the unique shortest path, so equality in pins the path to that straight line, traversed once, no backtracking.
Displacement can be larger in magnitude than the distance travelled.
False — the straight line is the shortest route, so can at most tie the distance, never exceed it.
Moving the origin changes the displacement.
False — ; the origin sits inside both and and cancels in the subtraction. The figure below shows this cancellation graphically. See Vector addition and subtraction.
Moving the origin changes the position vector.
True — is measured from the origin, so relocating O redraws the arrow. This is exactly why position depends on frame but displacement does not.
Distance can be negative in one-dimensional motion.
False — the "" you write in 1D encodes a direction on displacement; distance is a path length and stays always.
Displacement and distance are both scalars.
False — displacement is a vector (carries direction), distance is a scalar (a plain non-negative number).
Two people using different coordinate systems will report the same displacement for one journey.
True — displacement is origin-independent, so it is a physical quantity both observers agree on, even if their position vectors differ.
The magnitude of the position vector is the distance the object has travelled.
False — is the straight gap from the origin to the point right now; distance is the length of the whole path taken to get anywhere.
Why the origin cancels — see it, don't just say it. Two observers pick different origins and for the same journey A→B:
Both draw different position arrows, but the difference arrow (the red displacement A→B) is identical — the origin-to-origin shift is added and subtracted, leaving nothing behind.
Spot the error
The mislabelled walk below is the single most common trap. Study the sketch, then read the corrections:
"The runner did one lap of a 400 m track, so her displacement is 400 m."
Wrong — one full lap returns to the start, so displacement is ; the 400 m is the distance.
"She walked 3 m east then 4 m north, so her distance is m."
Wrong — 5 m is the displacement magnitude (straight A→C hypotenuse in the figure). The distance is m, the sum of the two green legs walked.
"Displacement is m, so the object is 5 m below zero distance."
Wrong — the "" is a direction along the axis, not a signed distance. Distance is m; there is no such thing as negative distance.
"Because m, the displacement must be zero."
Wrong — equal magnitudes only mean both points sit the same distance from O (on a circle of radius 5). Displacement is , which can be a long chord unless the points actually coincide.
"To find displacement I add the position vectors: ."
Wrong — displacement is the difference . Addition would give a meaningless origin-dependent arrow, not the start→end change.
"The path bends, but since it starts and ends at the same height, distance = displacement."
Wrong — equality needs a single-direction straight line. A bending path already breaks "shortest route", so distance strictly exceeds .
"Distance is a vector because it can point in any direction along the road."
Wrong — the road bends, so no single direction describes it; distance discards direction entirely and is a scalar.
Why questions
Why does the origin cancel in displacement but not in position?
Position is one arrow from O, so O defines it; displacement subtracts two such arrows and the shared O term drops out ( has no O left) — exactly the identical red arrow in the two-origin figure above.
Why is the straight-line path always the shortest between two points?
Any detour adds sideways travel that must later be undone, so the total length only grows — the straight segment is the one path with no wasted motion, giving ; the detour figure below shows the wasted sideways bulge.
Why do we bother with a reference point at all?
A raw coordinate like "at 3" is meaningless without stating 3 from where, in which direction; the origin and axes (Coordinate systems and unit vectors) supply that anchor.
Why can two very different journeys share the same displacement?
Displacement remembers only the two endpoints, so any pair of paths with the same start and end give identical , however differently the feet moved.
Why does average velocity use displacement while average speed uses distance?
Velocity is a vector built from the net directed change (displacement) over time; speed is a scalar built from total road covered (distance) over time, so a round trip has zero average velocity but nonzero average speed.
Why is computed with Pythagoras theorem and not by adding components directly?
The components are perpendicular legs; the straight arrow is their hypotenuse, so , not .
Why detours must lengthen — the sideways-bulge picture. A straight shortcut versus a bulging detour between the same two dots:
The detour spends length going sideways (orange) that the straight route never pays for; every unit of sideways bulge must eventually be walked back, so the green path can only be longer.
Edge cases
An object stays perfectly still for an hour. What are its distance and displacement?
Both are zero — no path was walked (distance ) and start equals end (displacement ).
An object walks 5 m east, then 5 m back west to the start. Distance and displacement?
Distance m (both legs count); displacement (endpoints coincide). This is the cleanest case where they disagree maximally.
On a purely straight one-way walk in 1D, how do distance and compare?
They are exactly equal — no bending, no reversing, so the inequality becomes equality.
An object ends at the origin after a long journey — is its final position vector zero?
Yes, (it sits on the origin); its distance travelled is nonzero, and its displacement is nonzero unless the start was also the origin, in which case too.
Can displacement be nonzero while every individual coordinate change looks "cancelled" in one axis?
Yes — e.g. return to the same but a different : but , so . Cancellation in one axis does not zero the whole vector.
A drone flies in full 3D from to m. What is its displacement magnitude?
Use all three axes: , so m — Pythagoras theorem extends unchanged into the third dimension.
In polar coordinates a point sits at radius , angle . Does displacement care which coordinate system you used to describe the endpoints?
No — displacement is the physical start→end arrow; Cartesian or polar are just two languages for the same two points, so (and its length) come out identical either way.
A particle moves along a semicircle of radius from one end of the diameter to the other. Distance vs displacement?
Distance (arc length); displacement (the straight diameter). Since , the inequality holds strictly, as it must for a curved path.
What is the smallest possible distance for a given nonzero displacement?
Exactly itself — achieved only by walking the straight line directly; any other path is longer.