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1.1.14 · D5Measurement, Vectors & Kinematics

Question bank — Average velocity vs instantaneous velocity

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Before you start, two words we will lean on constantly, in plain language:

  • Displacement = the straight arrow from where you started to where you ended (a vector). See Distance vs displacement.
  • Distance = the total length of the wiggly path you actually walked (a plain positive number).

Average velocity is built from displacement; average speed is built from distance. Keep that split in your head and most traps collapse.

True or false — justify

Every prompt is a claim. Decide true/false, then give the reason — a bare verdict scores nothing.

If a car returns to its exact starting point, its average velocity over the trip is zero.
True. Average velocity is , and returning home means , so the whole ratio is zero no matter how fast you drove.
If average velocity over an interval is zero, the object never moved.
False. Zero average velocity only means net displacement is zero. The object can travel far and return; its average speed (distance over time) is then large.
Average speed can be smaller than the magnitude of average velocity.
False. Path length is always at least as long as the straight-line displacement, so average speed — never smaller.
The magnitude of average velocity equals the average speed for a ball thrown straight up and caught at the same height.
False. The ball reverses direction, so displacement (zero) is far shorter than distance (twice the peak height); but average speed is positive.
Instantaneous velocity is just average velocity computed over a very short time.
Partly, but the word "very short" is a trap. It is the limit as — no finite window gives it exactly. Any real interval still smears fast and slow bits together.
On a position–time graph, the secant line and the tangent line always have different slopes.
False. For motion at constant velocity the graph is a straight line, so every secant is the tangent — same slope everywhere. See Position-time graphs.
For uniform acceleration, average velocity over equals the instantaneous velocity at the midpoint time.
True. The position graph is a parabola; its chord slope matches the tangent slope at the midpoint — a genuine theorem, not a coincidence (see Equations of motion under uniform acceleration).
For non-uniform acceleration, average velocity still equals the instantaneous velocity at the midpoint.
False. The midpoint rule is a special property of quadratics. With a cubic or jerky motion the chord slope generally matches the tangent at some other interior time, not the midpoint.
Average velocity is the ordinary average of the start and end velocities.
False in general. That formula only holds under uniform acceleration, and even then it averages by time. The true definition is always .
If two trips share the same start and end points and the same duration, they share the same average velocity.
True. Average velocity depends only on and , both fixed by endpoints and time — the wiggly path between them is irrelevant.

Spot the error

Each line contains a piece of reasoning with a hidden flaw. Name the flaw.

"The car's average velocity was 60 km/h, so at the halfway time it was going 60 km/h."
Wrong for general motion — average velocity need not be hit at the halfway time. It is guaranteed only at the midpoint time and only under uniform acceleration.
"Speed is the size of velocity, so average speed must be the size of average velocity."
The error swaps the order of "take size" and "average". Average speed averages the sizes of instantaneous velocity (distance/time); takes the size after averaging the vector — they differ whenever direction reverses.
"At the top of a throw the ball is momentarily still, so its acceleration there is zero too."
Confuses velocity with its rate of change. At the peak , but gravity still acts, so . See Acceleration as derivative of velocity.
"To find instantaneous velocity I plugged into ."
You cannot divide by zero. The instantaneous value is the limit as ; you simplify the ratio first (while ), then let it shrink.
"The particle covered 200 m of road, so its displacement is 200 m."
Distance and displacement conflated. 200 m is path length; displacement is the straight arrow start-to-end and may be shorter (even zero).
"I averaged the speedometer readings taken every second and called that the average velocity."
Two errors: speedometer gives speed (no direction), and averaging readings weights by time samples, not displacement. Average velocity is strictly .
"The slope of the chord between two graph points gives instantaneous velocity there."
The chord (secant) slope gives average velocity over the interval. Instantaneous velocity is the tangent slope at a single point — the limit as the chord's ends slide together. See Slope and tangent in calculus.
"Since the object sped up then slowed to the same speed, its average velocity over that stretch is that common speed."
Ignores direction and the displacement definition. Average velocity is displacement over time; matching end speeds tells you nothing about it.

Why questions

Answer the "why" with mechanism, not restatement.

Why do we divide displacement by rather than just report the displacement?
To get a rate — position change per unit time — so trips of different durations become comparable. A big displacement over a long time can be a slow trip.
Why must average speed be greater than or equal to ?
The path length (numerator of speed) is a sum of little step lengths that can never be shorter than the straight arrow connecting the endpoints — the triangle-inequality idea.
Why does shrinking reduce the "blur" in a velocity estimate?
A long window lumps fast and slow moments into one average, hiding variation. A tiny window contains almost one instant, so the ratio reports the motion there rather than a mixture.
Why is instantaneous velocity a vector even though a speedometer shows only a number?
Because it is the derivative of the position vector, it inherits a direction — the way the object is heading at that instant. The speedometer just drops the direction and keeps the magnitude (speed).
Why does the secant line "become" the tangent as the two points merge?
As the interval shrinks, the chord pivots to rest along the curve at one point; the limit of the secant slopes is precisely the tangent slope, which is the derivative.
Why can a graph with a flat spot (horizontal tangent) mean the object is momentarily at rest?
A flat tangent has slope zero, and slope on a position–time graph is velocity — so there.
Why does average velocity ignore how wild the path was in between?
It uses only the endpoints via ; the interior of the journey never enters the formula.

Edge cases

The scenarios the definitions quietly cover — check you agree.

An object stays perfectly still for 10 s. What are its average and instantaneous velocities?
Both are zero throughout: gives , and the flat position graph has zero tangent slope everywhere.
An object at rest for 10 s — is its average speed also zero?
Yes. Distance travelled is zero, so distance over time is zero as well; here speed and agree because there is no reversal.
A particle moves right, stops instantaneously, then returns. What is at the exact turning instant?
Zero — the tangent to the position graph is horizontal there, even though the particle is only momentarily at rest and acceleration is nonzero.
What is the instantaneous velocity at a sharp corner (a kink) in the position–time graph?
Undefined — the left and right tangent slopes disagree, so no single derivative exists. Physically this means an instantaneous jump in velocity, which real bodies avoid.
Over a round trip taking time , what does the ratio (average speed)/ equal?
It is undefined — while average speed is positive, so the ratio divides a nonzero number by zero.
If displacement is nonzero but the object also backtracked partway, how do average speed and compare?
Average speed is strictly larger, because backtracking adds path length that does not add to the net displacement.
As the interval start and end times approach each other on a smooth graph, what does average velocity approach?
The instantaneous velocity at that shared instant — this limiting process is the very definition of the derivative .

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