Average velocity vs instantaneous velocity: Worked examples

Before we compute anything, one reminder in plain words. Position $r (t)$ = "where the thing is at time $t$ ." In one dimension we write just $x (t)$ , a signed number: positive means one way (say, right/east), negative means the other way. Displacement $Δ x = x (end) - x (start)$ is the net arrow from start to finish. Two operations give us velocity:

Speed is a different animal, and because Cases C, D, H, I lean on it, we pin it down now:

The scenario matrix

Every velocity problem is one (or a blend) of these case classes. The table lists them; each row is claimed by an example below.

#	Case class	What makes it tricky	Covered by
A	Motion in one direction, positive $x$	baseline — build intuition	Ex 1
B	Motion in one direction, negative velocity	sign of the answer	Ex 2
C	Reversal (turns around)	$∥ v_{a v g} ∥ \neq =$ avg speed; $v = 0$ at turn	Ex 3
D	Zero displacement (round trip)	$v_{a v g} = 0$ but avg speed $\neq = 0$	Ex 4
E	Degenerate: interval shrinks to a point	why the limit is needed at all	Ex 5
F	Instant $v$ from the definition (no shortcut)	proving the derivative	Ex 6
G	Midpoint theorem for uniform acceleration	$v_{a v g} = v (t_{mi d})$	Ex 7
H	Real-world word problem (multi-leg trip)	translating words → $Δ x, Δ t$	Ex 8
I	Exam twist: given $v_{a v g}$ , find unknown time	reverse-solving	Ex 9
J	Turning-point case, graph-read	flat tangent, sign flip of $v$	Ex 10

We lean on Slope and tangent in calculus, Position-time graphs, Distance vs displacement and Equations of motion under uniform acceleration throughout.

Example 1 — Case A: steady motion, positive direction

Figure — the position line, its secant over $[0, 5]$ (dashed coral), and the note that here secant = tangent, slope $4$ m/s:

Figure — Average velocity vs instantaneous velocity

Example 2 — Case B: negative velocity

Example 3 — Case C: motion that reverses

Worked example Statement

$x (t) = t^{2} - 4 t$ (metres) on $[0, 4]$ . Find $v_{a v g}$ on $[0, 4]$ , the total distance, the average speed, and the instant the particle turns around.

Forecast: the parabola dips down then comes back. Guess: will $∣ v_{a v g} ∣$ equal the average speed here?

$v (t) = \frac{d}{d t} (t^{2} - 4 t) = 2 t - 4$ . Why first? The turning point is where $v = 0$ .
$v = 0 \Rightarrow 2 t - 4 = 0 \Rightarrow t = 2$ s. Why? At the reversal the tangent is flat (see Slope and tangent in calculus).
Endpoints: $x (0) = 0$ , $x (4) = 16 - 16 = 0$ . So $Δ x = 0$ . Why does this matter? Net displacement is zero even though it moved a lot.
$v_{a v g} = \frac{0}{4} = 0$ m/s. Why zero? It ended where it started.
Distance: from $t = 0$ to $2$ it goes to $x (2) = 4 - 8 = - 4$ (moved $4$ m). From $t = 2$ to $4$ it returns to $0$ (another $4$ m). Total $= 8$ m. Why split at $t = 2$ ? Distance must add path lengths of each unidirectional leg.
Average speed $= \frac{8}{4} = 2$ m/s (using the definition above). Why differs from $v_{a v g}$ ? Speed uses path length; velocity uses net arrow.

Verify: $∣ v_{a v g} ∣ = 0 \leq 2 =$ avg speed ✓ (an instance of the universal law $∣ v_{a v g} ∣ \leq$ avg speed). Turning point at $t = 2$ : $x = - 4$ is the lowest point of the parabola ✓.

Figure — the parabola dipping to $x = - 4$ , the zero-slope secant over $[0, 4]$ (dashed), the flat tangent (mint) at the turn $t = 2$ , and the two equal $4$ -m legs (butter):

Example 4 — Case D: pure round trip (zero displacement)

Example 5 — Case E: the degenerate interval (why we need the limit)

Worked example Statement

For $x (t) = t^{2}$ , try to get the velocity at $t = 3$ by using $v_{a v g}$ on the interval $[3, 3]$ . What happens, and how does the limit rescue us?

Forecast: guess what $\frac{Δ x}{Δ t}$ does when the two times are the same.

Try $[3, 3]$ : $Δ x = x (3) - x (3) = 0$ and $Δ t = 3 - 3 = 0$ . So $v_{a v g} = \frac{0}{0}$ . Why is this useless? $0/0$ is undefined — a single instant has no interval to average over.
Instead shrink a tiny window $[3, 3 + h]$ , where $h$ is our small time gap (same role as $Δ t$ ). $x (3 + h) - x (3) = (3 + h)^{2} - 9 = 6 h + h^{2}$ . Why keep $h \neq = 0$ ? Only then can we divide.
$v_{a v g} = \frac{6 h + h ^{2}}{h} = 6 + h$ . Why cancel $h$ ? Legal because $h \neq = 0$ before the limit.
Let $h \to 0$ : $v (3) = 6$ m/s. Why now allowed to set $h = 0$ ? The awkward $0/0$ is gone; $6 + h$ is perfectly defined at $h = 0$ .

Verify: Shortcut $v (t) = 2 t$ , so $v (3) = 6$ ✓. The limit is precisely the tool that turns a forbidden $0/0$ into a real number — the whole reason instantaneous velocity is a derivative.

Figure — the parabola $x = t^{2}$ with three secants from $t = 3$ (coral $h = 1.2$ , butter $h = 0.6$ , mint $h = 0.2$ ) collapsing onto the dashed tangent of slope $6$ :

Example 6 — Case F: instantaneous velocity straight from the definition

Example 7 — Case G: the midpoint theorem (uniform acceleration)

Worked example Statement

Under constant acceleration $x (t) = 2 t^{2} + t$ (so $a$ is constant). Show that $v_{a v g}$ on $[2, 6]$ equals $v$ at the midpoint $t = 4$ .

Forecast: the parent note claimed this is a theorem, not luck. Predict $v_{a v g}$ and $v (4)$ separately, then compare.

$x (2) = 8 + 2 = 10$ , $x (6) = 72 + 6 = 78$ . Why? Endpoints.
$v_{a v g} = \frac{78 - 10}{6 - 2} = \frac{68}{4} = 17$ m/s. Why divide by 4? $Δ t = 4$ s.
$v (t) = \frac{d}{d t} (2 t^{2} + t) = 4 t + 1$ . Why linear? Constant acceleration $\Rightarrow$ velocity grows linearly.
Midpoint $t_{mi d} = \frac{2 + 6}{2} = 4$ . $v (4) = 16 + 1 = 17$ m/s. Why check the midpoint? For a linear $v (t)$ , its average over an interval sits exactly at the middle time.

Why the midpoint rule holds in general. Whenever velocity is linear, $v (t) = m t + c$ (here $m = 4$ , $c = 1$ ). Why start from a linear $v$ ? Constant acceleration means $v$ changes by equal amounts in equal times — that is precisely the graph of a straight line. Average velocity is the total displacement over $Δ t$ ; why bring in displacement? because $v_{a v g} = Δ x /Δ t$ by definition, and total displacement is the area under the $v (t)$ graph (this area-equals-displacement fact comes from Equations of motion under uniform acceleration — velocity accumulated over time is distance covered). So over $[p, q]$ , $v_{a v g} = \frac{1}{q - p} \int_{p}^{q} (m t + c) d t = \frac{1}{q - p} [\frac{m}{2} t^{2} + c t]_{p}^{q} = \frac{m}{2} (p + q) + c .$ Why this average-value formula? Dividing the area $\int_{p}^{q} v d t$ by the width $q - p$ is exactly "total displacement over total time" — the definition of $v_{a v g}$ written with an integral. Meanwhile $v$ at the midpoint is $v (\frac{p + q}{2}) = m \cdot \frac{p + q}{2} + c = \frac{m}{2} (p + q) + c$ — identical. So the equality is forced for any linear $v (t)$ , i.e. any uniform acceleration.

Verify: $17 = 17$ ✓, and the general formula gives $\frac{4}{2} (2 + 6) + 1 = 16 + 1 = 17$ ✓.

Example 8 — Case H: real-world multi-leg word problem

Worked example Statement

A commuter drives $12$ km east in $15$ min, stops $5$ min at a light (no motion), then $8$ km west in $10$ min. Find (a) average velocity and (b) average speed for the whole journey, in km/h.

Forecast: the stop adds time but no distance. Guess whether it raises or lowers the averages.

Net displacement $= + 12 - 8 = + 4$ km (east). Why signed sum? East $+$ , west $-$ .
Total time $= 15 + 5 + 10 = 30$ min. Why include the stop? Time keeps ticking even at rest.
Convert to hours by the ratio-of-units method: $30 min \times \frac{1 h}{60 min} = \frac{30}{60} h = 0.5 h$ . Why this trick? Multiplying by a fraction equal to $1$ ( $\frac{1 h}{60 min}$ ) cancels the "min" unit and leaves "h."
$v_{a v g} = \frac{4 km}{0.5 h} = 8$ km/h east. Why the direction? $Δ x > 0$ means net eastward.
Total distance $= 12 + 8 = 20$ km. Why add magnitudes? Path length (average-speed definition).
Average speed $= \frac{20}{0.5} = 40$ km/h.

Verify: $8 \leq 40$ ✓ (an instance of $∣ v_{a v g} ∣ \leq$ avg speed). The $5$ -min stop lowered both averages by inflating $Δ t$ while adding zero distance and zero displacement.

Example 9 — Case I: exam twist, solve for the unknown

Worked example Statement

A train has constant $v_{a v g} = 25$ m/s over a straight track and covers a displacement of $1500$ m. It then, on a second leg, must average $30$ m/s over $600$ m. Find each leg's time and the overall average velocity.

Forecast: overall average is not $(25 + 30) /2$ . Guess whether the true answer is above or below $27.5$ .

Leg 1 time: $t_{1} = \frac{Δ x}{v _{a v g}} = \frac{1500}{25} = 60$ s. Why rearrange? From $v = Δ x /Δ t$ , isolate $Δ t$ .
Leg 2 time: $t_{2} = \frac{600}{30} = 20$ s. Why same formula? Same definition, second leg.
Total displacement $= 1500 + 600 = 2100$ m; total time $= 60 + 20 = 80$ s. Why add each? Both legs run in the same direction, so their displacement arrows point the same way and simply add (no cancellation); time always accumulates leg after leg.
Overall $v_{a v g} = \frac{2100}{80} = 26.25$ m/s. Why not $27.5$ ? Averaging velocities is wrong; you must divide total displacement by total time, which weights each leg by the time it took, not equally.

Verify: $26.25$ m/s lies between $25$ and $30$ , closer to $25$ because more time was spent on the slower leg ( $60$ s vs $20$ s) ✓. This is exactly the [!mistake] the parent note warned about.

Example 10 — Case J: turning point read off a graph

Worked example Statement

A ball thrown up has $x (t) = 20 t - 5 t^{2}$ (metres, taking up as $+$ ). Find when $v = 0$ (the top), the velocity just before and just after, and confirm $v_{a v g}$ over the whole flight from launch to landing.

Forecast: at the very top the ball is momentarily still. Guess whether $v_{a v g}$ over the full up-and-down flight is zero.

$v (t) = \frac{d}{d t} (20 t - 5 t^{2}) = 20 - 10 t$ . Why? Slope of the position parabola.
Top: $v = 0 \Rightarrow 20 - 10 t = 0 \Rightarrow t = 2$ s. Why? Flat tangent = the reversal instant.
Just before ( $t = 1.9$ ): $v = 20 - 19 = + 1$ m/s (still rising). Just after ( $t = 2.1$ ): $v = 20 - 21 = - 1$ m/s (falling). Why signs flip? Direction reverses through the turning point, even though at $t = 2$ exactly $v = 0$ .
Landing: $x = 0 \Rightarrow 20 t - 5 t^{2} = 0 \Rightarrow t (20 - 5 t) = 0 \Rightarrow t = 4$ s. Why the other root? $t = 0$ is launch; $t = 4$ is landing.
Full-flight displacement $= x (4) - x (0) = 0 - 0 = 0$ , so $v_{a v g} = \frac{0}{4} = 0$ m/s.

Verify: $v$ changes sign $+ \to -$ passing through $0$ at $t = 2$ ✓ (dispelling the myth " $v$ can't be $0$ while moving"). $v_{a v g} = 0$ over the round-trip flight, matching Case D ✓.

Figure — height $x (t) = 20 t - 5 t^{2}$ vs time, the flat top marked at $t = 2$ ( $v = 0$ ), rising leg labelled $v > 0$ , falling leg $v < 0$ , launch and landing at $x = 0$ :

Recall Which cell trips people up most?

Case I (exam twist) ::: because the temptation to write $(v_{1} + v_{2}) /2$ is strongest there; the fix is total displacement over total time, which time-weights the legs. In Case C and Case J, what is the instantaneous velocity at the turning instant? ::: exactly $0$ (flat tangent), even though the object is otherwise moving. State the universal law relating the two averages. ::: $∣ v_{a v g} ∣ \leq$ average speed, because path length $\geq ∣ displacement ∣$ ; equality only when motion never reverses.

Connections

Average velocity vs instantaneous velocity (parent)
Position and displacement vectors
Distance vs displacement
Average speed vs instantaneous speed
Acceleration as derivative of velocity
Slope and tangent in calculus
Position-time graphs
Equations of motion under uniform acceleration