Average velocity vs instantaneous velocity: Visual walkthrough

Step 1 — What a position–time graph even is

WHAT. Before we talk about slopes, we draw the thing whose slope we want. We put time $t$ on the horizontal axis and position $x$ on the vertical axis. A single dot on this picture says: "at this time, the object was at this place."

WHY. We plot position against time because velocity is "how much position changes per second." A graph of position vs time turns that idea — change-per-second — into something you can literally see as steepness. Steep = position changing fast = moving fast. Flat = position barely changing = slow or stopped.

PICTURE. Below, the warm curve is $x (t) = 3 t^{2} - 2 t$ . Read it left to right like a story: it dips slightly below zero early (the particle drifts backward first), reaches a lowest point, then climbs. Two moments are marked: $t = 1$ (position $x = 1$ ) and $t = 3$ (position $x = 21$ ).

Figure — Average velocity vs instantaneous velocity

See Position-time graphs for how these are read in general.

Step 2 — Pick two instants and draw the chord

WHAT. Choose two times, $t_{1} = 1$ and $t_{2} = 3$ . Mark the two curve points $P = (1, 1)$ and $Q = (3, 21)$ . Draw the straight line through them. That straight line is called the secant line (secant just means "a line cutting through two points of a curve").

WHY. Average velocity only ever cares about two moments — the start and the end. It never looks at what happened in between. A straight line between the two points is the perfect visual: it forgets the wiggle and keeps only "where you started, where you ended, how long it took."

PICTURE. The orange chord joins $P$ and $Q$ . Notice it slices right through the curve; it does not hug it.

Step 3 — The slope of that chord is average velocity

WHAT. Slope means "rise over run": how much you go up ( $Δ x$ , change in position) divided by how much you go across ( $Δ t$ , change in time).

v_{a v g} = rise over run \frac{Δ x}{Δ t} = \frac{x ( t _{2} ) - x ( t _{1} ) up: metres gained}{across: seconds passed t _{2} - t _{1}}

Term by term:

$Δ x = x (t_{2}) - x (t_{1})$ — the vertical gap between $P$ and $Q$ ; how far the position moved.
$Δ t = t_{2} - t_{1}$ — the horizontal gap; how long that took.
their ratio — metres per second — is exactly steepness, and steepness of an $x$ – $t$ line is velocity.

WHY this ratio and not something else? Because "velocity" literally means position-change per time. Rise ( $Δ x$ ) is the position change; run ( $Δ t$ ) is the time. Divide them and the units force the answer to be $m/s$ . No other combination gives a rate.

PICTURE. We draw the rise $Δ x = 20$ as a vertical plum bar and the run $Δ t = 2$ as a horizontal teal bar, making a right triangle whose hypotenuse is the chord. The slope of the chord = (plum bar) / (teal bar).

v_{a v g} = \frac{21 - 1}{3 - 1} = \frac{20}{2} = 10 m/s

Step 4 — Slide $Q$ toward $P$ : the chord starts to tilt

WHAT. Keep $P$ fixed at $t_{1} = 1$ . Now move the second point $Q$ closer: try $t_{2} = 2.5$ , then $2$ , then $1.5$ . Each new $Q$ gives a new, shorter chord with its own slope.

WHY. A chord over a wide interval blurs together fast and slow parts of the motion — its slope is an average that hides the journey. To find the velocity at the single instant $t_{1} = 1$ , we must stop letting the far-away endpoint pollute the answer. So we drag $Q$ toward $P$ and watch what the slope settles down to.

PICTURE. Three chords are drawn from the same $P$ , each ending at a closer $Q$ . As $Q$ slides in, the chords swing and their slopes change. Their slopes (computed with $v_{a v g} = Δ x /Δ t$ ) are:

$t_{2}$	$Δ t$	$x (t_{2})$	$Δ x$	slope $= Δ x /Δ t$
$3$	$2$	$21$	$20$	$10$
$2.5$	$1.5$	$13.75$	$12.75$	$8.5$
$2$	$1$	$8$	$7$	$7$
$1.5$	$0.5$	$3.75$	$2.75$	$5.5$

The slopes are marching downward toward a target value. Watch them: $10, 8.5, 7, 5.5, \dots$

Step 5 — The limit: the chord becomes the tangent

WHAT. We do algebraically what the eye already sees. Let the second time be $t + Δ t$ and let $Δ t$ shrink toward $0$ . This is the definition of the derivative — the limit of the secant slope.

v (t) = Δ t \to 0 lim \frac{x ( t + Δ t ) - x ( t ) shrinking rise}{shrinking run Δ t}

WHY a limit and not just "plug in $Δ t = 0$ "? If we set $Δ t = 0$ directly we get $\frac{0}{0}$ — meaningless, because both the rise and the run vanish together. The limit asks a subtler question: "what value is the ratio heading toward as the window closes?" That target exists even though the endpoint $Δ t = 0$ itself is forbidden.

Work it out for $x (t) = 3 t^{2} - 2 t$ :

$x (t + Δ t) = 3 (t + Δ t)^{2} - 2 (t + Δ t) = 3 t^{2} + 6 t Δ t + 3 Δ t^{2} - 2 t - 2 Δ t$ .
Subtract $x (t) = 3 t^{2} - 2 t$ : the numerator collapses to $6 t Δ t + 3 Δ t^{2} - 2 Δ t$ .
Divide by $Δ t$ (allowed because $Δ t \neq = 0$ before the limit): $6 t + 3 Δ t - 2$ .
Let $Δ t \to 0$ : the $3 Δ t$ term dies, leaving $v (t) = 6 t - 2$

Term by term in $6 t - 2$ : the $6 t$ says the velocity itself grows with time (the particle keeps speeding up); the $- 2$ is the leftover backward drift from the start.

At our point $t = 1$ : $v (1) = 6 (1) - 2 = 4 m/s$ — exactly the number the shrinking slopes $10, 8.5, 7, 5.5, \dots$ were closing in on.

PICTURE. The limiting line no longer cuts through the curve at two points — it just kisses it at $P$ . That kissing line is the tangent. Its slope is $v (1) = 4$ .

Step 6 — Degenerate case A: a straight-line graph (constant velocity)

WHAT. Suppose the position graph is already a straight line, e.g. $x (t) = 2 t$ . Then every secant, no matter how wide or narrow, lies exactly on that same line.

WHY it matters. This is the edge case where average and instantaneous velocity are identical for every interval. There is no wiggle to smear, so shrinking the window changes nothing. Secant slope $=$ tangent slope $= 2 m/s$ everywhere.

PICTURE. One clean line; a wide chord and a narrow chord drawn on it are indistinguishable in slope.

Recall When are average and instantaneous velocity equal for

all intervals? Exactly when the $x$ – $t$ graph is a straight line — i.e. constant velocity. ::: The secant and tangent coincide everywhere.

Step 7 — Degenerate case B: a flat tangent (a turning point)

WHAT. At the very bottom of our curve $x = 3 t^{2} - 2 t$ the graph momentarily stops going down and starts going up. Set $v (t) = 6 t - 2 = 0 \Rightarrow t = \frac{1}{3} s$ . At that instant the tangent is horizontal.

WHY it matters. A horizontal tangent has slope $0$ , so instantaneous velocity is zero even though the particle is only pausing, not permanently stopped. This is the "turning point" from the parent's mistakes list — the ball at the top of a throw. The particle was moving backward before, forward after, and for one instant its velocity is genuinely $0$ .

PICTURE. A flat teal tangent line resting on the lowest point of the curve; velocity $= 0$ there.

The one-picture summary

Everything at once: the fixed point $P$ , a family of secants collapsing onto the tangent, the little rise/run triangle for the average, and the flat tangent marking the turning point. Read it left to right as secant → shrink → tangent.

secant slope = v_{a v g} \frac{Δ x}{Δ t} slide Q into P Δ t \to 0 tangent slope = v (t) \frac{d x}{d t}

Recall Feynman: the whole walkthrough in plain words

Draw where you are against the clock — that's the curve. Pick two moments and join them with a ruler: the tilt of that ruler is your average velocity, rise (metres gained) over run (seconds passed). But that ruler cheats — it only looks at the start and end and ignores the middle. So keep the first pinned dot still and slide the second dot toward it. The ruler swings and its tilt settles onto one number. Slide all the way in and the ruler stops cutting the curve and just touches it — that touching line is the tangent, and its tilt is your speed at that exact tick: the instantaneous velocity, $\frac{d x}{d t}$ . Two special cases keep you honest: if the curve was a straight line all along, the ruler never had to swing — average equals instant everywhere. And if the touching line comes out perfectly flat, your speed is zero for one instant — that's the top of a throw, where you turn around.

Connections

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