Graphs — x-t, v-t, a-t; areas and slopes meaning: Visual walkthrough

Step 0 — The three words we will use (drawn first)

Before any graph, let us fix three plain-English quantities and give them short names.

We build everything from one assumption: the acceleration $a$ stays constant the whole time. That is the case of uniformly accelerated motion — a ball rolling down a straight ramp, or a stone in free fall.

Step 1 — Draw the acceleration graph (the flattest possible picture)

Figure — Graphs — x-t, v-t, a-t; areas and slopes meaning

WHY we start here: acceleration is the simplest of the three when it's constant — a flat line. From a flat line we can climb up the ladder $a \to v \to x$ using areas, exactly as the parent's mnemonic promised.

PICTURE: the burnt-orange line is dead level. Under it, from time $0$ to time $t$ , sits a shaded rectangle. Notice its two sides: width $t$ (across), height $a$ (up). Hold that rectangle in mind — it is the whole of Step 2.

Step 2 — The rectangle's area is the velocity gained

WHAT we did: computed the area of the flat rectangle (area of a rectangle = height × width).

WHY this equals velocity gained: the parent note showed area under $a$ – $t$ accumulates the change in $v$ . Here is the intuition — each second you add $a$ metres/second to your speed; after $t$ seconds you have added it $t$ times, giving $a t$ . ( $Δ$ is just shorthand for "the change in".)

WHY that tool and not another? We chose area (not slope) because we are going up the ladder $a \to v$ . Slopes only take us down. The area of a flat rectangle needs no calculus at all — just length × width.

Now, "velocity gained" plus "velocity we started with" gives the velocity right now:

$v = start speed u + gained so far a t ⟹ v = u + a t$

PICTURE: the rectangle from Step 1 is now stacked on top of the starting level $u$ . The total height of the shaded region at time $t$ is exactly $v$ .

Step 3 — Turn "velocity now" into the velocity graph

WHY a straight line and not a curve: $v$ increases by the same amount ( $a$ ) every second — equal steps make a straight ramp. The slope of this line is $a$ (rise $Δ v$ over run $Δ t$ equals $a$ ), which is a nice self-check: sliding down the ladder $v \to a$ by taking the slope gives back our constant $a$ . Everything is consistent.

PICTURE: teal line climbing from $u$ on the left to $v = u + a t$ on the right. The gap between the line and the time axis is the region whose area we attack next — that area is the distance travelled.

Step 4 — The distance is the area under the velocity line

The clean way to get a trapezium's area is average height × width:

$Δ x = average of the two ends (\frac{u + v}{2}) \times width t$

Term by term: $u$ is the height on the left edge, $v$ the height on the right edge; their average is the typical height of the sloped roof; multiplying by width $t$ sweeps out the whole area.

$x = \frac{u + v}{2} t$

PICTURE: the trapezium is shaded, with its left side labelled $u$ , right side labelled $v$ , and a dashed line marking the average height halfway between them.

Step 5 — Cut the trapezium into a rectangle + a triangle (the punchline)

WHY this cut: it separates "distance you'd cover coasting at your start speed" from "extra distance because you sped up." Each piece is a shape we can measure with primary-school geometry.

The rectangle — bottom slab: $area_{rect} = height u \times width t = u t$ This is where you'd be if acceleration were zero — pure coasting.

The triangle — top wedge. Its base is $t$ ; its height is the velocity gained, which we found in Step 2 to be $a t$ : $area_{tri} = \frac{1}{2} \times base t \times height = Δ v a t = \frac{1}{2} a t^{2}$

Add the two pieces: $x = coasting part u t + speeding-up part \frac{1}{2} a t^{2} ⟹ x = u t + \frac{1}{2} a t^{2}$

PICTURE: the same trapezium, now split — orange rectangle below, plum triangle above, each labelled with its area. This single cut is the derivation. (This equals Step 4's answer once you substitute $v = u + a t$ ; see the integration viewpoint for the calculus route to the same shapes.)

Step 6 — Edge case A: starting from rest ( $u = 0$ )

$x = u t^{0} + \frac{1}{2} a t^{2} = \frac{1}{2} a t^{2}$

WHY it matters: this is the free-fall-from-rest formula (with $a = g$ ). The picture makes it obvious — no bottom slab, pure triangle. PICTURE: a single plum triangle rising from the origin; the missing rectangle is drawn as a faint dotted outline to show what vanished.

Step 7 — Edge case B: deceleration, then reversing ( $a < 0$ )

WHY the sign bookkeeping: area above the axis is positive displacement (going up); area below is negative displacement (coming down). The very same formula $x = u t + \frac{1}{2} a t^{2}$ handles this automatically, because $a$ carries a minus sign and eventually the negative $\frac{1}{2} a t^{2}$ overwhelms the positive $u t$ .

Sign rule (all cases): the object is slowing down exactly when $v$ and $a$ have opposite signs, and speeding up when they share a sign — see signs of vector components. This is why "negative velocity" ≠ "slowing down."

PICTURE: teal line starting positive, crossing zero, going negative; the upper triangle shaded orange (positive area) and the lower triangle shaded plum (negative area). If the two triangles are equal, net displacement is zero — the ball is back where it started, though the total distance (sum of the two areas' magnitudes) is not zero. That distance-vs-displacement split is the heart of Distance vs Displacement.

Step 8 — Numeric check on the pictures

The one-picture summary

Recall Feynman retelling — the walkthrough in plain words

Picture the speedometer of a car whose driver presses the pedal by the same tiny amount every second. On the acceleration graph that steady press is a flat ceiling; the box under it, so many seconds wide and so tall, is the extra speed you've picked up — that's why $v = u + a t$ . Now draw the speed itself: because it grows in equal steps it draws a straight ramp starting at your opening speed $u$ . The land under that ramp is the ground you cover. Slide a knife across the ramp at height $u$ : underneath sits a plain rectangle — the distance you'd have coasted at your starting speed, $u t$ — and on top rides a triangle — the bonus distance from speeding up, $\frac{1}{2} a t^{2}$ . Stack the two, $x = u t + \frac{1}{2} a t^{2}$ . Start from rest and the rectangle disappears, leaving only the triangle $\frac{1}{2} a t^{2}$ . Throw the ball upward and the ramp dips below the line: the land above counts as going up, the land below as coming down, and if they balance you land back where you began — having travelled plenty, displaced nothing.

Connections

Equations of Uniformly Accelerated Motion — the two boxed results derived here.
Differentiation and Integration in Physics — the calculus behind "area = integral".
Instantaneous vs Average Velocity — the average height $\frac{u + v}{2}$ is the average velocity.
Free Fall and Projectile Motion — Step 7's upward-throw picture with $a = - g$ .
Distance vs Displacement — why signed area ≠ total area.
Vectors — Components and Signs — reading the $+ / -$ of $v$ and $a$ .