Equations of motion (SUVAT) — derivations from calculus: Visual walkthrough

Step 1 — What a velocity–time graph even is

WHAT. Draw two number lines crossing at a corner. The horizontal one measures time $t$ (seconds, running left to right). The vertical one measures velocity $v$ (metres per second, running upward). A single dot on this grid says "at this time, the object was moving at this speed".

WHY. Before we can talk about "area under a graph", we need the graph itself. Velocity is the thing we can draw as a curve because it changes as time passes — and a picture of that change is exactly a velocity–time graph.

PICTURE. In the figure, the blue dot sits at height $v_{0}$ above time $t_{0}$ . Read it like a map: go right to the time, go up to the speed.

Figure — Equations of motion (SUVAT) — derivations from calculus

Step 2 — Why constant acceleration draws a straight line

WHAT. Acceleration $a$ is "how much velocity you gain each second". Constant $a$ means you gain the same chunk of velocity every single second. Plot that: start at height $u$ (the initial velocity, our speed at $t = 0$ ), then each second the dot rises by $a$ . Connect the dots — you get a straight, tilted line.

WHY. A straight line is the simplest possible graph, and its steepness is the acceleration. We use $a$ = constant precisely because it makes the graph a line, and the area under a line is a shape we already know from primary school (rectangles and triangles). No line, no easy area.

PICTURE. The line starts at the yellow dot $(0, u)$ on the vertical axis and climbs steadily. Its steepness — how much it rises for each step right — is $a$ . A steeper line means bigger acceleration.

Step 3 — The claim: area under the line = displacement

WHAT. Displacement $s$ is how far the object has moved (with direction). The claim we will build, not assume, is: the shaded region between the line and the time axis equals $s$ .

WHY. Distance = speed × time. If speed never changed, this would just be one rectangle (speed tall, time wide) — and a rectangle's area is height × width = speed × time = distance. So area already is distance when speed is constant. The whole trick below is to handle the case where speed keeps changing.

PICTURE. The region under the sloped line is shaded green. We are going to measure its area two different ways and show both give the same $s$ .

Step 4 — Why "area = distance" even for changing speed (the thin-strip idea)

WHAT. Chop the time axis into very thin vertical strips, each of width $d t$ (a tiny sliver of time — so small the speed barely changes across it). Over one sliver the speed is basically constant at height $v$ , so that sliver's distance is height × width $= v d t$ . Add up every sliver to get the total.

WHY. This is the single idea behind integration: a curvy area is impossible to grab all at once, but a rectangle is trivial. So we cut the impossible shape into millions of trivial rectangles and sum them. Adding infinitely many infinitely thin strips is exactly what the integral sign $\int$ means.

PICTURE. One highlighted red strip of width $d t$ sits under the line at height $v$ . Its tiny area $v d t$ is one crumb of the total distance; the sum of all crumbs fills the green region of Step 3.

Step 5 — Split the region into a rectangle + a triangle

WHAT. Instead of summing strips one by one, notice the shaded region is made of two simple pieces stacked:

a rectangle sitting on the bottom, height $u$ , width $t$ ;
a triangle perched on top, filling the gap between the flat top of the rectangle and the sloped line.

WHY. We already know these two shapes' areas by heart. This turns an integral into schoolroom geometry — and it makes the mysterious $\frac{1}{2} a t^{2}$ term appear as literally a triangle's area, which is the payoff of this whole page.

PICTURE. The rectangle is shaded blue (the "if speed had stayed $u$ forever" distance). The triangle is shaded yellow (the "extra distance the acceleration added"). Together they tile the green region exactly.

Step 6 — Measure each piece and add

WHAT. Now just plug in the dimensions from the picture.

Rectangle: height $u$ , width $t$ → area $= u \times t = u t$ .
Triangle: its base runs along the time axis with width $t$ ; its height is how much the velocity rose over that time, which is $a t$ (from $v = u + a t$ , the line climbed by $a t$ ). Triangle area $= \frac{1}{2} \times base \times height = \frac{1}{2} t (a t) = \frac{1}{2} a t^{2}$ .

WHY. Adding the two areas gives the total shaded region, which by Step 3 is the displacement $s$ . No calculus symbols needed once the shapes are named — the integral of Step 4 and the geometry of Step 5 give identical answers.

PICTURE. Each side of each shape is labelled with its length right where it lives: the rectangle's $u$ and $t$ , the triangle's base $t$ and height $a t$ .

Step 7 — Every case, checked on the picture

The picture must survive all inputs, so let us push every knob.

WHY do this. A derivation you can only trust for one friendly example is worthless. We test the degenerate and sign-flipped cases directly on the shapes.

Case $a = 0$ (no acceleration). The line goes flat, the triangle collapses to zero height. Only the blue rectangle survives: $s = u t$ . Correct — coasting at $u$ .

Case $u = 0$ (dropped from rest). The rectangle vanishes (zero height). Only the yellow triangle remains: $s = \frac{1}{2} a t^{2}$ . This is free fall with $a = g$ .

Case $a < 0$ (braking). The line slopes down. The triangle now sits below the flat top and its area is subtracted — it eats into the rectangle. So $s$ is less than $u t$ : you travelled less than if you'd held your speed. If you brake long enough the line crosses the axis; area below the axis is negative displacement (moving backward).

Case very large $t$ . The $\frac{1}{2} a t^{2}$ triangle grows like $t^{2}$ while the rectangle grows only like $t$ . Eventually the triangle dominates — acceleration always wins over head-start given enough time.

The one-picture summary

One figure holds the whole story: the tilted line $v = u + a t$ , the blue rectangle $u t$ , and the yellow triangle $\frac{1}{2} a t^{2}$ stacked to fill the area that is the displacement.

Recall Feynman retelling — the walkthrough in plain words

Draw how fast something is going, second by second, going up the page as time runs to the right. If it's pushed evenly, that drawing is a straight slanted line. Now here's the magic: the amount of floor space underneath that line is exactly how far the thing travelled. Why? Slice the floor into skinny vertical splinters — each splinter is "how fast × a blink of time" = a tiny distance — and pile them all up.

That floor space is just a box with a triangle sitting on it. The box is "where you'd be if you never sped up" = starting-speed times time = $u t$ . The triangle on top is the bonus from speeding up; a triangle is half a rectangle, its width is the time $t$ and its height is the extra speed gained $a t$ , so its area is $\frac{1}{2} t \cdot a t = \frac{1}{2} a t^{2}$ . Add box plus triangle: $s = u t + \frac{1}{2} a t^{2}$ . If there's no push, the triangle disappears. If you start from rest, the box disappears. If you're braking, the triangle turns into a bite taken out of the box. Same picture, every time.

Recall

On a $v$ – $t$ graph, area equals what? ::: Displacement $s$ . Why does the extra-distance term have $t^{2}$ ? ::: It's a triangle whose base and height both grow with $t$ , so its area grows as $t^{2}$ . What does the blue rectangle represent physically? ::: Distance covered at the constant starting speed $u$ , i.e. $u t$ . With $u = 0$ , what does $s = u t + \frac{1}{2} a t^{2}$ become and why? ::: $s = \frac{1}{2} a t^{2}$ — the rectangle has zero height, only the triangle remains. With $a = 0$ , what survives on the graph? ::: The line is flat, the triangle vanishes, leaving $s = u t$ .

Connections

Equations of motion (SUVAT) — derivations from calculus — the parent; this is its picture-proof.
Velocity-time graphs — the whole page lives on one.
Differentiation and Integration — thin strips summing to an area is integration.
Free fall and g — the $u = 0$ case, $s = \frac{1}{2} g t^{2}$ .
Projectile motion — apply this per-axis.
Vectors — remember $s, u, v, a$ carry sign/direction.