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Sketching motion graphs by hand

What you will learn: turn a story into x–t, v–t and a–t sketches step by step, go backwards from a graph, and dodge the usual mistakes.

Some books call position d (distance along the track). It works the same as x here.

1. The three graphs and how they link

Slope of x–t = v. Steep = fast. Going up = moving in the + direction.
Slope of v–t = a.
Area under v–t = displacement (change in x). Area below the t axis counts as negative.
Area under a–t = change in v.

Memory trick: going down the stack (x → v → a) you take the slope. Going up the stack you take the area.

2. From a story to three sketches

Story: “A car starts from rest, speeds up for 3 s, cruises, then brakes to a stop.”

Pick a + direction and a start point. Write it on the page: “+ = forward, x = 0 at the start.”
Split time into intervals. Start a new interval each time the push changes: start, cruise, brake, stop. Draw light dashed lines down through all three graphs.
Decide a first for each interval. Sign: which way is v changing? Size: harder push or steeper ramp = bigger a. No numbers given? Pick easy ones. Draw a–t as flat steps.
Then v. Start at v₀. Slope of each piece = that a. a = 0 → flat. a + → rising. a − → falling.
Then x. Start at x₀. Slope = v. Pick the shape (see the key below): straight, concave up, or concave down.
Check continuity. x and v never jump (no gaps, no vertical lines). x–t has no sharp corners. Only a can jump.
Label key points. Times at the dashed lines, top speed, where v = 0, final x. Check one number with an area.

Interval	What happens	a	v	x shape
0–3 s	speeds up from rest	+ (pick 2 m/s²)	0 → 6 m/s	concave up
3–7 s	cruises	0	flat at 6 m/s	straight, slope 6
7–9 s	brakes	− (pick −3 m/s²)	6 → 0 m/s	concave down
9–10 s	stopped	0	0	flat

Braking in 2 s is harder than speeding up over 3 s, so the − step is taller than the + step.

Area checks: on the v–t graph the hatched triangle is ½ × 3 × 6 = 9 m, which matches x at 3 s. On the a–t graph the hatched box is 2 × 3 = 6 m/s, which matches v at 3 s.

Shape key for x–t

Ramps (no friction). On a straight ramp, a points down the slope and stays the same, so a–t is a flat step. Steeper ramp → bigger step. Flat floor → a = 0. A bend in the ramp → a jumps to a new step, but v and x carry on with no jump.

3. Going backwards: from v–t to a–t and x–t

Given the v–t graph in the middle, build the other two.

Mark split times where the v line changes slope, and where it crosses v = 0.
a–t: each straight piece of v becomes a flat step at its slope (rise ÷ run). Flat v → a = 0.
x–t direction: v above the axis → x goes up. v below → x goes down. v = 0 → x has a flat top or bottom (turnaround).
x–t bend: big |v| = steep. v heading toward 0 → x flattens out. v moving away from 0 → x gets steeper. (Bend up if a is +, down if a is −.)
x numbers: area under v in each interval = change in x. Add them up as you go.

Reading this v–t graph:

0–2 s: moving +, slowing down.
2 s: v = 0, turns around (x is at its top, 4 m).
2–4 s: moving −, speeding up. Back at x = 0.
4–6 s: steady −4 m/s.
6–8 s: moving −, slowing to a stop at −12 m.

Areas: +4, −4, −8, −4 m → x = 4, 0, −8, −12 m.

4. Common mistakes

5. Practice grid

Name:Date:

For each story: write your + direction, mark the intervals with dashed lines, then sketch a → v → x. Pick easy numbers if none are given.

Show answer key (shapes only)

a: 0, then a big + step, then a smaller + step. v: flat at 2 m/s, rises steeply, then rises gently. x: straight, then concave up, then still concave up but bending less. No jumps, no corners.
a: one negative step for the whole 6 s. It does not touch 0 at the top. v: one straight line from +3 down through 0 (at 3 s) to −3 m/s. x: a concave down hump, flat at the top at 3 s, back to the start at 6 s.
a: 0, then + (slowing a leftward puck), then 0. v: flat at −4 m/s, then rising to 0 at 4 s, then 0. x: straight line down from 10 m to 2 m, then concave up flattening out at −2 m, then flat.

Try this

A ball rolls down a ramp that gets less steep halfway down. What happens to the a–t graph at the bend? To v–t?
On a v–t graph the line is below the t axis and moving up toward zero. Is the object speeding up or slowing down? What is the sign of a?
A v–t graph is a straight line from 0 at t = 0 up to 8 m/s at t = 4 s. How far did the object move?