Physics I: Mechanics › Kinematics › full formula sheet

Δx = x − x₀Say it: displacement equals final position minus initial position — the net change, not the path taken

Displacement

Where you ended up relative to where you started — the single quantity every constant-acceleration equation is built on.

Δx is displacement (m), x the final position, x₀ the starting position. In the constant-acceleration form, v₀ is the initial velocity, a the acceleration, t the time. Signs follow your coordinate system.

Before this lesson: Velocity

Where it comes from

Walk 3 km east to a café, then 2 km west to a library. How far did you walk? 5 km — that’s distance. But where are you, relative to home? 1 km east — that’s displacement. Physics problems about motion ask the second question far more often than the first, because displacement is what the kinematic equations predict.

Before reading on: a hiker walks 4 km north, camps, then walks 4 km south back to the trailhead. What is the distance hiked? The displacement? Which one can be zero while the other isn’t?
distance
=
4 + 4 = 8 km
Every step counts. Distance never decreases along a trip.
Δx
=
x − x₀ = 0 − 0 = 0 km
Start and end coincide: net change zero. Displacement is blind to the path — it only sees endpoints.

Displacement is a vector in one dimension: a signed number. And when the motion is accelerated, displacement gains a second, richer form — the one that predicts where something will be after t seconds.

Derivation

The definition Δx = x − x₀ needs no proof — but the constant-acceleration form Δx = v₀t + ½at² earns its keep. It falls out of two facts we already own: displacement is average velocity times time, and under constant acceleration the average velocity is the midpoint of v₀ and v.

Δx
=
v̄ · t
Step 1 — displacement = average velocity × time. This is the definition of v̄ rearranged; it holds for any motion.
v̄
=
(v₀ + v) / 2
Step 2 — the midpoint. Constant acceleration means velocity rises in a straight line, so its average is exactly halfway between start and finish values. (This step fails if acceleration varies.)
Δx
=
½(v₀ + v)t = ½(v₀ + v₀ + at)t
Step 3 — substitute v = v₀ + at. The velocity-formula page gave us v = v₀ + at; put it in for the final v.
=
v₀t + ½at²
Step 4 — expand. ½·2v₀t = v₀t, ½·at·t = ½at². Done: displacement = “coasting distance” (v₀t) + “speeding-up bonus” (½at²). ∎

Sanity checks: if a = 0, we get Δx = v₀t (constant velocity — correct). If v₀ = 0, Δx = ½at² (Galileo’s famous t² law for falling bodies). Both collapse correctly.

How to use it

The procedure, every time:

  1. Choose: definition or motion form? “Positions given” → Δx = x − x₀. “v₀, a, t given; find where it ends up” → Δx = v₀t + ½at².
  2. Set the coordinate system first. Origin (x = 0) and positive direction; then x, x₀, v₀, a all get signs from it.
  3. Plug in with parentheses. The ½at² term: square t first, then multiply — ½·a·(t²), not (½at)².
  4. Read the sign of the answer. Negative Δx means “ended up on the negative side of the start” — it is information, not an error.

The three collapse cases

Δx = v₀t   (a = 0: steady cruising)no acceleration, no bonus term
Δx = ½at²   (v₀ = 0: starts from rest)Say it: displacement equals one half a t squared — Galileo’s falling-body law
Δx = v₀t + ½at²   (general)coasting distance plus the speeding-up bonus
Common mistake: using the position x where the formula wants displacement Δx = x − x₀. If the motion starts at x₀ = 5 m, “x = 12 m” gives Δx = 7 m, not 12 m. Always subtract where it started.
Common mistake: computing ½·a·t and then squaring everything: (½at)². The exponent belongs to t alone — square first, multiply after.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: v₀ = 2 m/s, a = 1 m/s², t = 4 s

  1. List givens. v₀ = 2 m/s, a = 1 m/s², t = 4 s. Constant acceleration ✓
  2. Coasting term: v₀t = 2 · 4 = 8 m.
  3. Bonus term: ½at² = ½ · 1 · 16 = 8 m.
  4. Add. Δx = 8 + 8 = 16 m. Half the displacement came from starting motion, half from speeding up.
Common mistake: writing ½·1·4 = 2 m — forgetting to square t. The t² is the whole reason falling objects accelerate through distance, not just across it.
Your turn — v₀ = 3 m/s, a = 2 m/s², t = 5 s. Displacement?

Answer: 40 m. v₀t = 15 m; ½at² = ½ · 2 · 25 = 25 m; total 40 m.

Example 2 — Galileo’s case: dropped from rest, t = 2 s

  1. List givens. v₀ = 0 (dropped, not thrown), a = g = 9.8 m/s² downward, t = 2 s.
  2. Collapse to Δx = ½at². v₀ = 0 kills the coasting term.
  3. Compute. ½ · 9.8 · 4 = 19.6 m.
  4. Sense-check. After 1 s it would be 4.9 m — four times the time gives four times the distance, the t² signature ✓.
Common mistake: using v₀ = 9.8 (“gravity starts it moving”). Dropped means released from rest: v₀ = 0. Gravity is the acceleration a, not a starting velocity.
Your turn — dropped from rest; how far in 3 s?

Answer: 44.1 m. ½ · 9.8 · 9 = 44.1 m — note 9× the 1-second distance, t² at work.

Example 3 — braking: v₀ = 20 m/s, a = −5 m/s², t = 4 s

  1. List givens. v₀ = 20 m/s, a = −5 m/s² (opposes motion), t = 4 s.
  2. Coasting term: 20 · 4 = 80 m — where it would have gone unbraked.
  3. Bonus term: ½ · (−5) · 16 = −40 m — distance the braking erased.
  4. Add. Δx = 80 − 40 = 40 m. Check with v = 20 − 5·4 = 0 m/s — it stopped exactly at 4 s, so 40 m is the full stopping distance ✓.
Common mistake: taking a = +5 (“it’s just 5”). Then Δx = 80 + 40 = 120 m — a car that travels farther because it braked. The sign on a is the physics.
Your turn — v₀ = 15 m/s, a = −3 m/s², t = 4 s. Displacement?

Answer: 36 m. 15 · 4 = 60 m; ½ · (−3) · 16 = −24 m; 60 − 24 = 36 m.

Before reading on: a ball is thrown straight up at 14.7 m/s. After 3 s, is it above or below the launch point — and by how much? Guess the sign before computing.

Example 4 — the symmetry surprise: thrown up at 14.7 m/s, t = 3 s

  1. List givens (up positive). v₀ = +14.7 m/s, a = −9.8 m/s², t = 3 s.
  2. Coasting term: 14.7 · 3 = 44.1 m upward.
  3. Bonus term: ½ · (−9.8) · 9 = −44.1 m.
  4. Add. Δx = 44.1 − 44.1 = 0 m — it is exactly back at the launch point. Not a coincidence: 14.7 = 1.5 × 9.8, so up-time equals down-time and the trip is symmetric.
Common mistake: answering “44.1 m above the ground” — confusing displacement with height gained. Δx = 0 means returned to start; the peak was halfway through the trip (we’ll compute peaks in the max-height lesson).
Your turn — thrown up at 19.6 m/s; displacement after 4 s?

Answer: 0 m. 19.6 · 4 = 78.4 m; ½ · (−9.8) · 16 = −78.4 m; sum zero — symmetric flight again (19.6 = 2 × 9.8).

Memorization tips

  • Say it aloud: “displacement is final position minus initial position.” Endpoints only — the path is invisible.
  • Read Δx = v₀t + ½at² as a story: “coasting distance plus the speeding-up bonus.” Each term has a job.
  • Collapse check: a = 0 → Δx = v₀t; v₀ = 0 → Δx = ½at². If your setup doesn’t collapse, recheck.
  • Square t first, multiply after: ½·a·(t²), never (½at)².
  • The 4× rule: double the time, quadruple the ½at² part. t² grows fast — falling for 3 s covers 9× the first second’s distance.
  • Displacement ≠ distance: 5 km walked can be 1 km of displacement. Use Δx when equations ask; use distance only for odometer questions.

Final challenge

Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and Displacement is yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What is displacement?

Displacement is the net change in position: delta-x = x - x0, final minus initial. Unlike distance, it is signed and path-independent — a round trip has zero displacement.

How is displacement different from distance?

Distance counts every step along the path (always positive, always growing). Displacement sees only endpoints: walk 5 km and end 1 km east of home, and your displacement is 1 km east.

When do I use delta-x = v0 t + 1/2 a t^2?

For motion with constant acceleration when you know the starting velocity, the acceleration, and the time — and want the net position change. If acceleration varies, this form fails.

Why does time appear squared?

Because velocity itself grows with time under constant acceleration: in each successive second the object covers more ground. The derivation shows it: substituting v = v0 + at into delta-x = 1/2(v0 + v)t produces the t^2 term.

Can displacement be negative?

Yes — the sign tells you which side of the start the object ended on, relative to your chosen positive direction. Negative displacement is information, not an error.

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