Physics I: Mechanics › Kinematics › full formula sheet

v = dx/dtSay it: velocity is the rate of change of position — how fast position changes, and in which direction

Velocity

Speed tells you how fast; velocity tells you how fast and where to — and the difference breaks naive answers.

v is velocity (m/s), x position (m), t time (s). The average over an interval is v̄ = Δx/Δt; the instantaneous value dx/dt is the limit as Δt shrinks to zero. Speed = |v|, the magnitude without direction.

Before this lesson: Average acceleration

Where it comes from

Your car’s odometer and speedometer disagree about what “how fast” means. The odometer accumulates distance — every meter counts, there and back. The speedometer reports velocity — how fast your position changes, with direction attached. Drive to the store and back home: the odometer says 6 km, but your velocity averaged over the trip is zero, because your position didn’t change.

Before reading on: you walk 3 km east, then 1 km west, in 1 hour. What is your average speed? Your average velocity? Which one can be zero?
average speed
=
distance / time = (3 + 1) km / 1 h = 4 km/h
Distance counts every step — the odometer’s view.
average velocity
=
Δx / Δt = (3 − 1) km / 1 h = 2 km/h east
Displacement is net position change: +3 − 1. The 1 km west undid part of the east.

That distinction — displacement vs. distance — is the entire content of average velocity. And the speedometer’s number is the next idea: what happens when the interval shrinks until “average” stops being an average at all.

Derivation

Start with the average over a finite interval, then shrink the interval toward zero. The limit is the instantaneous velocity:

v̄
=
Δx / Δt = [x(t + Δt) − x(t)] / Δt
Step 1 — the average. Net position change divided by elapsed time. This is exact for any interval, no approximations.
v
=
limΔt→0 [x(t + Δt) − x(t)] / Δt = dx/dt
Step 2 — shrink Δt. As the interval collapses to an instant, the average becomes the instantaneous velocity — what a perfect speedometer reads. This limit is exactly the derivative of position.

Geometrically: on an x–t graph, v̄ is the slope of the secant line between two points, and v is the slope of the tangent line at one point. Steeper tangent = faster motion right now.

How to use it

The procedure, every time:

  1. Decide: average or instantaneous? “Over the trip / between t = 2 and t = 5” → average, v̄ = Δx/Δt. “At t = 3 s / right now” → instantaneous, differentiate x(t).
  2. Compute displacement, not distance. Δx = xfinal − xinitial, with signs from your coordinate system.
  3. Attach the direction. In 1-D that is a sign; in words, “east” or “upward.” A bare number without direction is speed, not velocity.
  4. For x(t) given: v(t) = dx/dt — differentiate term by term.

Average velocity from a position function

x(t) = 4t + 3t²  ⇒  v(t) = 4 + 6tdifferentiate: d/dt[4t] = 4, d/dt[3t²] = 6tSay it: the velocity at any instant is the derivative of the position function

At t = 2 s: v = 4 + 6·2 = 16 m/s. The 3t² term makes velocity grow with time — accelerated motion, as we’ll see next.

Common mistake: answering a velocity question with distance/time and calling it done. If any backtracking happened, distance > |displacement|, and the answer is wrong in magnitude and missing direction.
Common mistake: reporting a negative velocity as a negative speed. Speed is |v| — always non-negative. v = −6 m/s means “6 m/s in the negative direction,” and the speed is 6 m/s.

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: 120 m east in 15 s

  1. List givens. Δx = +120 m (east positive), Δt = 15 s.
  2. Apply v̄ = Δx/Δt. = 120 / 15.
  3. Compute. = 8 m/s east.
  4. Check units and sense. m/s ✓; a brisk sprint, plausible.
Common mistake: writing “8 m/s” with no direction and calling it velocity. Direction is part of the quantity — without it you reported speed.
Your turn — a runner covers 90 m north in 18 s. Average velocity?

Answer: 5 m/s north. 90/18 = 5 m/s, direction north.

Example 2 — the round-trip trap: 50 m east, then 50 m west, in 20 s

  1. Displacement. Δx = +50 + (−50) = 0 m. You ended where you started.
  2. Average velocity. v̄ = 0 / 20 = 0 m/s.
  3. Average speed. distance/time = 100/20 = 5 m/s — emphatically not zero.
  4. The lesson. Zero average velocity does not mean “stood still” — it means the position came back. Speed and velocity split exactly here.
Common mistake: answering “5 m/s east” by dividing distance by time. That is the speedometer’s average, not the velocity — and it cannot be zero for a trip that moved.
Your turn — you swim 40 m north then 40 m south in 16 s. Average velocity? Average speed?

Answer: 0 m/s; 5 m/s. Δx = 0 so v̄ = 0; distance 80 m / 16 s = 5 m/s.

Example 3 — from a position function: x(t) = 4t + 3t², at t = 2 s

  1. Differentiate. v(t) = dx/dt = 4 + 6t. (Why 6t? d/dt[3t²] = 3·2t = 6t — the power rule pulls the 2 down.)
  2. Evaluate at t = 2. v = 4 + 6·2 = 4 + 12 = 16 m/s.
  3. Interpret. At exactly t = 2 s the object’s position is changing at 16 m/s — the tangent slope of the x–t curve there.
  4. Check against the average. From t = 0 to t = 2: x(2) − x(0) = (8 + 12) − 0 = 20 m, so v̄ = 10 m/s. The instantaneous 16 m/s is larger — velocity is growing, which the +6t term already told us.
Common mistake: plugging t = 2 into x(t) (= 20 m) and calling it velocity. Position and velocity have different units (m vs m/s) — the unit check catches this instantly.
Your turn — x(t) = 5t − 2t². Velocity at t = 1 s?

Answer: 1 m/s. v(t) = 5 − 4t; at t = 1: 5 − 4 = 1 m/s.

Before reading on: a car’s velocity is −9 m/s (east positive). Is it moving east or west? What is its speed?

Example 4 — signs carry the direction: v = −6 m/s

  1. Read the sign. East is positive, so −6 m/s means 6 m/s west.
  2. Speed. |v| = 6 m/s — the minus never survives into speed.
  3. The trap this dodges. Two cars at +6 and −6 m/s have the same speed but opposite velocities — they are approaching each other at 12 m/s, not parked.
Common mistake: “the velocity is −6, so it’s slower than a car at +3.” Speeds are 6 and 3 — the negative one is faster. Compare magnitudes, not signed values, when you mean speed.
Your turn — with east positive, a car has v = −9 m/s. Direction and speed?

Answer: 9 m/s west. The minus flips the direction to west; speed = |−9| = 9 m/s.

Memorization tips

  • Say it aloud: “velocity is displacement over time, with direction attached.” If your answer has no direction, you found speed.
  • The round-trip test: any there-and-back has zero average velocity. If your computation says otherwise, you divided distance.
  • Speed = |v|. The absolute value strips direction; it can never be negative.
  • Average ↔ endpoints only: v̄ needs just the start and end positions. Everything in between is invisible to it.
  • Instantaneous ↔ derivative: see x(t)? Differentiate. The tangent slope is the velocity.
  • Signs are the steering wheel: in 1-D, + and − are the entire direction system. Set positive before you compute.

Final challenge

Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and Velocity 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 the difference between velocity and speed?

Velocity is displacement over time with direction attached (a vector, m/s); speed is distance over time with no direction (a scalar). A round trip has zero average velocity but nonzero average speed.

What is instantaneous velocity?

The velocity at a single instant: v = dx/dt, the limit of average velocity as the time interval shrinks to zero. It is what a perfect speedometer reads, and geometrically the slope of the tangent to the position-time graph.

Can average velocity be zero while moving?

Yes — whenever the trip ends where it started. Average velocity depends only on net displacement: there-and-back gives delta-x = 0, so v-bar = 0, no matter how fast you went.

Why is velocity negative?

The sign marks direction relative to your chosen coordinate system, not 'slowness'. v = -6 m/s means 6 m/s toward the negative axis. Speed, the magnitude, is always positive.

How do I get velocity from a position function?

Differentiate: v(t) = dx/dt. For x(t) = 4t + 3t^2, v(t) = 4 + 6t. Evaluate at the time of interest for the instantaneous velocity.

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