Visual walkthrough — Power — average and instantaneous, units

This page is the visual companion to the parent Power note. If you have not met the dot product yet, keep Dot product of vectors open in another tab — we rebuild the piece we need here anyway.

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Step 1 — What "power" is asking

WHAT. Before any symbols: power is the answer to one question — how fast is energy being moved from one place to another?

WHY. Two movers carry identical loads up identical stairs. They do the same work (the same job). But if one takes 5 seconds and the other 50 seconds, we feel that the fast one is "stronger." That feeling is exactly what power measures — not the size of the job, but the speed of the job.

PICTURE. In the figure the two movers climb the same height. The magenta mover finishes in a short time; the violet mover dawdles. Same work, different power.

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Step 2 — Turning "per unit time" into a fraction

WHAT. We give names to the two things in our question. Let $W$ be the work done (energy handed over), measured in joules ($\text{J}$). Let $\Delta t$ be the time interval it took, measured in seconds ($\text{s}$). The little triangle $\Delta$ (Greek "delta") simply means "a chunk of" — here, a chunk of time.

WHY. "Per" in everyday English becomes "divide by" in maths. Miles per hour is miles $÷$ hours. Energy per second is energy $÷$ seconds. So the rate is a fraction.

$$P_{\text{avg}}=\underset{\text{energy per second}=\text{power}}{\underbrace{\frac{\stackrel{\text{energy handed over, in J}}{\overbrace{W}}}{\underset{\text{time it took, in s}}{\underbrace{\Delta t}}}}}$$

PICTURE. The bar shows the same energy $W$ split over a long interval (small height = low power) versus a short interval (tall spike = high power). Same area of energy, very different rate.

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Step 3 — From the whole interval to a single instant

WHAT. $P_{\text{avg}}$ smears out the answer over the whole interval. But power can change mid-journey — an accelerating car delivers more power at high speed than at low speed. To catch the value at one instant, we shrink the interval until it is practically a point.

WHY this tool — the limit. We use a limit because "at an instant" means "over an interval of length zero," and you cannot literally divide by zero. The limit is the trick that lets us ask "what value is this fraction heading toward as $\Delta t$ gets tinier and tinier?" It answers exactly the question a plain fraction cannot. (Same move you saw for instantaneous speed in Velocity and instantaneous rate (Kinematics).)

$$P_{\text{inst}}=\underset{\Delta t\to 0}{\lim }\frac{\Delta W}{\Delta t}=\underset{\text{"rate of work right now"}}{\underbrace{\frac{dW}{dt}}}$$

Here $\Delta W$ is the tiny sliver of work done during the tiny time $\Delta t$. When both shrink to zero together, we write them as $dW$ and $dt$ — the symbol $\frac{dW}{dt}$ just is "the instantaneous rate of $W$."

PICTURE. The work-versus-time curve bends. A wide secant slice gives the average slope; as we zoom the slice shut, it snaps onto the tangent line at that instant — its steepness is $P_{\text{inst}}$.

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Step 4 — Bringing in force and displacement

WHAT. We now open up the $dW$ hiding in $\frac{dW}{dt}$. From Work — definition and W = F·d cosθ, the tiny work done when a force pushes a thing through a tiny displacement is:

$$dW=\vec{F}\cdot d\vec{s}$$

WHY. The arrow on $\vec{F}$ and $d\vec{s}$ means these are vectors — they have a direction, not just a size. The dot $\cdot$ is the dot product, which we unpack next. We choose $dW$ (a tiny work) because we are about to divide by a tiny time $dt$, and only tiny-over-tiny survives the limit cleanly.

PICTURE. A force arrow $\vec{F}$ acts on a block; the block slides a small step $d\vec{s}$. The angle between them is $\theta$ (Greek "theta" — our name for that angle).

• $\vec{F}$ — the push, in newtons ($\text{N}$), with a direction.
• $d\vec{s}$ — the tiny step the object moves, in metres ($\text{m}$), with a direction.
• $\theta$ — the angle between the push and the step.

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Step 5 — What the dot product actually does

WHAT. The dot product $\vec{F}\cdot d\vec{s}$ means: take only the part of the force that lies along the direction of motion, and multiply it by the length of the step.

WHY this tool — the dot product. Only the component of the force along the motion transfers energy. A sideways push does no work. The dot product is the exact machine that keeps the useful (along-motion) part and throws away the sideways part. In symbols:

$$\vec{F}\cdot d\vec{s}=\underset{\text{size of push}}{\underbrace{F}}\underset{\text{length of step}}{\underbrace{(ds)}}\underset{\text{keeps only the along-motion fraction}}{\underbrace{\cos \theta }}$$

The $\cos \theta$ is the "fraction that points the right way." When $\theta =0$ the push is dead-on the motion, $\cos 0=1$, all of it counts. When $\theta =90^{\circ }$ the push is sideways, $\cos 90^{\circ }=0$, none of it counts.

PICTURE. We drop the force arrow onto the direction of motion (dashed projection). The magenta shadow is $F\cos \theta$, the only part that does work.

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Step 6 — Divide by time: the master formula appears

WHAT. Take the tiny-work equation $dW=\vec{F}\cdot d\vec{s}$ and divide both sides by the tiny time $dt$:

$$\frac{dW}{dt}=\vec{F}\cdot \frac{d\vec{s}}{dt}$$

WHY. The left side we already named: it is $P_{\text{inst}}$ from Step 3. On the right, $\frac{d\vec{s}}{dt}$ is "tiny step per tiny time" — that is exactly the definition of velocity $\vec{v}$. The force $\vec{F}$ is not changing during this instant, so it rides along outside untouched.

$$P_{\text{inst}}=\vec{F}\cdot \vec{v}$$

Term by term:

$$P_{\text{inst}}=\underset{\text{the push (N)}}{\underbrace{\vec{F}}}\cdot \underset{\text{how fast + which way it moves (m/s)}}{\underbrace{\vec{v}}}=\underset{\text{scalar form}}{\underbrace{Fv\cos \theta }}$$

PICTURE. Force arrow and velocity arrow on the moving block, with the projected magenta component $F\cos \theta$ and the resulting power read-out.

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Step 7 — Every case of $\cos \theta$ (this is where formulas trap you)

WHAT. The whole personality of the formula lives in $\cos \theta$. We must walk every angle so no scenario surprises you.

WHY. $\cos \theta$ ranges from $+1$ down through $0$ to $-1$, and each value tells a different physical story — energy in, none, or energy out.

$\theta$	$\cos \theta$	What it means physically
$0^{\circ }$	$+1$	Push fully along motion → maximum power in (engine driving)
between $0$ and $90^{\circ }$	between $0$ and $1$	Partial power in
$90^{\circ }$	$0$	Push perpendicular → zero power (circular motion, carrying a bag level)
between $90^{\circ }$ and $180^{\circ }$	negative	Power out — force drains energy
$180^{\circ }$	$-1$	Push exactly opposes motion → maximum power drained (brakes, friction)

PICTURE. Four blocks, one per case: aligned ($+$), perpendicular ($0$), opposing ($-$), and the degenerate $v=0$ case.

Negative power is not a mistake — it means energy is leaving the object (a car braking: friction points backwards relative to velocity, $\theta =180^{\circ }$, power is negative, kinetic energy drains, see Kinetic Energy and Work-Energy Theorem).

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The one-picture summary

Everything above collapses into a single flow: energy per time → shrink to an instant → open up the work as force-through-displacement → the displacement-per-time is velocity → the master formula, whose sign is governed by $\cos \theta$.

Recall Feynman retelling — say it to a 12-year-old

Power is "how fast you spend energy." Write it as a fraction: energy on top, seconds on the bottom. If the fast-or-slow changes as you go, zoom in on one split-second — that zoomed-in fraction is the tangent slope of the work-vs-time curve. Now unpack the tiny bit of energy: it's the push times the tiny step you moved, but only the part of the push that points where you're going (that's the $\cos \theta$ shadow). Divide by the split-second: the tiny step per split-second is just your speed. So power = push × speed × (how aligned they are). If you push a wall, your speed is zero, so no matter how hard you push, power is zero. If you push backwards against motion (like brakes), the alignment goes negative and you're pulling energy out.

Recall Rebuild the formula from scratch

Start from average power ::: $P_{\text{avg}}=W/\Delta t$ Shrink to an instant with a limit ::: $P_{\text{inst}}={\lim }_{\Delta t\to 0}\Delta W/\Delta t=dW/dt$ Substitute tiny work ::: $dW=\vec{F}\cdot d\vec{s}$ Divide by $dt$ and recognise velocity ::: $P=\vec{F}\cdot (d\vec{s}/dt)=\vec{F}\cdot \vec{v}$ Scalar form ::: $P=Fv\cos \theta$

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Connections

• Work — definition and W = F·d cosθ — supplies $dW=\vec{F}\cdot d\vec{s}$, the seed of Step 4.
• Dot product of vectors — the machine in Step 5 that keeps only the along-motion push.
• Velocity and instantaneous rate (Kinematics) — the limit trick and $d\vec{s}/dt=\vec{v}$.
• Kinetic Energy and Work-Energy Theorem — negative power drains kinetic energy (braking case).
• Energy conservation and efficiency — power in vs power out over a full path.

Concept Map