2.2 — Instantaneous velocity and speed

Average velocity (§2.1) is a good summary of "how fast, overall" — but if your speed changes along the way, that number hides the details. A driver who covers 1000 m in 60 s might have:

All three have the same average velocity but very different behavior. To answer "how fast were you going right at t = 15 s?", we need a new concept: instantaneous velocity.

The idea of a limit: smaller, smaller, smallest

Suppose a small teaching robot moves along a track and we know its position (measured by high-speed encoders):

\[ x(t) = 2.5\, t^2\quad (\mathrm{m},\ t\ \text{in seconds}) \]

We want to know "how fast is it at t = 4 s?". Start by estimating with average velocity, and shrink the time interval:

interval [4, 4+Δt] Δt x(4) x(4+Δt) Δx v_avg = Δx/Δt
[4, 5] 1 40 62.5 22.5 22.5 m/s
[4, 4.1] 0.1 40 42.025 2.025 20.25 m/s
[4, 4.01] 0.01 40 40.2003 0.2003 20.03 m/s
[4, 4.001] 0.001 40 40.02003 0.02003 20.003 m/s

The pattern is clear: as Δt shrinks, average velocity converges to a fixed number: 20 m/s. That is the instantaneous velocity at t = 4 s.

Mathematically, this is the limit of average velocity as Δt → 0:

\[ v(t) = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \lim_{\Delta t \to 0} \frac{x(t+\Delta t) - x(t)}{\Delta t} \]

Same definition as the derivative! So:

\[ v(t) = \frac{dx}{dt} \]

Instantaneous velocity is the derivative of position with respect to time.

Using derivative rules

For our robot with x(t) = 2.5 t², use the power rule (d/dt [t^n] = n t^{n-1}):

\[ v(t) = \frac{d}{dt}(2.5\, t^2) = 2.5 \cdot 2 t = 5t\ \mathrm{m/s} \]

If the position equation is more elaborate, say x(t) = 5t + 3t² − 0.1 t³:

\[ v(t) = \frac{dx}{dt} = 5 + 6t - 0.3 t^2\ \mathrm{m/s} \]

Instantaneous speed

Instantaneous speed is the absolute value of instantaneous velocity:

\[ s(t) = |v(t)| \]

Example: for the third equation above (v(t) = 5 + 6t - 0.3 t²), at t = 30 s:

Both are correct — one says "how fast and in which direction", the other says "how fast, period".

Graphical interpretation: slope of the tangent

On the x-t plot:

As the second point approaches the first (shrinking Δt), the chord rotates and approaches the tangent. That is the geometric picture of the limit.

Note: on a smooth curve (no corners), there is a unique tangent at every point — meaning v(t) has exactly one value at every instant.

When is instantaneous velocity "undefined"?

If position has a corner (say the robot instantly reverses direction), the slope from the left and from the right differ — no unique tangent exists:

\[ x(t) = \begin{cases} 2t & \text{if } t < 3 \\ 12 - 2t & \text{if } t \geq 3 \end{cases} \]

Physically that means "instantaneous reversal" — an idealization. In the real world this reversal always happens over a small but positive time window (a real robot can't jump from +2 to −2 in zero seconds — it would take infinite acceleration).

A few notes and common mistakes

1. Instantaneous velocity ≠ average velocity over a very small window. Do not conflate them: v_avg = Δx/Δt for a specific small Δt is an estimate that approaches the true value as Δt shrinks. v(t) = dx/dt is the limit of that estimate — the exact value at an instant.

2. "Velocity at an instant" is a real physical concept. Some people wonder, "how can there be velocity in a zero-second instant, when no distance was traveled?" Answer: instantaneous velocity is not like a distance — it is an instantaneous rate of change. Like a heart rate: at "this moment", it has meaning.

3. Units stay put. Derivatives don't change units. If x is in m and t in s, then v = dx/dt is in m/s. If you're given an equation where v is in km/h, convert first (§1.9), then work.

4. Derivative of a constant = zero. If x(t) = 5 m (constant), then v = 0. The object is at rest. Trivial, but easy to forget in compound problems.

5. If x(t) is linear, v is constant. x(t) = a + b t → v = b for all time. So instantaneous velocity = average velocity for this special case (uniform motion). We hinted at this in §2.1.

6. The sign of velocity matters. v = −20 m/s means the object is moving in the negative direction of the axis — not that it is "slower". Its speed is 20 m/s.

The inverse: from v(t) to x(t) (preview)

If we know velocity and want to recover position, we need the inverse of differentiation — the integral:

\[ x(t) = x(0) + \int_0^t v(t')\, dt' \]

Geometrically, Δx between t=0 and t=T equals the area under the v−t curve between those two times (positive area if v > 0; signed negative area if v < 0).

We'll return to this in §2.6; for now just remember: differentiation and integration are inverses of each other.

What you should be able to do

After this section, you should be able to:

Preview of §2.3

Instantaneous velocity tells us "how fast at this instant". Next question: how does velocity itself change? The rate of change of velocity is called acceleration:

\[ a(t) = \frac{dv}{dt} = \frac{d^2 x}{dt^2} \]

Acceleration is the second derivative of position — and the heart of every dynamics problem (because from Newton's second law, F = ma, acceleration is directly tied to force).

📚 See also: Halliday Vol 1, Ch 2, §2.2 — Instantaneous Velocity and Speed. 📖 Open reference: OpenStax University Physics Vol 1 — Chapter 3.2: Instantaneous Velocity and Speed. 📖 Feynman Lectures Vol I — Ch 8: Motion §8-2 (deep discussion of "speed at an instant" and related paradoxes).

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