Graphical analysis and motion integration

The §2.4 kinematic equations were valid only for constant acceleration. For variable acceleration (air drag, springs, variable-power engines), we need a more general tool. That tool is the integral — or its geometric equivalent: area under a curve.

📖 Prerequisite: if the definite integral isn’t yet solid for you, first read The definite integral for physicists — Riemann-sum definition, the power rule, and physical applications.

The inverse of differentiation: from velocity to position

From §2.2: v = dx/dt. Velocity is the derivative of position. To reverse this operation, we take an integral:

$$
x(t) – x(0) = \int_0^t v(t’)\, dt’
$$

Geometric interpretation: the integral of a function is the area under its curve. Hence:

$$
\Delta x = \text{area under the } v-t \text{ curve between two times}
$$

The inverse of differentiation: from acceleration to velocity

Similarly, from §2.3: a = dv/dt. Thus:

$$
v(t) – v(0) = \int_0^t a(t’)\, dt’
$$

$$
\Delta v = \text{area under the } a-t \text{ curve between two times}
$$

Two relationships: ∫a → Δv, then ∫v → Δx. For any acceleration function a(t), these two successive integrals fully describe the motion.

Recovering the §2.4 equations (constant acceleration)

Let’s check that this new tool reproduces the old equations. With a(t) = a constant:

$$
\Delta v = \int_0^t a\, dt’ = a t \Rightarrow v(t) = v_0 + a t
$$

$$
\Delta x = \int_0^t (v_0 + a t’)\, dt’ = v_0 t + \frac{1}{2} a t^2
$$

Same as equations 1 and 2. So §2.6 is a generalization of §2.4 — not a separate thing.

Example 1 — triangular acceleration (linearly increasing a)

Scenario: a rocket that burns its fuel at a linear rate has linearly increasing acceleration:

$$
a(t) = k t \quad \text{with}\ k = 2\ \mathrm{m/s^3}
$$

Starting from rest (v_0 = 0, x_0 = 0).

Velocity at t = 5 s:

$$
\Delta v = \int_0^5 2 t\, dt = 2 \cdot \frac{t^2}{2}\Big|_0^5 = 25\ \mathrm{m/s}
$$

Geometric interpretation: a(t) = 2t is linearly rising. The area under the curve from t=0 to t=5 is a triangle with base 5 and height 2 \cdot 5 = 10. Triangle area = ½ \cdot 5 \cdot 10 = 25 m/s. ✓

Position at t = 5 s: with v(t) = t²:

$$
\Delta x = \int_0^5 t^2\, dt = \frac{t^3}{3}\Big|_0^5 = \frac{125}{3} \approx 41.7\ \mathrm{m}
$$

Example 2 — falling with air resistance (numerical)

Scenario: a skydiver in free fall. Acceleration:

$$
a(t) = -g + \alpha v^2\quad (\text{with}\ \alpha > 0)
$$

where the second term is air drag (proportional to v²). This is hard to solve analytically, but we can integrate numerically:

This is Euler’s method — the simplest numerical integrator. Shrinking Δt gives a more accurate answer.

Area density — “the exact integral”

Mathematically, a definite integral is a limit:

$$
\int_a^b f(t)\, dt = \lim_{n \to \infty} \sum_{i=1}^{n} f(t_i)\, \Delta t
$$

with Δt = (b-a)/n and t_i sample points. We approximate the area under the curve by infinitely many very thin rectangles. This is called a Riemann sum.

If f(t) is a simple function (polynomial, sin, cos, exp), a closed form exists. If it’s complicated, the only route is numerical.

Example 3 — read directly from a graph

Suppose you have a v-t plot of a car:

Total distance = total area:

No equations needed — just the geometry of the plot. That’s the power of graphical analysis.

Reading the x-t plot inversely

If you have x(t), its derivative gives v(t) (tangent slope, §2.2) and its second derivative gives a(t) (curvature, §2.3). The reverse also holds: taking the area under x makes no sense (it’s already a length, its area would be “length × time” — not a physical quantity).

So the golden rule:

From Operation You get
x(t) derivative v(t)
v(t) derivative a(t)
a(t) integral Δv
v(t) integral Δx

A few notes and common mistakes

1. Area under a curve can be negative.
If v(t) < 0 (curve below the axis), area is signed negative. This gives backward displacement. For total distance, sum the absolute areas — not the signed area.

2. Integral means “net amount”, not “average”.
For average velocity, v_avg = Δx / Δt = ∫v dt / Δt. Average is area divided by interval. That’s different from the area itself.

3. It’s v-t not x-t.
Note that the area under v-t gives displacement, not the area under x-t. Picking the wrong plot is a common mistake.

4. Units should work out.
Area under v-t: units m/s × s = m (meters). ✓
Area under a-t: units m/s² × s = m/s. ✓
If units don’t add up, something’s wrong.

5. For constant acceleration, the §2.4 equations still work.
Integration isn’t compulsory — just if a varies or you’re reading from a graph. For constant acceleration, the kinematic equations are faster.

What you should be able to do

After this section, you should be able to:

Preview of §2.7 and beyond

Chapter 2 is nearly done — 1D kinematics is now complete. §2.7 onward has worked problems and practice problems that test all these tools across diverse scenarios. After Chapter 2, we move to Chapter 3 (2D kinematics — projectiles and circular motion).

📚 See also: Halliday Vol 1, Ch 2, §2.6 — Graphical Integration.
📖 Open reference: OpenStax University Physics Vol 1 — Chapter 3.4.5: Kinematics of Free Fall (graphical view of displacement from velocity).

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