Kinematics with calculus (a → v → s)

Why integrating acceleration gives velocity and integrating velocity gives displacement, where the suvat formulas actually come from, and why they silently break the moment acceleration varies.

Mechanics · Kinematics

Kinematics with calculus (a → v → s)

Why integrating acceleration gives velocity and integrating velocity gives displacement, where the suvat formulas actually come from, and why they silently break the moment acceleration varies.

Why it works

Velocity is defined as the rate of change of displacement, and acceleration as the rate of change of velocity:

v = \frac{ds}{dt}, \qquad a = \frac{dv}{dt}

Differentiation takes you down that chain (

s \to v \to a

: "how fast is this changing?"). So going back up the chain — from how the motion is changing to the motion itself — is integration:

v = \int a\,dt, \qquad s = \int v\,dt

Each integration introduces a

+C

, and in mechanics the constant isn't abstract — it's physics. Integrating acceleration tells you how velocity changes, but not what it started at; the constant is the initial velocity. Same again for displacement: the constant is where the particle began. That's why exam questions always hand you facts like "initially at rest" or "starts at the origin" — they are the values of your constants of integration.

Now the "ohhh": the suvat formulas are not separate physics — they're what you get by integrating a constant acceleration. Watch:

v = \int a\,dt = at + C, \quad \text{and } v = u \text{ at } t=0 \implies C = u

That's

v = u + at

. Integrate again and the constant is the starting position — that's

s = ut + \tfrac{1}{2}at^2

. You've been using integration results all along without being told.

Which exposes the trap: that derivation *only worked because

a

came out of the integral as a constant*. The moment acceleration depends on time —

a = 4 - 2t

, say — suvat's assumptions are gone and its formulas quietly give wrong answers. Variable acceleration isn't a harder case of suvat; it's the general case, and integration is the only honest tool for it.

One more distinction that earns marks: integrating

v

over a time interval gives displacement (signed — where you ended up relative to where you started). If the particle doubled back, the distance travelled is bigger: split the integral where

v = 0

and add the sizes of the pieces — exactly the signed-area idea from definite integration, wearing a mechanics shirt.A velocity–time graph makes it visible: this particle's velocity crosses zero at

t = 1

, so on

0 \le t \le 2

part of the area sits above the axis and part below. Add them with their signs and you get displacement; add their sizes and you get distance travelled.