M11.4MechanicsStretch

Using Integration

Integration reverses differentiation, so it steps back up the kinematics chain: integrate acceleration to get velocity, and velocity to get displacement. The constant of integration is found from initial conditions.

25 min Video by Zeeshan Zamurred Variable Acceleration

Edexcel Mechanics Y1 — Variable Acceleration playlist (Zeeshan Zamurred)Watch the full walkthrough before the notes below.

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What you'll be able to do

Integrate acceleration to find velocity
Integrate velocity to find displacement
Use initial conditions to find the constant
Find distance travelled with definite integrals

The integration chain

Integration goes the opposite way to differentiation: integrate $a$ to get $v$ , and integrate $v$ to get $s$ . Each integration introduces a constant, found from a known value (e.g. initial velocity).

v = \int a d t, s = \int v d t

Integrate up:

a \to v \to s

(with

+ c

v = \int 6 t d t = 3 t^{2} + c

2At

t = 0

v = 4

, so

c = 4

Answer

v = 3 t^{2} + 4

Using initial conditions

Each $+ c$ is found by substituting a known value — usually the value at $t = 0$ . Without this step the answer is incomplete.

Tip — Every integration needs its + c pinned down by an initial condition.

Distance with definite integrals

The distance travelled between two times is the definite integral of velocity: $\int_{t_{1}}^{t_{2}} v d t$ . (If the velocity changes sign, split the integral at $v = 0$ and add the magnitudes, as with areas under curves.)

s = \int_{t_{1}}^{t_{2}} v d t

Definite integral gives displacement over the interval.

Formula recap

v = \int a d t

Velocity from acceleration.

s = \int v d t

Displacement from velocity.

s = \int_{t_{1}}^{t_{2}} v d t

Displacement over an interval.

Common mistakes to avoid

Forgetting the constant of integration.

Find + c from the initial conditions every time.

Integrating to go from velocity to acceleration.

Integration goes a → v → s; differentiation goes the other way.

Key takeaways

Integrate to step up: a → v → s.
Use initial conditions to find each constant of integration.
Distance over an interval = definite integral of velocity.

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