3.02gMechanicsStretch

Integrating Vectors

Integration reverses differentiation for vectors too. From an acceleration vector you integrate to velocity, and again to position — using initial conditions to fix the vector constants.

30 min Kinematics

The big picture

Integrate acceleration to obtain velocity and use the initial velocity to determine a vector constant. Integrate velocity to obtain position and use the initial position to determine a second vector constant. A known position alone does not determine initial velocity. For each component, track the units and apply the same time origin. A scalar integration constant cannot represent two independently specified initial components.

Edexcel A Level Maths: 8.5 Integrating VectorsCurated by Kepler, not produced by us. Watch it, then work through the notes below.

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What you'll be able to do

Integrate a vector component-wise
Find velocity from acceleration
Find position from velocity
Use initial conditions for the constant vector

Component-wise integration

Integrate each component and add a constant $\textbf{vector}$ . $\mathbf{v} = \int \mathbf{a}\,dt + \mathbf{c}$ , with $\mathbf{c}$ found from the velocity at a known time.

\mathbf{v} = \int \mathbf{a}\,dt + \mathbf{c}

Integrate each component; add a constant vector.

Example 1If

\mathbf{a} = 2\mathbf{i}

and

\mathbf{v} = 3\mathbf{j}

t = 0

, find

\mathbf{v}(t)

\int 2\mathbf{i}\,dt = 2t\mathbf{i} + \mathbf{c}

2At

t=0

\mathbf{v} = 3\mathbf{j}

, so

\mathbf{c} = 3\mathbf{j}

Answer

\mathbf{v} = 2t\mathbf{i} + 3\mathbf{j}

Tip — The constant of integration is a VECTOR — find it from the initial velocity or position.

Two components need two constants

For each component, track the units and apply the same time origin. A scalar integration constant cannot represent two independently specified initial components.

Example 1Given

\mathbf a=6t\mathbf i-4\mathbf j

\mathbf v(0)=2\mathbf i+5\mathbf j

and

\mathbf r(0)=\mathbf i-3\mathbf j

, find the position.

1Integrating acceleration:

\mathbf v=(3t^2+C_1)\mathbf i+(-4t+C_2)\mathbf j

2Initial velocity gives

C_1=2

C_2=5

3Integrate again:

\mathbf r=(t^3+2t+D_1)\mathbf i+(-2t^2+5t+D_2)\mathbf j

4Initial position gives

D_1=1

D_2=-3

Answer

\mathbf r=(t^3+2t+1)\mathbf i+(-2t^2+5t-3)\mathbf j

Think like an examiner

Write the defining equations and show the substitution before evaluating.
Check the domain, units and interpretation of the final result.

Common misconceptions

Adding a scalar constant of integration.

The constant is a vector (it has i and j parts).

Skipping the initial conditions.

Use the given velocity/position to find the constant vector.

Key results to remember

\mathbf{v} = \int \mathbf{a}\,dt + \mathbf{c}

Velocity from acceleration.

\mathbf{r} = \int \mathbf{v}\,dt + \mathbf{c}

Position from velocity.

Key takeaways

Integrate each component, add a constant vector.
v = ∫a dt + c; r = ∫v dt + c.
Use initial conditions to find the constant vector.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

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