Radford Mathematics IB Mathematics resources • AA SL

IB Mathematics: Analysis and Approaches SL — Topic 5 Calculus

Kinematics

Move fluently between displacement, velocity and acceleration and distinguish displacement from total distance.

AA SL · SL 5.9

Learning goal

Solve one-dimensional kinematics problems using differentiation and integration.

Syllabus link

AA SL: SL 5.9 · Topic 5 Calculus.

Big idea

Velocity is signed. Integrating velocity gives displacement; integrating its magnitude gives total distance travelled.

Key relationship

\(v=\frac{ds}{dt}\)

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Core idea

Velocity is signed. Integrating velocity gives displacement; integrating its magnitude gives total distance travelled.

Solve one-dimensional kinematics problems using differentiation and integration.

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Key results and method

\[v=\frac{ds}{dt}\]
\[a=\frac{dv}{dt}=\frac{d^2s}{dt^2}\]
\[\Delta s=\int_{t_1}^{t_2}v(t)\,dt\]
\[\text{distance}=\int_{t_1}^{t_2}|v(t)|\,dt\]
  1. Differentiate position to obtain velocity and acceleration.
  2. For distance, find times when velocity changes sign.
  3. Split the integral or integrate \(|v|\).
  4. Check units: m, m/s, m/s².
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Worked example

Question

If \(v(t)=t-2\) for \(0\le t\le4\), find displacement and distance.

Solution
  1. \(\int_0^4(t-2)dt=0\), so displacement is zero.
  2. Velocity changes sign at \(t=2\).
  3. Distance is the sum of the two triangular areas: \(2+2=4\).
Answer: Displacement \(0\); distance \(4\) units.
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Practice

Try these questions before moving on.

  1. Find acceleration from a velocity function.
  2. Find total distance when velocity changes direction twice.
  3. Given initial position, recover the position function from velocity.

Search the worksheet library for more practice →

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Tutorials and premium resources

▶ Displacement integrals — Part 1 ▶ Distance from velocity — Part 2 ▶ Kinematics exam walkthrough
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