Kinematics (AA HL)
Kinematics uses differentiation and integration to describe motion: displacement, velocity and acceleration are all linked by derivatives, and going backwards from acceleration to velocity to position uses integration instead. The one genuine trap is that distance travelled and displacement aren't the same thing whenever the direction of motion changes. It's part of the broader Kinematics & Related Rates topic.
25 questions on this sub-topic.
The key relationships
Covered under IB syllabus reference SL5.9: kinematic problems involving displacement \(s\), velocity \(v\), acceleration \(a\) and total distance travelled. Both relationships below are in the formula booklet.
Velocity and acceleration
\(v = \dfrac{ds}{dt}, \quad a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}\)
Differentiate to move from position to velocity to acceleration.
Displacement and distance
Displacement \(=\displaystyle\int_{t_1}^{t_2}v(t)\,dt,\quad\) Distance \(=\displaystyle\int_{t_1}^{t_2}|v(t)|\,dt\)
Integrate to go the other way. Distance needs the modulus of \(v(t)\), split at every sign change.
Need the full syllabus wording and formula-booklet reference table? See Kinematics & Related Rates, including the GDC guidance there.
Worked examples
A particle has displacement \(s=2t^2+t\) metres.
Find the average velocity over \([1,3]\).
Worked solution
\(s(1) = 3,\ s(3)\) M1
\(= 21.\) A1
Average velocity \(= \dfrac{21 - 3}{3 - 1}\) M1
\(= 9\) m/s. A1
A particle has velocity \(v(t)=\sin t\) m/s for \(0\le t\le2\pi\).
Find the total distance travelled.
Worked solution
\(v = 0\) at \(t\) M1
\(= 0, \pi, 2\pi.\) A1
Distance \(= \left|\int_0^{\pi}\sin t\,dt\right| + \left|\int_{\pi}^{2\pi}\sin t\,dt\right|\) M1 A1
\(= 2 + 2 = 4\) m. A1
A particle has acceleration \(a = 6t\) m/s².
Find the change in velocity between \(t = 1\) and \(t = 3.\)
Worked solution
\(\Delta v = \int_1^3 6t\,dt\) M1
\(= [3t^2]_1^3\) A1
\(= 27 - 3\) M1 A1
\(= 24\) m/s. A1
Common mistakes
- Treating distance and displacement as the same thing. If \(v(t)\) stays one sign throughout the interval they do agree, but the moment velocity changes sign, integrating \(v(t)\) directly only gives displacement - distance needs \(|v(t)|\) and a split at the sign change.
- Assuming \(v(t)=0\) always means a change of direction. A particle is only momentarily at rest at that instant - it's changing direction only if the sign of \(v(t)\) actually flips either side of it, so check both sides before concluding anything.
- Mixing up which derivative gives which quantity. Velocity is the derivative of displacement and acceleration is the derivative of velocity - going the other way needs integration, not differentiation, so keep straight which direction you're moving through \(s\), \(v\) and \(a\).
Ready to practise properly?
25 kinematics questions, marked instantly like the real exam.
Quick answers
What's the difference between displacement and distance travelled?
Displacement is the signed integral of velocity over the interval, so a particle moving backwards subtracts from the total. Distance travelled integrates the modulus of velocity instead, so every part of the journey adds positively - you need to split the integral wherever the velocity changes sign.
How are displacement, velocity and acceleration related?
Velocity is the derivative of displacement with respect to time, and acceleration is the derivative of velocity: \(v=\dfrac{ds}{dt}\) and \(a=\dfrac{dv}{dt}=\dfrac{d^2s}{dt^2}\). Going the other way, integrating velocity gives displacement and integrating acceleration gives velocity.