Kinematics and Rates of Change (AA SL)
Kinematics is where Differentiation meets motion in a straight line: displacement, velocity and acceleration are all connected by derivatives (and, in reverse, by integrals). Once you can read a question and know which of the three quantities you're given and which you need, the mechanics are just differentiation or integration you already know.
18 questions on this sub-topic.
Displacement, velocity and acceleration
Covered under IB syllabus reference SL5.9. These relationships aren't printed in the formula booklet as a labelled block - they follow directly from the differentiation and integration you already use elsewhere.
Kinematics
Velocity is the derivative of displacement, \(v=\dfrac{ds}{dt}\), and acceleration is the derivative of velocity, \(a=\dfrac{dv}{dt}\). "At rest" means \(v=0\).
Going the other way
Since \(v=\dfrac{ds}{dt}\), integrating velocity recovers displacement: \(s(t)=\displaystyle\int v(t)\,dt.\) Use any given initial condition (often \(s(0)\)) to find the constant of integration.
Need the full syllabus wording and the rest of differentiation? See Differentiation.
Worked examples
A particle moves along a line so that its displacement at time \(t\) seconds is \(s(t) = t^2 - 4t + 3\) metres, \(t \ge 0\).
Find the velocity at time \(t\), and find the value of \(t\) when the particle is at rest.
Worked solution
\(v(t) = s'(t) = 2t - 4.\) A1
At rest when \(v = 0\): \(2t - 4 = 0 \Rightarrow t\) M1
\(= 2\) A1
A particle moves in a straight line with velocity \(v(t) = 4t - t^2\) ms\(^{-1}\), \(t \ge 0\). At \(t = 0\) the particle is at the origin.
(a) Find \(s(t)\).
(b) Find the displacement when the particle is next at rest.
Worked solution
(a) \(s(t) = \int(4t - t^2)\,dt = 2t^2 - \dfrac{t^3}{3} + C.\) M1
\(s(0) = 0 \Rightarrow C = 0.\) So \(s(t) = 2t^2 - \dfrac{t^3}{3}.\) A1
(b) At rest: \(v = 0 \Rightarrow t(4-t) = 0 \Rightarrow t\) M1 \(= 4.\) A1
\(s(4) = 32 - \dfrac{64}{3} = \dfrac{32}{3}\) m. A1
Common mistakes
- Confusing "at rest" with "at the origin". At rest means \(v(t)=0\), not \(s(t)=0\). Solve the velocity equation, not the displacement equation, to find when a particle stops.
- Dropping the constant of integration. When recovering \(s(t)\) from \(v(t)\), the \(+C\) matters - you need the given initial condition (usually \(s(0)\)) to pin it down, or your final displacement will be wrong even if the method is right.
- Reporting a negative time as the answer. When solving a factorised velocity equation like \(t(4-t)=0\), always reject any solution with \(t<0\) since motion is only defined for \(t \ge 0\).
Ready to practise properly?
18 kinematics questions, marked instantly like the real exam.
Quick answers
How are velocity and acceleration related to displacement?
Velocity is the derivative of displacement, \(v = \dfrac{ds}{dt}\), and acceleration is the derivative of velocity, \(a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}\). Going the other way, displacement is the integral of velocity, \(s(t)=\int v(t)\,dt\).
What does it mean for a particle to be at rest?
A particle is at rest when its velocity is zero, \(v(t) = 0\). Solve this equation for \(t\), then substitute back into \(s(t)\) to find its position at that instant. A GDC's graph or solver tools can check this too - see using your GDC.