Kinematics & Related Rates (AA HL)

Q: When is a particle "at rest" vs "changing direction"?

A particle is momentarily at rest whenever v(t) = 0, but that only means it's changing direction if the sign of v(t) actually flips either side of that instant. If the velocity touches zero and returns to the same sign, the particle pauses but keeps moving the same way.

Kinematics applies differentiation and integration to motion: displacement, velocity and acceleration are all connected by derivatives, and distance travelled needs a slightly different calculation from displacement. Related rates of change extends the same ideas to any two connected quantities changing over time, using implicit differentiation to link their rates together.

Practise this topic → Exam-style questions →

What the syllabus says

This topic maps onto two points in the official IB Analysis & Approaches syllabus, one core and one HL extension.

Code	Syllabus content
SL5.9	Kinematic problems involving displacement \(s\), velocity \(v\), acceleration \(a\) and total distance travelled, where \(v=\dfrac{ds}{dt}\), \(a=\dfrac{dv}{dt}\), displacement from \(t_1\) to \(t_2\) is \(\int_{t_1}^{t_2}v(t)\,dt\), and distance is \(\int_{t_1}^{t_2}\|v(t)\|\,dt.\)
AHL5.14	Implicit differentiation. Related rates of change. Optimisation problems, including cases where the optimum solution is at an endpoint.

SL5.9 is core AA content examinable at both levels; the related rates content in AHL5.14 is HL-only.

Key terms

Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.

What is displacement?

Displacement is a particle's signed position relative to a starting point - positive in one direction, negative in the other. It's found by integrating velocity, and unlike distance, it can decrease if the particle moves backwards.

e.g. If \(v(t)=3t^2+2\) and \(s(0)=0\), then \(s(t)=t^3+2t\), so \(s(4) = 64+8 = 72\) m.

What is velocity?

Velocity is the rate of change of displacement with respect to time, \(v=\dfrac{ds}{dt}.\) It's a signed quantity: positive velocity means moving in the positive direction, negative means moving backwards, and \(v=0\) means momentarily at rest.

e.g. If \(s(t)=t^3-6t^2+9t\), then \(v(t)=3t^2-12t+9\), so \(v(0) = 9\) m/s.

What is the difference between distance and displacement?

Displacement is the net signed change in position, \(\int v(t)\,dt.\) Distance travelled adds up every bit of movement regardless of direction, \(\int|v(t)|\,dt\) - so if a particle reverses direction, distance keeps growing while displacement can shrink back towards zero.

e.g. For \(v=t^2-4t\) on \([0,5]\), displacement \(\approx-8.33\) m but total distance travelled \(=13\) m.

What is a related rate of change?

A related rate connects how two (or more) quantities change with time through a fixed equation, such as Pythagoras' theorem or a volume formula. Differentiating that equation with respect to time links the rates together, so you can find one rate from the other.

e.g. For \(x^2+y^2=25\): \(2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0\), so \(\dfrac{dy}{dt}=-\dfrac{x}{y}\dfrac{dx}{dt}.\)

What does "momentarily at rest" mean?

A particle is momentarily at rest at any instant where its velocity is exactly zero, \(v(t)=0.\) This doesn't necessarily mean it changes direction - only check that if you need to, by seeing whether the sign of \(v(t)\) actually flips either side.

e.g. For \(v=12t-3t^2\): \(v=0\) at \(t=0\) and \(t=4\), and the maximum velocity of \(12\) m/s occurs at \(t=2\).

Key formulas

A small set of relationships covers every kinematics and related rates question. The tables below summarise them at a glance - the explanations underneath go into more depth on each one.

Formula reference

The three kinematics relationships are on the official formula booklet; setting up and differentiating a related rates equation is a technique rather than a listed formula.

Formula	Used for	Booklet?
\(v=\dfrac{ds}{dt},\quad a=\dfrac{dv}{dt}\)	Velocity and acceleration from displacement	✓ Yes
Displacement \(=\displaystyle\int_{t_1}^{t_2}v(t)\,dt\)	Net position change	✓ Yes
Distance \(=\displaystyle\int_{t_1}^{t_2}\|v(t)\|\,dt\)	Total distance travelled	✓ Yes
Differentiate a connecting equation implicitly w.r.t. \(t\)	Related rates of change	Not in booklet - technique

Displacement vs distance travelled

These two quantities come from the same velocity function but answer different questions - one cares about direction, the other doesn't.

Feature	Displacement	Distance travelled
Formula	\(\int_{t_1}^{t_2}v(t)\,dt\)	\(\int_{t_1}^{t_2}\|v(t)\|\,dt\)
Can it be negative?	Yes	No, always \(\ge0\)
Direction changes	Backward movement subtracts	Backward movement still adds
When they're equal	Only if \(v(t)\) never changes sign on the interval

Kinematics relationships

Displacement, velocity and acceleration form a chain of derivatives - and integrating in reverse recovers the function above.

Velocity from displacement

\[v(t) = \dfrac{ds}{dt}\]

Differentiate the displacement function once to get velocity.

✓ In the formula booklet

Acceleration from velocity

\[a(t) = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}\]

Differentiate velocity once more, or displacement twice, to get acceleration.

✓ In the formula booklet

Displacement from velocity

\[s(t_2)-s(t_1) = \int_{t_1}^{t_2}v(t)\,dt\]

Integrate velocity to recover the net change in position over an interval.

✓ In the formula booklet

Working with distance travelled

Whenever velocity changes sign, distance travelled requires splitting the integral at each sign change.

Find where velocity changes sign

Solve \(v(t)=0\) within the given interval - these are the points where the particle may reverse direction.