Revise Kinematics and related rates of change for IB Mathematics: Applications and Interpretation (5.7)

We work with motion along a straight line, measured from a fixed origin O. Displacement $s$ is the position relative to O — a signed quantity, positive on one side and negative on the other. Velocity $v$ is the rate of change of displacement: how fast, and in which direction, the position is changing. Acceleration $a$ is the rate of change of velocity. Two words matter for the exam: velocity is a vector (it carries a sign), while speed is its magnitude $|v|$; and displacement (net change of position) is not the same as the total distance travelled (the length of the whole journey).

Going forward, differentiate with respect to time:

Velocity: $v(t) = \dfrac{ds}{dt} = s'(t)$

Acceleration: $a(t) = \dfrac{dv}{dt} = v'(t) = \dfrac{d^2s}{dt^2} = s''(t)$

Differentiation runs displacement to velocity to acceleration. Integration runs the chain in reverse: velocity is the integral of acceleration, and displacement is the integral of velocity. Each integration introduces a constant, and each constant is fixed by an initial condition given in the problem — 'starts from rest', 'initial velocity 8 m/s', 'starts at the origin'. Integrate twice and you need two conditions, so use a fresh letter for the second constant.

Going back, integrate with respect to time:

Velocity from acceleration: $v(t) = \displaystyle\int a(t)\,dt + C$

Displacement from velocity: $s(t) = \displaystyle\int v(t)\,dt + K$

Sometimes two quantities that both change over time are tied together by an equation — the volume and radius of a balloon, the area and radius of a spreading circle, the height and volume of liquid in a cone. If you know the rate at which one is changing, you can find the rate of the other. The tool is the chain rule: differentiate the linking equation with respect to time, treating every variable as a function of $t$. The order of operations matters — differentiate first, substitute the instant's numbers last.

The chain rule for related rates: $\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt}$

Equivalently, to make an unknown rate the subject: $\dfrac{dx}{dt} = \dfrac{dy/dt}{dy/dx}$

In an exam the marks are analytic: each is tied to a specific line of working. A method mark (M) rewards a correct approach even if the arithmetic later slips; an accuracy mark (A) rewards a correct result and is usually dependent on the method mark being earned. Follow-through (FT) means a correct final step performed on your own earlier (wrong) value still scores, ISW ('ignore subsequent working') means a correct answer is not un-marked by a later fumble, and equivalent correct forms are accepted. All of that protection exists only if your method is on the page. Study how each mark below is earned by a specific line.

Your GDC is indispensable here. Use its numerical derivative to check a velocity or acceleration at a point, and its equation solver to find where $v(t)=0$ — write the equation down first, then state that you solved it. For total distance travelled, enter the definite integral of $|v(t)|$ directly using the absolute-value function, which handles the sign changes for you. For related rates, the GDC only helps with the final arithmetic: the differentiation of the linking equation must be shown by hand, because that is where the method marks live.

Everything here is one pair of reversible ideas — differentiate to step forward through position, velocity and acceleration; integrate (with an initial condition) to step back — plus the chain rule that ties two changing quantities together. That same toolkit powers the modelling and optimisation questions across Paper 2, where a quantity changing in time is exactly the situation you now know how to analyse. Master the habit — read the calculus question the words are really asking, show the derivative or integral, then interpret with units — and the rest of the calculus of change becomes variations on a method you already own.