Kinematics in O-Level Additional Mathematics (4049): displacement, velocity and acceleration for motion in a straight line, moving between them with calculus, and solving applied motion problems

Calculus in motion

Kinematics is where the Calculus strand of Additional Mathematics (SEAB 4049) meets the real world: the motion of a particle along a straight line. The three quantities, displacement, velocity and acceleration, are linked by differentiation in one direction and integration in the other, so kinematics is really an application of everything in the differentiation and integration modules. The student who keeps the chain $s \to v \to a$ (differentiate) and $a \to v \to s$ (integrate) clearly in mind rarely gets lost.

This module covers three outcomes that build in sequence: the language and signs of displacement, velocity and acceleration; moving between them with calculus; and applied problems on maximum displacement, total distance and direction changes. This overview ties the three dot points together; each has its own worked answers and practice.

The language of motion

The displacement, velocity and acceleration outcome defines the three quantities and their signs. Displacement $s$ is the signed position from a fixed origin, velocity $v$ is how fast and in which direction the position is changing, and acceleration $a$ is how velocity is changing. A positive value points in the chosen positive direction; a negative value points the other way. You also read motion from displacement-time and velocity-time graphs.

Moving between them with calculus

The kinematics using calculus outcome is the engine of the topic. Differentiating moves you down the chain, $v = \dfrac{ds}{dt}$ and $a = \dfrac{dv}{dt}$, while integrating moves you back up, $v = \displaystyle\int a\,dt$ and $s = \displaystyle\int v\,dt$. Each integration introduces a constant, which you fix using the initial conditions stated in the question (for example $s = 0$ when $t = 0$).

Applied motion problems

The applications of kinematics outcome brings the skills together in practical problems. Maximum or minimum displacement occurs where $v = 0$, and maximum or minimum velocity where $a = 0$. For total distance travelled you find the times when $v = 0$, work out the displacement over each leg, and add the magnitudes, because distance ignores direction.

How this module is examined

• Both papers, all questions. Paper 1 and Paper 2 (each 2 hours 15 minutes, 90 marks, 50 percent) cover the full syllabus, and you answer every question.
• Use initial conditions. When you integrate, find the constant from the stated condition before evaluating anything else; an unfixed constant invalidates the result.
• Split at direction changes for total distance. Find where $v = 0$, compute each leg separately, and add the magnitudes; never just take the final displacement as the distance.

Check your knowledge

Short questions across the three outcomes. Work them with full method, then check the solutions.

1. A particle has displacement $s = 2t^3 - 3t^2$ m. Find its velocity $v$ as a function of $t$. (2 marks)
2. For the same particle, find its acceleration when $t = 2$ s. (2 marks)
3. A particle has velocity $v = 4 - 2t$ m s$^{-1}$. Find the time at which it is instantaneously at rest. (2 marks)
4. A particle has acceleration $a = 6t$ m s$^{-2}$ and velocity $v = 1$ m s$^{-1}$ when $t = 0$. Find $v$ as a function of $t$. (3 marks)