Calculus for Kinematics | CIE IGCSE Additional Maths Revision Notes 2025

Differentiation for Kinematics

How is differentiation used in kinematics?

Displacement, velocity and acceleration are related by calculus
In terms of differentiation and derivatives
- velocity is the rate of change of displacement
  - $v = \frac{d s}{d t}$ or $v (t) = s' (t)$
- acceleration is the rate of change of velocity
  - $a = \frac{d v}{d t}$ or $a (t) = v' (t)$
- so acceleration is also the second derivative of displacement
  - $a = \frac{d^{2} s}{d t^{2}}$ or $a (t) = s'' (t)$

Worked example

The displacement, $x$ m, of a particle at $t$ seconds, is modelled by the function $x (t) = 2 t^{3} - 27 t^{2} + 84 t$ .

Find expressions for $x' (t)$ and $x'' (t)$ , and state what these expressions represent.

$x = 2 t^{3} - 27 t^{2} + 84 t$

$x' (t)$ means the same as $\frac{d x}{d t}$

$x' (t) = 6 t^{2} - 54 t + 84 x' (t) = 6 (t^{2} - 9 t + 14)$

$x' (t) = 6 (t - 2) (t - 7)$
This expression represents the velocity of the particle.
Answers may or may not need to be factorised, depending on the question

$x'' (t)$ means the same as $\frac{d^{2} x}{d t^{2}}$

$x'' (t) = 12 t - 54$

$x'' (t) = 6 (2 t - 9)$
This expression represents the acceleration of the particle.

Integration for Kinematics

How is integration used in kinematics?

Since velocity is the derivative of displacement ( $v = \frac{d s}{d t}$ ) it follows that

$s = \int v d t$

Similarly, velocity will be an antiderivative of acceleration

$v = \int a d t$

How would I find the constant of integration in kinematics problems?

A boundary or initial condition would need to be known
- phrases involving the word “initial”, or “initially” are referring to time being zero, i.e. $t = 0$
- you might also be given information about the object at some other time (this is called a boundary condition)
- substituting the values in from the initial or boundary condition would allow the constant of integration to be found

How are definite integrals used in kinematics?

Definite integrals can be used to find the displacement of a particle between two points in time
- $\int_{t_{1}}^{t_{2}} v (t) d t$ would give the displacement of the particle between the times $t = t_{1}$ and $t = t_{2}$
  - This can be found using a velocity-time graph by subtracting the total area below the horizontal axis from the total area above
  - You could think of this as the "net area" e.g. 4 m - 1 m = 3 m
- $\int_{t_{1}}^{t_{2}} |v (t)| d t$ gives the distance a particle has travelled between the times $t = t_{1}$ and $t = t_{2}$
  - This can be found using a velocity velocity-time graph by adding the total area below the horizontal axis to the total area above
  - You could think of this as the "gross area" e.g. 4 m + 1 m = 5 m

Exam Tip

Sketching the velocity-time graph can help you visualise the distances travelled using areas between the graph and the horizontal axis

Worked example

A particle moving in a straight horizontal line has velocity $v$ ms^-1 at time $t$ seconds modelled by $v (t) = 8 t^{3} - 12 t^{2} - 2 t$ .

Given that the initial position of the particle is at the origin, find an expression for its displacement from the origin at time

t

seconds.

The integral of the velocity gives the displacement

s (t) = \int v (t) d t = \int (8 t^{3} - 12 t^{2} - 2 t) d t s (t) = 2 t^{4} - 4 t^{3} - t^{2} + c

"Initial" means when

t = 0

and "at the origin" means

s = 0

Substitute these values in to find

c

0 = 2 {(0)}^{4} - 4 {(0)}^{3} - {(0)}^{2} + c c = 0

s (t) = 2 t^{4} - 4 t^{3} - t^{2}

Find the displacement of the particle from the origin after the first five seconds of its motion.

To find the displacement after 5 seconds, integrate the velocity between 0 and 5

\int_{0}^{5} 8 t^{3} - 12 t^{2} - 2 t d t {[2 t^{4} - 4 t^{3} - t^{2}]}_{0}^{5} = [2 {(5)}^{4} - 4 {(5)}^{3} - {(5)}^{2}] - [2 {(0)}^{4} - 4 {(0)}^{3} - {(0)}^{2}] = 725 - 0

725 m

Explain why this value is not the same as the distance travelled during the first 5 seconds.

When finding the displacement, any negative displacements are taken into account. e.g. 3m - 1m = 2m. When finding the distance, we are looking for the total of all the areas, ignoring their direction/sign. e.g. 3m + 1m = 4m.
This function for velocity has some regions underneath the x-axis, and some regions above the x-axis between 0 and 5 seconds, so the displacement and distance covered in this time will be different.

Calculus for Kinematics (CIE IGCSE Additional Maths)

Revision Note

Differentiation for Kinematics

How is differentiation used in kinematics?

Worked example

Integration for Kinematics

How is integration used in kinematics?

How would I find the constant of integration in kinematics problems?

How are definite integrals used in kinematics?

Exam Tip

Worked example

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Author: Jamie W