Kinematics: motion with calculus

IB Mathematics Analysis and Approaches SL· 12 min read

1. Core Kinematic Relationships via Differentiation★★☆☆☆⏱ 10 min

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📘 Definition

Kinematic Derivative Relationships

Velocity is the first time derivative of displacement, and acceleration is the first time derivative of velocity, or the second time derivative of displacement.

v(t) = \frac{ds}{dt}, \quad a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}

📐 Worked Example

A particle moves along a straight line with displacement function $s (t) = 2 t^{3} - 9 t^{2} + 12 t$ for $t \geq 0$ , where $s$ is in metres and $t$ in seconds. Find the velocity and acceleration of the particle at $t = 2$ seconds.

1
Differentiate $s (t)$ once to get velocity:
2
$v(t) = \frac{ds}{dt} = 6t^2 - 18t + 12$
3
Substitute $t = 2$ to find velocity at 2 seconds:
4
$v(2) = 6(2)^2 - 18(2) + 12 = 24 - 36 + 12 = 0 \text{ m/s}$
5
Differentiate $v (t)$ once to get acceleration:
6
$a(t) = \frac{dv}{dt} = 12t - 18$
7
Substitute $t = 2$ to find acceleration at 2 seconds:
8
$a(2) = 12(2) - 18 = 24 - 18 = 6 \text{ m/s}^2$

✓ Quick check

Confirm you understand the derivative relationships:

If $s (t) = 5 sin (t)$ , what is $a (t)$ ?
- $5 cos (t)$
- $- 5 sin (t)$
- $- 5 cos (t)$
- $5 sin (t)$
Reveal answer
$-5\sin(t)$ —
Differentiate twice: first derivative is $5 cos (t)$ , second is $- 5 sin (t)$ .

2. Finding Displacement from Velocity via Integration★★★☆☆⏱ 12 min

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v(t) = \int a(t) dt + c_1, \quad s(t) = \int v(t) dt + c_2

📐 Worked Example

A particle has velocity $v (t) = 6 t^{2} - 18 t + 12$ for $t \geq 0$ . At $t = 0$ , the particle is at displacement $s = 0$ . Find the displacement function $s (t)$ and the net displacement between $t = 0$ and $t = 3$ .

1
Integrate $v (t)$ to find general displacement function:
2
$s(t) = \int (6t^2 - 18t + 12) dt = 2t^3 - 9t^2 + 12t + c$
3
Apply initial condition $s (0) = 0$ : $0 = 0 - 0 + 0 + c$ , so $c = 0$ .
4
Evaluate definite integral from $t = 0$ to $t = 3$ for net displacement:
5
$\int_0^3 (6t^2 - 18t + 12) dt = \left[2t^3 -9t^2 +12t\right]_0^3 = (54 - 81 + 36) - 0 = 9 \text{ m}$

3. Calculating Total Distance Travelled★★★★☆⏱ 15 min

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📐 Worked Example

Using the same velocity function $v (t) = 6 t^{2} - 18 t + 12$ , calculate the total distance travelled by the particle between $t = 0$ and $t = 3$ .

1
Find roots of $v (t) = 0$ : factor to get $6 (t - 1) (t - 2) = 0$ , so roots at $t = 1$ and $t = 2$ .
2
Test sign of $v (t)$ on each interval: positive 0<t<1, negative 1<t<2, positive 2<t<3.
3
Integrate absolute value of $v (t)$ over each interval:
4
$\int_0^1 v(t) dt = 5, \quad \int_1^2 -v(t) dt = 1, \quad \int_2^3 v(t) dt = 4$
5
Sum the three results to get total distance: $5 + 1 + 4 = 10$ m.

4. Graphical Interpretation of Kinematic Functions★★★☆☆⏱ 10 min

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Graph	Slope represents	Area under curve represents
Displacement vs time	Instantaneous velocity	No standard kinematic meaning
Velocity vs time	Instantaneous acceleration	Net displacement / total distance
Acceleration vs time	Rate of change of acceleration	Change in velocity

📐 Worked Example

A velocity vs time graph is a straight line from $v = 4$ m/s at $t = 0$ down to $v = - 2$ m/s at $t = 3$ . Use area calculation to find total distance travelled.

1
Find root of v(t) at t=2, where velocity crosses zero.
2
Calculate area of positive triangle from t=0 to t=2: $\frac{1}{2} \times 2 \times 4 = 4$ m.
3
Calculate area of negative triangle from t=2 to t=3: $\frac{1}{2} \times 1 \times 2 = 1$ m.
4
Sum areas for total distance: $4 + 1 = 5$ m.

5. Common Pitfalls

Wrong move:

Using a single definite integral of $v (t)$ to calculate total distance directly

Why:

Negative velocity regions cancel positive displacement values, returning net displacement not total path length

Correct move:

Find all roots of $v (t)$ in the interval, split the integral at these points, and sum absolute values of each sub-integral

Wrong move:

Treating acceleration as the first derivative of displacement

Why:

Mixing up the order of differentiation skips the velocity step, leading to wrong units and values

Correct move:

Follow the chain: differentiate displacement once for velocity, differentiate velocity once for acceleration

Wrong move:

Forgetting the constant of integration when finding $s (t)$ from $v (t)$

Why:

Indefinite integration produces an arbitrary constant that corresponds to the initial position of the particle

Correct move:

Use the given initial condition (e.g. $s (0) = 3$ ) to solve for the constant before finalizing your displacement function

Wrong move:

Assuming velocity is zero when acceleration is zero

Why:

Zero acceleration means velocity is constant, not that it equals zero

Correct move:

Set $v (t) = 0$ explicitly to find points where the particle is stationary or changes direction

Wrong move:

Stating acceleration units as m/s

Why:

Acceleration is the rate of change of velocity, so units are metres per second squared

Correct move:

Always confirm units match the quantity you are calculating to avoid losing method marks

6. Quick Reference Cheatsheet

Quantity	Mathematical Relationship	Standard Units
Displacement $s (t)$	$s (t) = \int v (t) d t + c$	m
Velocity $v (t)$	$v (t) = \frac{d s}{d t} = \int a (t) d t + c$	m/s
Acceleration $a (t)$	$a (t) = \frac{d v}{d t} = \frac{d ^{2} s}{d t ^{2}}$	m/s²
Net Displacement $t_{1}$ to $t_{2}$	$\int_{t_{1}}^{t_{2}} v (t) d t$	m
Total Distance $t_{1}$ to $t_{2}$	$\int_{t_{1}}^{t_{2}} ∣ v (t) ∣ d t$	m

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · Paper 2
Kinematics integration 6-mark question
2022 · Paper 1
Velocity differentiation 7-mark question
2021 · Paper 2
Total distance calculation 4-mark question

What's Next

Mastering kinematics with calculus gives you a high-yield skill that appears in almost every IB Math AA SL Paper 2, often worth 6-8 marks. This topic builds directly on your prior differentiation and integration fluency, and prepares you for applied calculus problems in related rates, motion modelling, and optimization contexts. You will frequently combine these kinematic rules with your knowledge of function roots, graph sketching, and definite integral evaluation to score full marks on extended response questions. This is one of the most commonly tested applied calculus topics on the syllabus, so consistent practice will directly boost your exam score.