Kinematic applications of calculus

IB Mathematics: Applications and Interpretation SL· 5.7 Kinematic applications of calculus· 35 min read

1. Relationships Between Kinematic Quantities★★☆☆☆⏱ 15 min

📘 Definition

Core Kinematic Relationships

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

For motion along a straight line, velocity is the first derivative of displacement with respect to time. Acceleration is the first derivative of velocity (or second derivative of displacement) with respect to time. To reverse, integrate acceleration to get velocity, and integrate velocity to get displacement.

All quantities are signed: positive values indicate motion in one direction, while negative values indicate motion in the opposite direction. This sign convention allows us to use calculus to describe direction changes naturally.

📐 Worked Example

The displacement of a particle is given by $s (t) = 2 t^{3} - 15 t^{2} + 36 t + 4$ , where $s$ is in metres and $t$ is in seconds, $t \geq 0$ . Find the velocity and acceleration functions.

1
Velocity is the first derivative of displacement:
2
$v(t) = \frac{d}{dt}\left(2t^3 - 15t^2 + 36t + 4\right) = 6t^2 - 30t + 36 = 6(t-2)(t-3)$
3
Acceleration is the first derivative of velocity:
4
$a(t) = \frac{d}{dt}\left(6t^2 - 30t + 36\right) = 12t - 30$
5
Final answer with units: $v (t) = 6 t^{2} - 30 t + 36 ms^{- 1}$ and $a (t) = 12 t - 30 ms^{- 2}$ .

2. Initial Conditions and Finding Unknown Quantities★★★☆☆⏱ 20 min

When integrating acceleration to get velocity or velocity to get displacement, you get a constant of integration. You need an initial condition (a known value of the quantity at a given time, usually $t = 0$ ) to solve for this constant. Common initial conditions are

Starts from rest: $v (0) = 0$
Starts at origin: $s (0) = 0$
Initial displacement: $s (0) = s_{0}$

📐 Worked Example

A particle starts from rest at the origin ( $s = 0, v = 0$ when $t = 0$ ) with acceleration $a (t) = 4 t - 6 ms^{- 2}$ . Find the displacement function $s (t)$ .

1
Integrate acceleration to get velocity, including the constant of integration:
2
$v(t) = \int (4t - 6) dt = 2t^2 - 6t + C$
3
Use initial condition $v (0) = 0$ to find $C$ :
4
$0 = 2(0)^2 - 6(0) + C \implies C = 0$
5
Integrate velocity to get displacement, with a new constant:
6
$s(t) = \int (2t^2 - 6t) dt = \frac{2}{3}t^3 - 3t^2 + D$
7
Use initial condition $s (0) = 0$ to find $D$ :
8
$0 = 0 - 0 + D \implies D = 0$
9
Final displacement function: $s (t) = \frac{2}{3} t^{3} - 3 t^{2}$ metres.

✓ Quick check

Test your understanding of constants:

A particle has $v (t) = 3 t^{2} + 2$ , and $s (1) = 5$ . What is the value of the integration constant $C$ when finding $s (t)$ ?
- 1
- 2
- 3
- 4
Reveal answer
2 —
Correct: $s (t) = t^{3} + 2 t + C$ , substitute $s (1) = 5$ : $5 = 1 + 2 + C ⟹ C = 2$

3. Calculating Total Distance Travelled★★★★☆⏱ 20 min

A very common exam question asks for total distance travelled, not displacement. Displacement only measures the net change in position from start to end, while total distance measures all movement, even when the particle reverses direction.

📘 Definition

Total Distance Travelled

For a time interval $[a, b]$ , total distance is the integral of the absolute value of velocity. A particle reverses direction when $v (t) = 0$ , so split the interval at all such points, then add the absolute value of the integral over each sub-interval.

\text{Total Distance} = \int_a^b |v(t)| dt

📐 Worked Example

For the particle with $v (t) = 6 t^{2} - 30 t + 36$ , find the total distance travelled between $t = 0$ and $t = 4$ seconds.

1
Find all points where $v (t) = 0$ in the interval:
2
$6t^2 - 30t + 36 = 6(t-2)(t-3) = 0 \implies t=2, t=3$
3
Check the sign of $v (t)$ on each sub-interval: $v > 0$ on $[0, 2]$ , $v < 0$ on $[2, 3]$ , $v > 0$ on $[3, 4]$
4
Set up the total distance integral:
5
$D = \int_0^2 v(t) dt + \int_2^3 -v(t) dt + \int_3^4 v(t) dt$
6
Evaluate each integral using the antiderivative $s (t) = 2 t^{3} - 15 t^{2} + 36 t + 4$ :
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$\int_0^2 v(t) dt = s(2)-s(0) = 32 - 4 = 32$
8
$\int_2^3 -v(t) dt = -(s(3)-s(2)) = -(31-32) = 1$
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$\int_3^4 v(t) dt = s(4)-s(3) = 36 - 31 = 5$
10
Add the results: $D = 32 + 1 + 5 = 38$ . Total distance is 38 metres.

4. Common Pitfalls

Wrong move:

Giving displacement instead of total distance when asked

Why:

Displacement is faster to calculate and the question wording is easy to misread

Correct move:

Always scan for the keywords "distance" vs "displacement" before starting your calculation

Wrong move:

Forgetting to solve for the constant of integration

Why:

It is common to assume the constant is zero without checking the given initial condition

Correct move:

Always write $+ C$ after integration, then explicitly solve for $C$ using the given initial condition

Wrong move:

Not splitting the integral at points where $v (t) = 0$

Why:

Direct integration of $v (t)$ over the whole interval cancels out positive and negative areas, giving displacement, not distance

Correct move:

Find all roots of $v (t)$ in the time interval first, then split the integral at every root

Wrong move:

Forgetting to include units in the final answer

Why:

Kinematic problems are applied, so units are required but easy to omit

Correct move:

Note the units given in the question, and add the correct units to your final answer

5. Quick Reference Cheatsheet

Quantity	Differential Relationship	Integral Relationship
Velocity	$v = \frac{d s}{d t}$	$v = \int a d t + C$
Acceleration	$a = \frac{d v}{d t} = \frac{d ^{2} s}{d t ^{2}}$	N/A
Displacement	N/A	$s = \int v d t + C$
Total Distance $[a, b]$	N/A	$\int_{a}^{b} ∣ v (t) ∣ d t$ , split at $v (t) = 0$

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 · 1
Find acceleration at given time
2022 · 2
Calculate total distance travelled
2021 · 1
Find displacement from acceleration

What's Next

Kinematic applications of calculus is a core applied calculus topic that is commonly tested in both Paper 1 and Paper 2 of IB AI SL. The concepts here connect directly to the core ideas of derivatives as rates of change and integrals as accumulators of change, reinforcing your understanding of core calculus principles. Mastering this topic will prepare you for other applied problems involving rates of change that appear in the exam, and builds on your basic kinematics knowledge from earlier in the course. It is also a popular question for extended response problems, so ensuring you are confident with all steps will help you earn full marks on these higher-weight questions.