Kinematics with calculus

IB Mathematics: Analysis and Approaches HL· Syllabus 5.9: Kinematic problems (displacement, velocity, acceleration; total distance travelled)· 16 min read

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

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For a particle moving along a straight line, the three core kinematic quantities are linked by rates of change. The constant-acceleration (SUVAT) equations do NOT apply here, because acceleration is generally a function of time. In HL you will meet motion functions that are polynomial, trigonometric, exponential, or combinations of these.

📘 Definition

Kinematic Derivative Relationships

Velocity is the first time derivative of displacement, and acceleration is the first time derivative of velocity, or equivalently 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, and state whether the particle is speeding up or slowing down at that instant.

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
$a(2) = 12(2) - 18 = 24 - 18 = 6 \text{ m/s}^2$
9
At $t = 2$ the particle is instantaneously at rest ( $v = 0$ ) and about to move in the positive direction, since $a (2) > 0$ . It is a turning point of the motion.

A particle is speeding up when velocity and acceleration have the same sign (their product $v \cdot a > 0$ ) and slowing down when they have opposite signs ( $v \cdot a < 0$ ). This is a common HL follow-up to a differentiation part.

✓ 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. Recovering Velocity and Displacement by Integration★★★☆☆⏱ 13 min

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Integration reverses differentiation: integrating acceleration gives velocity, and integrating velocity gives displacement. Each indefinite integration introduces a constant, which you determine from a given boundary (initial) condition. At HL the integrand is often trigonometric or exponential, so lean on your standard integrals and, where needed, substitution.

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

📐 Worked Example

A particle starts from rest and has acceleration $a (t) = 4 cos (2 t)$ m/s $^{2}$ for $t \geq 0$ . Given that at $t = 0$ the velocity is $v (0) = 1$ m/s and the displacement is $s (0) = 0$ , find $v (t)$ and $s (t)$ .

1
Integrate acceleration, using $\int cos (k t) d t = \frac{1}{k} sin (k t)$ :
2
$v(t) = \int 4\cos(2t)\,dt = 2\sin(2t) + c_1$
3
Apply $v (0) = 1$ : $2 sin (0) + c_{1} = 1$ , so $c_{1} = 1$ and $v (t) = 2 sin (2 t) + 1$ .
4
Integrate velocity, using $\int sin (k t) d t = - \frac{1}{k} cos (k t)$ :
5
$s(t) = \int \big(2\sin(2t) + 1\big)\,dt = -\cos(2t) + t + c_2$
6
Apply $s (0) = 0$ : $- cos (0) + 0 + c_{2} = 0 \Rightarrow - 1 + c_{2} = 0$ , so $c_{2} = 1$ .
7
$s(t) = -\cos(2t) + t + 1$

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

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To find total distance, first locate every time in the interval where $v (t) = 0$ (candidate points where the particle reverses direction). Split the interval at these roots, integrate $∣ v (t) ∣$ over each sub-interval, and sum. Equivalently, total distance $= \int_{t_{1}}^{t_{2}} ∣ v (t) ∣ d t$ . On Paper 2 this single integral of the absolute value can be evaluated directly on your GDC.

\text{Total distance} = \int_{t_1}^{t_2} |v(t)|\,dt

📐 Worked Example

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

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 the sign of $v (t)$ on each interval: positive on $0 < t < 1$ , negative on $1 < t < 2$ , positive on $2 < t < 3$ .
3
Integrate the absolute value of $v (t)$ over each interval:
4
$\int_0^1 v(t)\,dt = 5, \quad \int_1^2 -v(t)\,dt = 1, \quad \int_2^3 v(t)\,dt = 5$
5
Sum the three results to get total distance: $5 + 1 + 5 = 11$ m.

📐 Worked Example

A particle moves with velocity $v (t) = 2 cos (t)$ m/s for $0 \leq t \leq 2 π$ . Find the net displacement and the total distance travelled (Paper 2 / GDC method).

1
Net displacement is the signed integral of velocity over the whole interval:
2
$\int_0^{2\pi} 2\cos(t)\,dt = \big[2\sin(t)\big]_0^{2\pi} = 0 \text{ m}$
3
The particle returns to its start, so net displacement is 0, but it has clearly moved. Compute total distance from the absolute value:
4
$\int_0^{2\pi} |2\cos(t)|\,dt = 8 \text{ m}$
5
Enter $\int_{0}^{2 π} abs (2 cos (t)) d t$ directly on the GDC to obtain 8 m; no manual interval splitting is required.

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 (signed) / total distance (unsigned)
Acceleration vs time	Rate of change of acceleration	Change in velocity

📐 Worked Example

A velocity-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 the net displacement and the total distance travelled.

1
The line crosses $v = 0$ where the graph reaches zero; by similar triangles this is at $t = 2$ .
2
Positive triangle from $t = 0$ to $t = 2$ : area $= \frac{1}{2} \times 2 \times 4 = 4$ m.
3
Negative triangle from $t = 2$ to $t = 3$ : area $= \frac{1}{2} \times 1 \times 2 = 1$ m (below the axis).
4
Net displacement $= 4 - 1 = 3$ m; 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, returning net displacement not total path length

Correct move:

Integrate $∣ v (t) ∣$ : find all roots of $v (t)$ in the interval, split at these points and sum absolute values, or use the GDC abs-integral

Wrong move:

Treating acceleration as the first derivative of displacement

Why:

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

Correct move:

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

Wrong move:

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

Why:

Indefinite integration produces an arbitrary constant that fixes the initial velocity or position

Correct move:

Use the given boundary condition (e.g. $v (0) = 1$ , $s (0) = 3$ ) to solve for the constant before finalizing the function

Wrong move:

Dropping the $\frac{1}{k}$ factor when integrating $sin (k t)$ or $cos (k t)$

Why:

The chain rule on the inner function $k t$ scales the integral; ignoring it gives velocities and displacements that are off by a factor of $k$

Correct move:

Apply $\int cos (k t) d t = \frac{1}{k} sin (k t)$ and $\int sin (k t) d t = - \frac{1}{k} cos (k t)$ carefully

Wrong move:

Assuming velocity is zero when acceleration is zero

Why:

Zero acceleration means velocity is momentarily constant (a turning point of $v$ ), not that it equals zero

Correct move:

Set $v (t) = 0$ explicitly to find where the particle is instantaneously at rest 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: displacement m, velocity m/s, acceleration m/s $^{2}$

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
Speeding up / slowing down	$v \cdot a > 0$ speeding up; $v \cdot a < 0$ slowing down	—

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 with GDC, total distance, 7-mark question
2022 · Paper 1
Velocity and acceleration by differentiation, analytic, 8-mark question
2021 · Paper 2
Trigonometric velocity, direction changes and total distance, 6-mark question

What's Next

Mastering kinematics with calculus is a high-yield HL skill that appears in almost every IB Math AA HL Paper 2, often worth 6-9 marks, and in analytic form on Paper 1. It fuses your differentiation and integration fluency, including trigonometric and exponential integrals and substitution, with function-root analysis and definite-integral evaluation. Expect these ideas to reappear inside longer applied-calculus questions on related rates, optimization and modelling, and be ready to switch fluently between exact analytic working (Paper 1) and efficient GDC evaluation of the total-distance integral (Paper 2).