# Kinematics with calculus

> IB Mathematics: Analysis and Approaches HL · IB AA HL Unit 5: Calculus
> Source: https://www.owlsprep.com/study/ib-math-aa-hl-u5-kinematics-with-calculus/

This module connects HL differential and integral calculus to 1D kinematics: the derivative and integral relationships between displacement, velocity and acceleration, variable-acceleration problem solving that draws on trigonometric and exponential integration, total distance versus net displacement, and both analytic (Paper 1) and GDC (Paper 2) methods.

**Prerequisites:** [Differentiation of polynomial, trigonometric and exponential functions, including the chain rule](https://www.owlsprep.com/study/ib-math-aa-hl-u5-differentiation-rules/); [Definite and indefinite integration, including integration by substitution](https://www.owlsprep.com/study/ib-math-aa-hl-u5-integration-techniques-substitution-parts-partial/)

## Learning objectives

- Define displacement, velocity and acceleration as time-dependent 1D motion functions
- Relate the three kinematic quantities via differentiation and indefinite/definite integration
- Recover velocity from acceleration and displacement from velocity, integrating trigonometric and exponential motion functions and applying boundary conditions
- Solve exam-style problems for motion with non-constant acceleration, analytically (Paper 1) and with a GDC (Paper 2)
- Distinguish between net displacement and total distance travelled for changing-direction motion using total distance = the integral of |v(t)|

## Core Kinematic Relationships via Differentiation

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.

**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 $s(t) = 2t^3 - 9t^2 + 12t$ 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.

**Check your understanding**

Confirm you understand the derivative relationships:

1. If $s(t) = 5\sin(t)$, what is $a(t)$?

   - $5\cos(t)$
   - $-5\sin(t)$
   - $-5\cos(t)$
   - $5\sin(t)$

   *Why:* Differentiate twice: first derivative is $5\cos(t)$, second is $-5\sin(t)$.

*Calculator:* allowed

## Recovering Velocity and Displacement by Integration

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 $v(t) = 6t^2 - 18t + 12$ 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 $a(t) = 4\cos(2t)$ 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 $\int \cos(kt)\,dt = \frac{1}{k}\sin(kt)$:
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 $v(t) = 2\sin(2t) + 1$.
4. Integrate velocity, using $\int \sin(kt)\,dt = -\frac{1}{k}\cos(kt)$:
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. $$s(t) = -\cos(2t) + t + 1$$

> **tip**
>
> Net displacement can be negative, which means the particle ends up on the negative side of its starting position, not that it travelled a negative distance. Keep the sign; it carries direction information.

*Calculator:* allowed

## Total Distance Travelled

> **warning**
>
> Total distance travelled is NOT equal to net displacement if the particle changes direction during the time interval.

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 $= \displaystyle\int_{t_1}^{t_2} |v(t)|\,dt$. 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 $v(t) = 6t^2 - 18t + 12$, 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 $0 \leq t \leq 2\pi$. 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 $\int_0^{2\pi} \text{abs}(2\cos(t))\,dt$ directly on the GDC to obtain 8 m; no manual interval splitting is required.

**Exam command terms**

IB exam questions use specific command terms for kinematics:

- **Find the displacement** — Calculate the net signed change in position; a single definite integral of $v(t)$, no interval splitting needed

- **Find the total distance travelled** — Integrate $|v(t)|$: split at the $v(t)=0$ points and sum absolute values, or evaluate the abs-integral on the GDC

*Calculator:* allowed

## Graphical Interpretation of Kinematic Functions

| 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 $= \tfrac{1}{2} \times 2 \times 4 = 4$ m.
3. Negative triangle from $t=2$ to $t=3$: area $= \tfrac{1}{2} \times 1 \times 2 = 1$ m (below the axis).
4. Net displacement $= 4 - 1 = 3$ m; total distance $= 4 + 1 = 5$ m.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using a single definite integral of $v(t)$ to calculate total distance directly
  - Why it fails: Negative-velocity regions cancel positive displacement, returning net displacement not total path length
  - Correct: 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:** Treating acceleration as the first derivative of displacement
  - Why it fails: Mixing up the order of differentiation skips the velocity step, giving wrong units and values
  - Correct: Follow the chain: differentiate displacement once for velocity, then differentiate velocity once for acceleration
- **Wrong:** Forgetting the constant of integration when finding $v(t)$ or $s(t)$
  - Why it fails: Indefinite integration produces an arbitrary constant that fixes the initial velocity or position
  - Correct: Use the given boundary condition (e.g. $v(0)=1$, $s(0)=3$) to solve for the constant before finalizing the function
- **Wrong:** Dropping the $\tfrac{1}{k}$ factor when integrating $\sin(kt)$ or $\cos(kt)$
  - Why it fails: The chain rule on the inner function $kt$ scales the integral; ignoring it gives velocities and displacements that are off by a factor of $k$
  - Correct: Apply $\int\cos(kt)\,dt=\tfrac{1}{k}\sin(kt)$ and $\int\sin(kt)\,dt=-\tfrac{1}{k}\cos(kt)$ carefully
- **Wrong:** Assuming velocity is zero when acceleration is zero
  - Why it fails: Zero acceleration means velocity is momentarily constant (a turning point of $v$), not that it equals zero
  - Correct: Set $v(t)=0$ explicitly to find where the particle is instantaneously at rest or changes direction
- **Wrong:** Stating acceleration units as m/s
  - Why it fails: Acceleration is the rate of change of velocity, so units are metres per second squared
  - Correct: Always confirm units match the quantity: displacement m, velocity m/s, acceleration m/s$^2$

## Cheatsheet

| Quantity | Mathematical Relationship | Standard Units |
| --- | --- | --- |
| Displacement $s(t)$ | $s(t) = \int v(t)\,dt + c$ | m |
| Velocity $v(t)$ | $v(t) = \frac{ds}{dt} = \int a(t)\,dt + c$ | m/s |
| Acceleration $a(t)$ | $a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}$ | m/s² |
| Net Displacement $t_1$ to $t_2$ | $\int_{t_1}^{t_2} v(t)\,dt$ | m |
| Total Distance $t_1$ to $t_2$ | $\int_{t_1}^{t_2} \|v(t)\|\,dt$ | m |
| Speeding up / slowing down | $v\cdot a>0$ speeding up; $v\cdot a<0$ slowing down | — |

## 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).

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