# Kinematics — Motion in a Straight Line

> CIE IGCSE Additional Mathematics · 2025-2027
> Source: https://www.owlsprep.com/study/cie-0606-u14-kinematics-motion-in-a-straight/

This guide covers calculus applications for straight line kinematics, motion graph interpretation, and scalar vs vector quantity distinctions, aligned with CIE IGCSE Add Maths 0606 2025-2027 syllabus requirements.

**Prerequisites:** Basic differentiation of polynomial, exponential, and trigonometric functions; Basic indefinite and definite integration

## Learning objectives

- Relate displacement, velocity, and acceleration using differentiation and integration
- Distinguish between scalar (distance, speed) and vector (displacement, velocity) quantities for 1D motion
- Interpret and sketch displacement, velocity, speed, and acceleration-time graphs
- Solve kinematics problems with both constant and variable acceleration
- Apply sign conventions correctly for straight line motion

## Core Kinematic Relationships & Sign Conventions

**Kinematic Calculus Relationships** — For straight line motion, the three core quantities are linked by calculus: velocity is the first derivative of displacement, acceleration is the first derivative of velocity (or second derivative of displacement). The reverse relationships use integration, with constants of integration representing initial conditions.

*Notation:* $v = \frac{ds}{dt}$, $a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$, $s = \int v \, dt$, $v = \int a \, dt$

> **Sign Convention Rule**
>
> Choose a positive direction (e.g. right/up) at the start of every problem. Negative values for velocity or displacement indicate motion opposite to your chosen positive direction.

**Worked example:** A particle moves in a straight line with displacement $s = 2t^3 - 5t^2 + 3t$ meters from the origin at time $t$ seconds. Find the velocity of the particle at $t=2$s.

1. Differentiate the displacement function to get velocity:

   $$v = \frac{ds}{dt} = 6t^2 - 10t + 3$$
2. Substitute $t=2$ into the velocity function:

   $$v = 6(2)^2 - 10(2) + 3 = 24 - 20 + 3 = 7$$
3. Final velocity is $7$ m/s in the positive direction.

> **Exam tip:** Always state your chosen positive direction in written answers to avoid sign error mark deductions.

## Constant vs Variable Acceleration Problems

The 0606 syllabus tests both constant acceleration and variable acceleration, and in 0606 both are handled with calculus. Differentiate to move from displacement to velocity to acceleration ($v = \frac{ds}{dt}$, $a = \frac{dv}{dt}$), and integrate to move back the other way ($v = \int a \, dt$, $s = \int v \, dt$), using the initial conditions to fix each constant of integration. When the acceleration is constant, integrating a constant simply gives a velocity that is linear in $t$; the List of formulas provides no separate constant-acceleration formulas, so you integrate in every case.

**Worked example:** A particle has constant acceleration $a=4$ m/s², initial velocity $u=2$ m/s, and initial displacement $s=0$ at $t=0$. Find the displacement at $t=3$s.

1. Integrate acceleration to get velocity, add constant of integration $C$:

   $$v = \int 4 \, dt = 4t + C$$
2. Use initial condition $v=2$ at $t=0$ to find $C=2$, so $v=4t+2$
3. Integrate velocity to get displacement, add constant $D$:

   $$s = \int (4t + 2) \, dt = 2t^2 + 2t + D$$
4. Use initial condition $s=0$ at $t=0$ to find $D=0$, substitute $t=3$:

   $$s = 2(3)^2 + 2(3) = 18 + 6 = 24$$
5. Final displacement is $24$ m.

**Worked example:** A particle has variable acceleration $a = 2t + 1$ m/s², with initial velocity $0$ m/s. Find velocity at $t=2$s.

1. Integrate acceleration to get velocity:

   $$v = \int (2t + 1) \, dt = t^2 + t + C$$
2. Use initial condition $v=0$ at $t=0$ to find $C=0$, substitute $t=2$:

   $$v = (2)^2 + 2 = 6$$
3. Final velocity is $6$ m/s.

> **Exam tip:** Whether the acceleration is constant or variable, use differentiation and integration throughout — 0606 provides no ready-made kinematics formulas, so find every velocity or displacement by integrating and applying the initial conditions.

## Motion Graph Interpretation & Sketching

| Graph Type | Gradient Represents | Area Under Graph Represents |
| --- | --- | --- |
| Displacement-Time ($s$-$t$) | Velocity ($v$) | No physical meaning |
| Velocity-Time ($v$-$t$) | Acceleration ($a$) | Change in displacement (signed area) |
| Speed-Time (speed-$t$) | Magnitude of acceleration | Total distance travelled (sum of positive areas) |
| Acceleration-Time ($a$-$t$) | Not required in 0606 | Change in velocity |

**Worked example:** A $v$-$t$ graph has a straight line from $(0,0)$ to $(2,4)$, a horizontal line to $(5,4)$, then a straight line down to $(7,0)$. Find the total distance travelled over 7 seconds.

1. Total distance is the total area under the $v$-$t$ graph (all values are positive, so no need for absolute values).
2. Area of first triangle: $0.5 \times 2 \times 4 = 4$ m
3. Area of rectangle: $3 \times 4 = 12$ m
4. Area of second triangle: $0.5 \times 2 \times 4 = 4$ m
5. Total distance: $4 + 12 + 4 = 20$ m

> **Exam tip:** For displacement from a $v$-$t$ graph, keep signs of areas below the $t$-axis. For distance, take absolute values of all areas.

## Scalar vs Vector Quantity Distinctions

A common exam trap is mixing scalar and vector quantities. Displacement and velocity are vectors (can be negative), while distance and speed are scalars (always non-negative). When calculating total distance over an interval, you must account for direction reversals.

**Worked example:** A particle's displacement is given by $s = t^2 - 4t$ meters, $t$ in seconds. Find the total distance travelled between $t=0$ and $t=3$s.

1. Find when velocity is zero (direction reversal point):

   $$v = \frac{ds}{dt} = 2t - 4 = 0 \implies t = 2$$
2. Calculate displacement at key points: $s(0)=0$, $s(2) = 4 - 8 = -4$ m, $s(3) = 9 - 12 = -3$ m
3. Calculate distance for each segment: $|s(2)-s(0)| = 4$ m, $|s(3)-s(2)| = 1$ m
4. Total distance: $4 + 1 = 5$ m

> **Exam tip:** Never calculate total distance as the difference between final and initial displacement, as this ignores direction reversals.

## Common pitfalls

- **Wrong:** Reaching for a ready-made constant-acceleration formula instead of integrating
  - Why it fails: 0606 lists no kinematics formulas, and such a shortcut fails as soon as the acceleration varies with time.
  - Correct: Always use differentiation and integration ($a = \frac{dv}{dt}$, $v = \int a \, dt$, $s = \int v \, dt$), fixing the constant of integration from the initial conditions — this works for constant and variable acceleration alike.
- **Wrong:** Calculating displacement from a speed-time graph
  - Why it fails: Speed is a scalar quantity, so the area under a speed-time graph gives total distance, not signed displacement.
  - Correct: Use a velocity-time graph (with sign values) to calculate displacement.
- **Wrong:** Omitting the constant of integration when solving kinematics integrals
  - Why it fails: The constant represents initial velocity or displacement, so omitting it leads to incorrect final values.
  - Correct: Use given initial conditions (e.g. $s=0$ at $t=0$) to solve for the constant every time you integrate.
- **Wrong:** Calculating total distance as $s_{final} - s_{initial}$
  - Why it fails: If the particle reverses direction, the net displacement is smaller than the total path length travelled.
  - Correct: Find all points where $v=0$, calculate displacement for each segment, and sum the absolute values of each segment's displacement.

## Cheatsheet

| Quantity | Type | Calculus Relationship | Graph Property |
| --- | --- | --- | --- |
| Displacement ($s$) | Vector | $s = \int v \, dt$ | Gradient of $s$-$t$ graph = $v$ |
| Velocity ($v$) | Vector | $v = \frac{ds}{dt} = \int a \, dt$ | Gradient of $v$-$t$ graph = $a$, Area under $v$-$t$ = $\Delta s$ |
| Acceleration ($a$) | Vector | $a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$ | Area under $a$-$t$ graph = $\Delta v$ |
| Distance | Scalar | Sum of absolute displacement segments | Area under speed-time graph |
| Speed | Scalar | Magnitude of velocity | Absolute value of $v$ on $v$-$t$ graph |

## What's next

Now that you have mastered straight line kinematics for CIE IGCSE Additional Mathematics 0606, you can move on to applying calculus to more advanced problems involving exponential and trigonometric motion functions, which are frequently tested in Paper 2. This topic is often combined with other calculus skills in longer structured questions, so make sure you are confident with differentiation and integration rules for all function types in the syllabus. Practice past paper kinematics questions to familiarize yourself with common exam phrasing and trap scenarios.

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