# Kinematics of a particle moving in a straight line

> Edexcel International A-Level Mathematics · IAL M1
> Source: https://www.owlsprep.com/study/edexcel-ial-math-m1-kinematics-of-a-particle-moving/

This guide covers Edexcel IAL M1 straight line kinematics with constant acceleration, including the full suvat formula set, 4 motion graph types, and exam-style problem solving aligned to the 2018 IAS specification.

**Prerequisites:** [Basic understanding of scalar and vector quantities](https://www.owlsprep.com/study/edexcel-ial-math-m1-introduction-to-mechanics/); Ability to calculate gradients and areas of linear graphs

## Learning objectives

- Recall and apply all 5 suvat formulae for constant acceleration straight line motion
- Interpret displacement-time, velocity-time, speed-time and acceleration-time graphs
- Calculate gradients (velocity/acceleration) and areas (displacement/distance) from motion graphs
- Solve Edexcel IAL M1 exam-style kinematics problems using algebraic and graphical methods
- Apply consistent sign conventions for vertical motion using g = 9.8 m s⁻²

## Suvat Notation and Formulae

**Constant acceleration motion** — Motion of a particle in a straight line where acceleration remains fixed in magnitude and direction for the full time interval. Described using the 5 suvat formulae.

*Example:* A car accelerating uniformly from rest to 20 m s⁻¹ over 10 seconds

Each suvat formula omits one of the 5 variables, so you can solve for an unknown if you know 3 other values. The full set of formulae are:

$$v = u + at \quad \text{(omits } s\text{)}$$

$$s = ut + \frac{1}{2} a t^2 \quad \text{(omits } v\text{)}$$

$$s = vt - \frac{1}{2} a t^2 \quad \text{(omits } u\text{)}$$

$$v^2 = u^2 + 2 a s \quad \text{(omits } t\text{)}$$

$$s = \frac{1}{2}(u + v) t \quad \text{(omits } a\text{)}$$

> **tip**
>
> List your 3 known variables and 1 target unknown first to select the formula that omits the unused variable, avoiding unnecessary simultaneous equations.

**Worked example:** A cyclist accelerates uniformly from rest to 12 m s⁻¹ over a distance of 40 m. Calculate the magnitude of the cyclist's acceleration, correct to 2 significant figures.

1. List known values: $u = 0$ m s⁻¹, $v = 12$ m s⁻¹, $s = 40$ m. Target unknown is $a$, unused variable is $t$, so use $v^2 = u^2 + 2as$.
2. $$12^2 = 0^2 + 2 \times a \times 40$$
3. $$144 = 80a$$
4. $$a = \frac{144}{80} = 1.8 \text{ m s}^{-2} \text{ (2 s.f.)}$$

> **Exam tip:** For vertical motion, explicitly state your chosen positive direction (usually upwards) to avoid sign errors with g.

*Calculator:* allowed

## Displacement-Time and Acceleration-Time Graphs

**Displacement-time graph** — Graph with time on the x-axis and displacement from a fixed origin on the y-axis. The gradient of any linear segment equals the constant velocity of the particle over that time interval.

For constant acceleration motion, displacement-time graphs are made of straight line segments: a flat line means the particle is stationary, a positive gradient means motion in the positive direction, and a negative gradient means motion in the reverse direction.

**Worked example:** A car travels at a constant 15 m s⁻¹ for 6 seconds, then remains stationary for 4 seconds. Calculate the total displacement of the car after 10 seconds.

1. First segment (0 to 6 s): gradient = 15 m s⁻¹, so displacement = $15 \times 6 = 90$ m.
2. Second segment (6 to 10 s): gradient = 0, so displacement does not change, remaining at 90 m.
3. Total displacement after 10 s = 90 m.

Acceleration-time graphs for constant acceleration motion are horizontal straight lines: a line at 0 means constant velocity, a positive horizontal line means constant positive acceleration, and a negative horizontal line means constant deceleration. The area under the line over a time interval equals the change in velocity over that interval.

> **Exam tip:** Displacement can be negative if the particle moves past the origin in the opposite direction, so graphs can fall below the x-axis.

*Calculator:* allowed

## Velocity-Time and Speed-Time Graphs

**Velocity-time graph** — Graph with time on the x-axis and velocity on the y-axis. The gradient of a linear segment equals constant acceleration, and the signed total area under the line equals total displacement.

> **note**
>
> Speed-time graphs plot the absolute value of velocity, so all values are positive. The total area under a speed-time graph equals total distance travelled, regardless of direction of motion.

**Worked example:** A lift accelerates uniformly from rest at 1 m s⁻² for 3 seconds, moves at constant velocity for 5 seconds, then decelerates uniformly to rest over 2 seconds. Calculate the total displacement of the lift using a velocity-time graph.

1. Calculate maximum velocity: $v = u + at = 0 + 1 \times 3 = 3$ m s⁻¹.
2. Total area under velocity-time graph = area of first triangle + area of rectangle + area of final triangle:
3. $$\text{Area} = (\frac{1}{2} \times 3 \times 3) + (3 \times 5) + (\frac{1}{2} \times 2 \times 3) = 4.5 + 15 + 3 = 22.5 \text{ m}$$

> **Exam tip:** For total distance travelled from a velocity-time graph, take the absolute value of any area below the x-axis before summing all areas.

*Calculator:* allowed

## Combined Suvat and Graph Problem Solving

**Exam command terms**

Common command terms for kinematics questions:

- **Show that** — Derive the given result using suvat or graph methods, show all working *(Show that the maximum height of the ball is 11 m)*

- **Sketch** — Draw a labelled graph with key values marked on axes, no graph paper required *(Sketch the velocity-time graph for the lift's motion)*

- **Find** — Calculate the unknown value, show intermediate steps for full marks

**Worked example:** A ball is thrown vertically upwards from ground level with initial speed 14.7 m s⁻¹. Air resistance is negligible, use g = 9.8 m s⁻². Find the maximum height reached by the ball, and the total time it is in the air before hitting the ground.

1. Define upwards as positive, so $a = -9.8$ m s⁻². At maximum height, $v = 0$. Target unknown: $s$, unused variable: $t$, so use $v^2 = u^2 + 2as$.
2. $$0 = 14.7^2 + 2(-9.8)s$$
3. $$s = \frac{216.09}{19.6} = 11.025 \approx 11 \text{ m (2 s.f.)}$$
4. To find total time in air, use $s = ut + \frac{1}{2} a t^2$, with $s = 0$ (ball returns to ground):
5. $$0 = 14.7t - 4.9 t^2$$
6. $$t(14.7 - 4.9t) = 0 \implies t = 0 \text{ (start)} \text{ or } t = 3 \text{ s}$$

> **Exam tip:** Use graphical methods to cross-check algebraic suvat answers, and vice versa, to catch sign or arithmetic errors in the exam.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using suvat formulae for motion with changing acceleration
  - Why it fails: Suvat only applies when acceleration is constant; variable acceleration requires calculus (covered in M2).
  - Correct: Confirm acceleration is constant before applying any suvat formula.
- **Wrong:** Treating area under a velocity-time graph as total distance travelled
  - Why it fails: Velocity can be negative, so areas below the x-axis subtract from displacement, rather than adding to distance.
  - Correct: Use signed area under velocity-time graphs for displacement, and area under speed-time graphs for total distance.
- **Wrong:** Taking $g = +9.8$ m s⁻² when upwards is defined as the positive direction
  - Why it fails: Gravity acts downwards, so it must have the opposite sign to your chosen positive direction.
  - Correct: State your positive direction explicitly at the start of vertical motion questions, and assign $g$ the correct sign.
- **Wrong:** Ignoring the vector nature of displacement, velocity and acceleration
  - Why it fails: Omitting negative signs for motion in the reverse direction leads to invalid calculation results.
  - Correct: Use consistent sign conventions for all vector quantities aligned to your chosen positive frame.
- **Wrong:** Selecting a suvat formula that includes an unused unknown variable
  - Why it fails: This creates unnecessary simultaneous equations and increases the risk of arithmetic error.
  - Correct: List known and target variables first, select the formula that omits the unused variable.

## Cheatsheet

| Formula/Graph | Use Case | Key Interpretation |
| --- | --- | --- |
| $v = u + at$ | Find $v/u/a/t$ when $s$ is unknown | Omits displacement |
| $s = ut + \frac{1}{2} a t^2$ | Find $s/u/a/t$ when $v$ is unknown | Omits final velocity |
| $v^2 = u^2 + 2as$ | Find $v/u/a/s$ when $t$ is unknown | Omits time |
| $s = \frac{1}{2}(u+v)t$ | Find $s/u/v/t$ when $a$ is unknown | Omits acceleration |
| Displacement-time graph | Find velocity over a time interval | Gradient = velocity |
| Velocity-time graph | Find acceleration and displacement | Gradient = acceleration; signed area = displacement |
| Speed-time graph | Find total distance travelled | Total area = total distance |
| Acceleration-time graph | Find change in velocity | Area under line = $\Delta v$ |

## What's next

Now that you have mastered straight line constant acceleration kinematics for Edexcel IAL M1, you are ready to progress to the next core Mechanics 1 topics. This kinematics knowledge forms the foundation for more advanced motion topics in later units, including variable acceleration (M2) and projectile motion (M2), but you will also apply these suvat and graph skills immediately in upcoming M1 topics such as forces and Newton's laws of motion, where you will calculate acceleration from force values before solving kinematics problems. Kinematics questions appear on almost every M1 exam paper, often as the first 5-8 mark question, so practicing past paper problems on this topic is an easy way to secure high marks early in the paper.

- [Vectors in Mechanics (M1)](https://www.owlsprep.com/study/edexcel-ial-math-m1-vectors-in-mechanics/)

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