Kinematics of a particle moving in a straight line

Q: What is the difference between area under a velocity-time and speed-time graph?

The area under a velocity-time graph gives displacement (areas below the x-axis are counted as negative for reverse motion). The area under a speed-time graph gives total distance travelled, as all areas are positive regardless of direction.

Edexcel International A-Level Mathematics· 2018 Issue 3 M1 §3.1· 25 min read

1. Suvat Notation and Formulae★★☆☆☆⏱ 8 min

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📘 Definition

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{)}

📐 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 $v^{2} = u^{2} + 2 a s$ .
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.

2. Displacement-Time and Acceleration-Time Graphs★★★☆☆⏱ 6 min

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📘 Definition

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.

3. Velocity-Time and Speed-Time Graphs★★★☆☆⏱ 7 min

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📘 Definition

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.

📐 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: $v = u + a t = 0 + 1 \times 3 = 3$ 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.

4. Combined Suvat and Graph Problem Solving★★★★☆⏱ 7 min

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📐 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 $v^{2} = u^{2} + 2 a s$ .
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 $s = u t + \frac{1}{2} a t^{2}$ , 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.

5. Common Pitfalls

Wrong move:

Using suvat formulae for motion with changing acceleration

Why:

Suvat only applies when acceleration is constant; variable acceleration requires calculus (covered in M2).

Correct move:

Confirm acceleration is constant before applying any suvat formula.

Wrong move:

Treating area under a velocity-time graph as total distance travelled

Why:

Velocity can be negative, so areas below the x-axis subtract from displacement, rather than adding to distance.

Correct move:

Use signed area under velocity-time graphs for displacement, and area under speed-time graphs for total distance.

Wrong move:

Taking $g = + 9.8$ m s⁻² when upwards is defined as the positive direction

Why:

Gravity acts downwards, so it must have the opposite sign to your chosen positive direction.

Correct move:

State your positive direction explicitly at the start of vertical motion questions, and assign $g$ the correct sign.

Wrong move:

Ignoring the vector nature of displacement, velocity and acceleration

Why:

Omitting negative signs for motion in the reverse direction leads to invalid calculation results.

Correct move:

Use consistent sign conventions for all vector quantities aligned to your chosen positive frame.

Wrong move:

Selecting a suvat formula that includes an unused unknown variable

Why:

This creates unnecessary simultaneous equations and increases the risk of arithmetic error.

Correct move:

List known and target variables first, select the formula that omits the unused variable.

6. Quick Reference Cheatsheet

Formula/Graph	Use Case	Key Interpretation
$v = u + a t$	Find $v / u / a / t$ when $s$ is unknown	Omits displacement
$s = u t + \frac{1}{2} a t^{2}$	Find $s / u / a / t$ when $v$ is unknown	Omits final velocity
$v^{2} = u^{2} + 2 a s$	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 = $Δ v$

7. Frequently Asked

Do I need to memorise suvat formulae for the exam?

Yes, the 5 suvat formulae are not included in the Edexcel IAL M1 formula booklet, so you must memorise all of them. You will also be expected to use g = 9.8 m s⁻² for all vertical motion questions.

What is the difference between area under a velocity-time and speed-time graph?

The area under a velocity-time graph gives displacement (areas below the x-axis are counted as negative for reverse motion). The area under a speed-time graph gives total distance travelled, as all areas are positive regardless of direction.

Going deeper

formula sheetEdexcel IAL Maths Formula BookletSuvat formulae are not provided: memorise all 5

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)