Kinematics of Motion in a Straight Line — A-Level Mathematics Mechanics Study Guide

For: A-Level Mathematics candidates sitting Paper 4 (Mechanics).

Covers: sign conventions for displacement, velocity and acceleration, suvat constant acceleration formulas, variable acceleration calculus, velocity-time graph interpretation, vertical motion under gravity, multi-stage motion problems, average vs instantaneous velocity, and return-to-origin/meeting point calculations.

You should already know: Basic differentiation and integration (P1 calculus), reading graphs.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Mathematics style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Kinematics of Motion in a Straight Line?

Kinematics of motion in a straight line (also called 1D linear kinematics) is the study of objects moving along a fixed straight path, describing their position, speed, and acceleration without analyzing the forces that cause their motion. It is the first topic in the A-Level Mathematics Mechanics syllabus, and forms the base for all future motion-related topics including projectiles, connected particles, and circular motion. All problems in this topic use a single coordinate axis aligned with the path of motion, simplifying vector calculations to signed scalar values.

2. Displacement, velocity, acceleration — definitions and signs

All three quantities are vectors, so their sign indicates direction relative to a pre-defined positive direction and fixed origin point:

• Displacement ($s$, units: m): The straight-line distance of the object from the origin, positive if in the positive direction, negative if in the opposite direction. This is not the same as distance, a scalar quantity that measures total path length regardless of direction.
• Velocity ($v$, units: m/s): The rate of change of displacement with respect to time: $v=\frac{\Delta s}{\Delta t}$. Positive if the object is moving in the positive direction, negative if moving backwards. Speed is the scalar magnitude of velocity, always positive.
• Acceleration ($a$, units: m/s²): The rate of change of velocity with respect to time: $a=\frac{\Delta v}{\Delta t}$. Positive if velocity increases in the positive direction, or decreases in the negative direction; negative if velocity decreases in the positive direction (deceleration) or increases in the negative direction.

Worked example

Take the origin as a bus stop, positive direction as north. If you walk 6m north then 2m south in 4 seconds: your displacement is $+4$ m, total distance is 8m, average velocity is $4/4=1$ m/s north, average speed is $8/4=2$ m/s. If you speed up from 1 m/s north to 3 m/s north in 1 second, your acceleration is $+2$ m/s²; if you then slow to a stop in 0.5 seconds, acceleration is $(0-3)/0.5=-6$ m/s².

3. Constant-acceleration formulas (suvat)

When acceleration is constant, we derive three core formulas from the basic definitions of velocity and acceleration, referred to as the suvat equations after their variable names: $s$ = displacement, $u$ = initial velocity, $v$ = final velocity, $a$ = constant acceleration, $t$ = time elapsed.

1. $v=u+at$: Derived directly from rearranging $a=\frac{v-u}{t}$
2. $s=ut+\frac{1}{2}a{t}^2$: Derived by multiplying average velocity $\frac{u+v}{2}$ by time, substituting $v$ from the first equation
3. $v^2=u^2+2as$: Derived by eliminating $t$ from the first two equations

Worked example

A van accelerates from rest at 2.5 m/s² for 6 seconds. Calculate its final velocity and total distance travelled. First list known values: $u=0$, $a=2.5$, $t=6$, unknowns $v$ and $s$.

• Final velocity: $v=u+at=0+(2.5)(6)=15$ m/s
• Distance: $s=ut+\frac{1}{2}a{t}^2=0+0.5(2.5)(36)=45$ m You can verify with the third equation: $15^2=0+2(2.5)(45)\to 225=225$, correct.

4. Variable acceleration — $v=\frac{ds}{dt}$, $a=\frac{dv}{dt}$, $s=\int vdt$

When acceleration is not constant (e.g. given as a function of time, or displacement is a non-linear function of time), suvat equations do not apply, and you use P1 calculus instead:

• To differentiate down from displacement: Velocity is the first derivative of displacement, acceleration is the first derivative of velocity (second derivative of displacement): $$v=\frac{ds}{dt},a=\frac{dv}{dt}=\frac{d^{2}s}{d{t}^2}$$
• To integrate up from acceleration: Velocity is the integral of acceleration, displacement is the integral of velocity. Always add a constant of integration, solved using initial conditions (e.g. $s=0$ at $t=0$): $$v=\int adt+u,s=\int vdt+s_0$$ where $u$ = initial velocity, $s_0$ = initial displacement.

Worked example

A particle's displacement is given by $s=t^3-3t^2+2t$, $t\ge 0$ (units: m, s). Find its acceleration at $t=2$s, and its displacement at $t=3$s if it starts at the origin.

• $v=\frac{ds}{dt}=3t^2-6t+2$
• $a=\frac{dv}{dt}=6t-6$, so at $t=2$, $a=12-6=6$ m/s²
• At $t=3$, $s=27-27+6=6$ m (no constant needed as initial displacement is 0)

5. Velocity-time graphs — area = displacement, gradient = acceleration

Velocity-time graphs plot velocity on the y-axis and time on the x-axis, with two core exam-tested rules:

1. Gradient = acceleration: The slope of the graph at any point equals instantaneous acceleration. For straight-line segments, gradient gives constant acceleration; for curved segments, draw a tangent to calculate gradient.
2. Signed area under graph = displacement: The area between the graph and the time axis gives displacement between two times. Area above the axis is positive displacement, area below is negative displacement. Total distance travelled is the sum of the absolute values of all areas.

Worked example

A v-t graph has a straight line from (0,0) to (5,10), then a horizontal line to (12,10), then a straight line to (16,0). Calculate total displacement and average acceleration over 16s.

• Area 1 (triangle): $0.5\times 5\times 10=25$ m
• Area 2 (rectangle): $7\times 10=70$ m
• Area 3 (triangle): $0.5\times 4\times 10=20$ m
• Total displacement: $25+70+20=115$ m
• Average acceleration: $\frac{\text{final velocity}-\text{initial velocity}}{\text{total time}}=\frac{0-0}{16}=0$ m/s²

6. Vertical motion under gravity (using $g=9.8$ or $10$ m/s²)

Objects moving vertically near the Earth's surface experience constant acceleration $g$ (gravitational acceleration) downwards, with value specified in the question as 9.8 or 10 m/s². The most common convention is to take upwards as positive, so acceleration $a=-g$. Key points:

• Maximum height occurs when instantaneous velocity $v=0$ (the object stops moving up before falling down)
• Return to the launch point occurs when displacement $s=0$

Worked example

A ball is thrown upwards from ground level with initial velocity 20 m/s, use $g=10$ m/s², upwards as positive. Find maximum height and time taken to hit the ground.

• Maximum height: $u=20$, $v=0$, $a=-10$. Use $v^2=u^2+2as$: $0=400+2(-10)s\to s=20$ m
• Time to hit ground: $s=0$, $u=20$, $a=-10$. Use $s=ut+\frac{1}{2}a{t}^2$: $0=20t-5t^2\to t(20-5t)=0\to t=0$ (launch) or $t=4$s.

7. Motion with two stages — joining sub-problems

Many exam problems involve motion with two different constant acceleration stages (e.g. a car accelerating then braking, a ball moving up then falling down). The key connection between stages is that the final velocity of the first stage is the initial velocity of the second stage. Total time and total displacement are the sum of values for each stage.

Worked example

A car accelerates from rest at 3 m/s² for 8 seconds, then brakes at 4 m/s² until it stops. Find total distance travelled.

• Stage 1 (acceleration): $u=0$, $a=3$, $t=8$. Final velocity $v=0+3\ast 8=24$ m/s, displacement $s_1=0+0.5\ast 3\ast 64=96$ m
• Stage 2 (braking): $u=24$, $a=-4$, $v=0$. Use $v^2=u^2+2as$: $0=576+2(-4)s_2\to s_2=72$ m
• Total distance: $96+72=168$ m

8. Average vs instantaneous velocity

• Instantaneous velocity: The velocity of the object at a single point in time, calculated as $\frac{ds}{dt}$ for variable acceleration, or from the suvat $v$ formula for constant acceleration at a given $t$.
• Average velocity: Total displacement divided by total time taken for the journey: $v_{avg}=\frac{\Delta s}{\Delta t}$. This is only equal to the average of initial and final velocity $\frac{u+v}{2}$ if acceleration is constant.

Worked example

A cyclist travels 120m at 6 m/s, then another 120m at 12 m/s. Calculate their average velocity for the whole journey.

• Time for first segment: $120/6=20$s, time for second segment: $120/12=10$s
• Total displacement: 240m, total time: 30s
• Average velocity: $240/30=8$ m/s (note this is not the average of 6 and 12, which is 9 m/s, because the cyclist spent more time at the lower speed)

9. Returning-to-origin and meeting-point problems

These are common high-mark exam questions:

• Returning to origin: Set displacement $s=0$ and solve for time, as shown in the vertical motion example earlier.
• Meeting point problems: Two particles moving along the same line meet when their displacements from the same shared origin are equal at the same time $t$. Adjust displacement equations for any head start in time or position.

Worked example

Particle A starts at the origin at $t=0$ moving at constant velocity 5 m/s. Particle B starts at the origin at $t=2$s with initial velocity 0, acceleration 2 m/s² in the same direction as A. Find the time when they meet.

• Displacement of A: $s_A=5t$
• Displacement of B for $t\ge 2$: $s_B=0+0.5\ast 2\ast (t-2{)}^2=(t-2{)}^2$
• Set equal: $5t=t^2-4t+4\to t^2-9t+4=0$
• Solve quadratic: $t=\frac{9\pm \sqrt{81-16}}{2}=\frac{9\pm \sqrt{65}}{2}$. The valid solution is $t\approx 8.53$s (the other solution is ~0.47s, before B starts moving).

10. Common Pitfalls (and how to avoid them)

• Wrong move: Using distance instead of displacement to calculate average velocity. Why: Students confuse scalar and vector quantities when objects reverse direction. Correct move: Always use signed displacement for average velocity, only use total path length for average speed.
• Wrong move: Applying suvat equations to variable acceleration problems. Why: Students forget suvat only works when acceleration is constant. Correct move: Check if acceleration is a function of time or displacement is non-linear in $t$: if yes, use calculus, not suvat.
• Wrong move: Using positive $g$ when taking upwards as positive. Why: Students default to positive values for acceleration without checking convention. Correct move: Explicitly write your sign convention at the start of every vertical motion question, assign $a=-g$ if up is positive.
• Wrong move: Forgetting constants of integration when solving variable acceleration problems. Why: Students focus on the integral calculation and ignore initial conditions. Correct move: Add $+c$ every time you integrate, then solve for $c$ immediately using given initial values.
• Wrong move: Assuming average velocity is the mean of initial and final velocity for all problems. Why: Students overapply the constant acceleration average rule. Correct move: Only use $\frac{u+v}{2}$ for constant acceleration, otherwise use total displacement over total time.

11. Practice Questions (Paper 4 Style)

Question 1

A particle moves in a straight line with velocity $v=2t^2-10t+8$ m/s, $t\ge 0$. (a) Find the acceleration of the particle at $t=3$s. [2 marks] (b) Find the total distance travelled by the particle in the first 3 seconds. [4 marks]

Worked Solution

(a) $a=\frac{dv}{dt}=4t-10$. At $t=3$, $a=12-10=2$ m/s². (b) First find when $v=0$ (direction reverses): $2t^2-10t+8=0\to t^2-5t+4=0\to t=1$s and $t=4$s. Only $t=1$s is in the first 3s. Displacement function: $s=\int vdt=\frac{2}{3}{t}^3-5t^2+8t+c$, $c=0$ as $s=0$ at $t=0$.

• At $t=1$: $s=\frac{2}{3}-5+8=\frac{11}{3}$ m
• At $t=3$: $s=18-45+24=-3$ m Total distance: $\left|\frac{11}{3}-0\right|+\left|-3-\frac{11}{3}\right|=\frac{11}{3}+\frac{20}{3}=\frac{31}{3}\approx 10.33$ m.

Question 2

A stone is dropped from the top of a 45m tall cliff. Use $g=10$ m/s², downwards as positive. (a) Find the speed of the stone when it hits the ground. [3 marks] (b) Find the time taken for the stone to fall the final 25m of its journey. [3 marks]

Worked Solution

(a) $u=0$, $a=10$, $s=45$. Use $v^2=u^2+2as=0+2\ast 10\ast 45=900$, so $v=30$ m/s (speed is magnitude, positive). (b) First find time to fall first $45-25=20$m: $s=20=0+0.5\ast 10\ast t_1^2\to t_1^2=4\to t_1=2$s. Total time to fall 45m: $45=0+5t_{total}^2\to t_{total}=3$s. Time for final 25m: $3-2=1$s.

Question 3

Two cars start at the same point on a straight road. Car P travels at constant speed 12 m/s. Car Q starts 4 seconds later, accelerating at 3 m/s² in the same direction. Find the distance from the start where Q catches P. [5 marks]

Worked Solution

Let $t$ = time after P starts. Displacement of P: $s_P=12t$. Displacement of Q for $t\ge 4$: $s_Q=0+0.5\ast 3\ast (t-4{)}^2=1.5(t-4{)}^2$. Set equal: $12t=1.5(t^2-8t+16)\to 12t=1.5t^2-12t+24\to 1.5t^2-24t+24=0\to t^2-16t+16=0$. Solve: $t=\frac{16\pm \sqrt{256-64}}{2}=8\pm 4\sqrt{3}$. Valid solution: $t\approx 14.93$s. Distance: $s=12\ast 14.93\approx 179.2$ m.

12. Quick Reference Cheatsheet

13. What's Next

This topic is the foundational building block for all subsequent mechanics topics in A-Level Mathematics Paper 4. You will apply the kinematics rules you learned here to 2D projectile motion later in the syllabus, use acceleration values derived from Newton's laws of motion to solve suvat problems for connected particles, and combine these concepts with work, energy and power calculations for moving objects. A strong grasp of 1D kinematics will make every later mechanics topic significantly easier to master, as all motion problems build on the core definitions and formulas covered in this guide.

If you are struggling with any of the concepts, worked examples, or practice questions in this guide, you can ask Ollie for personalized explanations, extra practice problems, or step-by-step walkthroughs tailored to your gaps at any time. Head to [the homepage](/ to get started, and check out our other A-Level Mathematics Paper 4 study guides for related topics like forces, Newton's laws, and projectiles.

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