Kinematics — A-Level Physics Study Guide

For: A-Level Physics candidates sitting A-Level Physics.

Covers: Displacement, velocity and acceleration definitions, equations of uniformly accelerated motion, motion graph gradient and area analysis, projectile motion component independence, and free fall with air resistance effects.

You should already know: IGCSE Physics, basic algebra and trigonometry.

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 Physics 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?

Kinematics is the branch of mechanics that describes the motion of objects without analyzing the forces that cause motion. It focuses on quantifying position, speed, and acceleration for point masses, extended objects, and moving systems, and accounts for 7-10% of total marks on A-Level Physics Paper 2 (AS Level) and Paper 4 (A Level). You may see it referenced as "motion analysis" or "trajectory calculation" in exam questions, and it forms the foundation for all subsequent mechanics topics in the syllabus.

2. Displacement, velocity, acceleration

All three core quantities are vector values, meaning they have both magnitude and direction, unlike scalar equivalents (distance, speed) that only have magnitude.

• Displacement ($\vec{s}$): The net change in position of an object relative to a fixed origin, measured in meters (m). It differs from distance, which is the total length of the path traveled. For example, if you walk 8m north then 3m south, your total distance is 11m, but your displacement is +5m (if north is defined as the positive direction).
• Velocity ($\vec{v}$): The rate of change of displacement over time, measured in $m{s}^{-1}$. Average velocity is given by $\bar{v}=\frac{\Delta s}{\Delta t}$, while instantaneous velocity is the derivative of displacement with respect to time: $v=\frac{ds}{dt}$. Speed is the scalar equivalent, measuring the rate of change of total distance.
• Acceleration ($\vec{a}$): The rate of change of velocity over time, measured in $m{s}^{-2}$. Average acceleration is $\bar{a}=\frac{\Delta v}{\Delta t}$, while instantaneous acceleration is $a=\frac{dv}{dt}=\frac{d^{2}s}{d{t}^2}$. Note that acceleration occurs even if speed is constant, for example in uniform circular motion where velocity direction changes continuously.

Exam tip: Examiners regularly ask for 1-mark definitions of these quantities, so always specify if they are scalar/vector and what quantity they measure the change of to get full marks.

3. Equations of uniformly accelerated motion

These equations (often called "suvat" equations for their variable labels) only apply when acceleration is constant, a common assumption in A-Level Physics questions unless explicitly stated otherwise. They are derived directly from the definitions of velocity and acceleration above:

1. Start with $a=\frac{v-u}{t}$ (average acceleration for constant a), rearrange to get the first equation: $$v=u+at$$ where $u$ = initial velocity, $v$ = final velocity, $t$ = time elapsed.
2. Average velocity for constant acceleration is $\frac{u+v}{2}$, so displacement is average velocity multiplied by time: $$s=\left(\frac{u+v}{2}\right)t$$
3. Substitute $v=u+at$ into the second equation to eliminate final velocity: $$s=ut+\frac{1}{2}a{t}^2$$
4. Eliminate time from the first two equations to get the fourth relation: $$v^2=u^2+2as$$

Worked Example: A van accelerates uniformly from rest at $3.2m{s}^{-2}$ for 10s. Calculate its final speed and total distance traveled.

• Known values: $u=0$, $a=3.2$, $t=10$
• Final speed: $v=0+(3.2\times 10)=32m{s}^{-1}$
• Distance: $s=(0\times 10)+\frac{1}{2}\times 3.2\times 10^2=160m$

Exam tip: Always list all known values first before selecting an equation, to avoid using a formula that includes an unknown you do not have. For example, if you have no value for time, use $v^2=u^2+2as$.

4. Graphs of motion — gradient and area

Motion graphs are a common 2-3 mark question topic, and questions test your ability to interpret gradients (slopes) and areas under the line for three core graph types:

1. Displacement-time (s-t) graph: The gradient equals instantaneous velocity. A flat line means the object is stationary, a straight sloped line means constant velocity, and a curved line means changing velocity (acceleration). The area under an s-t graph has no physical meaning at A-Level.
2. Velocity-time (v-t) graph: The gradient equals instantaneous acceleration. The total area under the line equals total displacement. A straight sloped line means constant acceleration, a flat line means constant velocity, and a curved line means changing acceleration.
3. Acceleration-time (a-t) graph: The area under the line equals the total change in velocity over the time interval. The gradient has no physical meaning for kinematics in the A-Level Physics syllabus.

Worked Example: A v-t graph shows a straight line from (0, 0) to (5, 15), then a flat line for 10s, then a straight line falling to (20, 0). Calculate total displacement and average acceleration over the 20s interval.

• Total area = area of first triangle (0-5s) + area of rectangle (5-15s) + area of second triangle (15-20s)
• $Totals=(\frac{1}{2}\times 5\times 15)+(10\times 15)+(\frac{1}{2}\times 5\times 15)=37.5+150+37.5=225m$
• Average acceleration = $\frac{finalv-initialv}{totalt}=\frac{0-0}{20}=0m{s}^{-2}$

5. Projectile motion — independent x and y components

The core principle of projectile motion is that horizontal and vertical motions are completely independent, as long as air resistance is negligible (the default assumption in all A-Level Physics projectile questions unless stated otherwise). The only link between the two axes is the total time of flight.

• Horizontal axis: No acceleration acts ($a_x=0$), so horizontal velocity remains constant for the full flight: $v_x=u_x$, and horizontal displacement is $s_x=u_{x}t$, where $u_x$ is the horizontal component of initial velocity.
• Vertical axis: Constant acceleration of $g=9.81m{s}^{-2}$ acts downwards, so you can use all four suvat equations for vertical motion. Always define a positive direction (up or down) at the start of calculations and assign signs consistently to velocity, displacement and acceleration.

For projectiles launched at an angle $\theta$ to the horizontal with initial speed $u$, split the velocity into components first: $u_x=u\cos \theta$, $u_y=u\sin \theta$.

Worked Example: A projectile is launched at 40° above the horizontal with initial speed $60m{s}^{-1}$. Calculate its maximum height.

• Vertical initial velocity: $u_y=60\times \sin (40°)\approx 38.6m{s}^{-1}$
• At maximum height, vertical velocity $v_y=0$, acceleration $a=-9.81m{s}^{-2}$ (upwards = positive)
• Use $v^2=u^2+2as$: $0=38.6^2+(2\times -9.81\times h)$
• Rearrange: $h=\frac{1489.96}{19.62}\approx 76m$

6. Free-fall and air resistance discussion

Free fall is defined as motion under the influence of gravitational force only, with no air resistance, so all objects in free fall accelerate at $g=9.81m{s}^{-2}$ towards the Earth, regardless of their mass. When air resistance is included, motion changes significantly:

• Air resistance (drag) is a force that opposes the direction of motion, and increases with the speed of the object and its cross-sectional area.
• For a falling object: Initially only weight acts, so acceleration = $g$. As speed increases, drag increases, so net force and net acceleration decrease. Eventually, drag equals the weight of the object, so net force = 0, acceleration = 0, and the object falls at a constant speed called terminal velocity.
• The v-t graph for an object falling with air resistance starts with a gradient equal to $g$, curves gradually, and becomes flat when terminal velocity is reached. When a skydiver opens their parachute, drag suddenly becomes larger than weight, so acceleration becomes upwards, speed decreases until a new lower terminal velocity is reached.

Exam tip: For 3-mark questions asking to explain terminal velocity, you must mention 1) the balance between drag and weight, 2) net force = 0, 3) acceleration = 0, to get full marks.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Using scalar distance instead of vector displacement in suvat equations, especially when objects reverse direction. Why it happens: Students confuse the two terms and forget to set a sign convention. Correct move: Always define a positive direction at the start of any calculation, and assign negative values to displacement, velocity or acceleration acting in the opposite direction.
• Wrong move: Using suvat equations when acceleration is not constant (e.g. when air resistance is significant). Why it happens: Students default to memorized equations without checking question context. Correct move: Only use suvat if the question explicitly states constant acceleration or negligible air resistance; for non-uniform acceleration, use motion graph analysis instead.
• Wrong move: Mixing up gradients and areas on motion graphs, for example calculating the area under an s-t graph to find velocity. Why it happens: Memorizing rules without understanding their origin. Correct move: Remember that gradient = $\frac{\Delta y}{\Delta x}$, so s-t gradient is $\frac{\Delta s}{\Delta t}=velocity$, v-t gradient is $\frac{\Delta v}{\Delta t}=acceleration$, and v-t area is $\sum v\times \Delta t=displacement$.
• Wrong move: Using the wrong trigonometric function for projectile velocity components, or forgetting to split velocity entirely. Why it happens: Rushing through questions without sketching the velocity vector. Correct move: Draw a quick right triangle for initial velocity, label the angle with the horizontal, so the adjacent side (x component) is $u\cos \theta$ and opposite side (y component) is $u\sin \theta$.
• Wrong move: Taking $g$ as positive when upwards is defined as the positive direction. Why it happens: Forgetting that gravity always acts downwards. Correct move: If you take up as positive, $g=-9.81m{s}^{-2}$, and note your sign convention in working so examiners follow your logic.

8. Practice Questions (A-Level Physics Style)

Question 1 (2 marks)

A hiker walks 5km west, then 12km north, then 5km east. State the total distance traveled and the magnitude of their total displacement.

Solution

• Total distance = sum of all path lengths = $5+12+5=22km$ (1 mark)
• East and west displacements cancel out, so net displacement is 12km north, magnitude = $12km$ (1 mark)

Question 2 (3 marks)

A car decelerates uniformly from $30m{s}^{-1}$ to rest over a distance of 75m. Calculate the magnitude of the deceleration and the time taken to stop.

Solution

• Known values: $u=30$, $v=0$, $s=75$
• Use $v^2=u^2+2as$ to find acceleration: $0=30^2+(2\times a\times 75)$ → $a=-\frac{900}{150}=-6m{s}^{-2}$, so magnitude of deceleration = $6m{s}^{-2}$ (2 marks)
• Use $v=u+at$ to find time: $0=30+(-6\times t)$ → $t=5s$ (1 mark)

Question 3 (5 marks)

A ball is thrown horizontally off a 78.4m high cliff with initial speed $15m{s}^{-1}$. Assume air resistance is negligible, $g=9.81m{s}^{-2}$. Calculate (a) the time taken for the ball to hit the ground, (b) the horizontal distance it travels before impact.

Solution

(a) Vertical motion: $u_y=0$, $s_y=78.4m$, $a=9.81m{s}^{-2}$ (down = positive)

• Use $s=ut+\frac{1}{2}a{t}^2$: $78.4=0+(\frac{1}{2}\times 9.81\times t^2)$
• Rearrange: $t^2=\frac{156.8}{9.81}\approx 16$ → $t\approx 4.0s$ (3 marks) (b) Horizontal motion: $u_x=15m{s}^{-1}$, $t=4.0s$
• $s_x=u_{x}t=15\times 4.0=60m$ (2 marks)

9. Quick Reference Cheatsheet

Core Formulas

Motion Graph Rules

Key Rules

• Air resistance is negligible unless stated otherwise for projectiles.
• Terminal velocity occurs when drag = weight, net force = 0, acceleration = 0.
• Always define a positive direction at the start of all calculation questions.

10. What's Next

Kinematics is the foundational building block for all mechanics topics in the A-Level Physics syllabus, so mastering this content is critical before moving on to dynamics (Newton's laws, force and momentum), work, energy and power, and circular motion, all of which rely on the definitions of displacement, velocity and acceleration you learned here. For A Level candidates, kinematics also extends to non-uniform acceleration using calculus in Paper 4, so familiarizing yourself with the derivative relationships between $s$, $v$ and $a$ will make that transition much smoother.

If you're stuck on any part of kinematics, from interpreting tricky motion graphs to solving complex projectile motion questions, you can ask Ollie, our AI tutor, for personalized explanations, extra practice questions, or step-by-step walkthroughs tailored to your weak spots. You can also find more topic guides and full past paper practice on the homepage to prepare for your A-Level Physics exams.

Aligned with the Cambridge International AS & A Level Physics 9702 syllabus. OwlsAi is not affiliated with Cambridge Assessment International Education.