IB Physics Topic 1.1: Kinematics | Notes & Flashcards | Aimnova

IB Physics — Kinematics

Topic 1.1 of IB Physics covers Kinematics, which is part of Unit 1: Space, time and motion. Students explore key concepts including Velocity and displacement, Acceleration, Displacement from a velocity–time graph, and more. A strong understanding of kinematics is essential for IB Physics exams and builds the foundation for connected topics across the syllabus.

Key concepts in Kinematics

Key Idea: Kinematics is the maths of motion — describing how things move with displacement, velocity and acceleration, without asking what causes it. It is the foundation of the whole course and is tested in every paper: quick one-step reads in Paper 1A (multiple choice), graph-plotting in Paper 1B, and full determine / show that / sketch questions in Paper 2.

📈 Reading a motion graph

Almost every kinematics graph question comes down to two readings: take the slope or take the area. Which one depends on which graph you are looking at.

On a v–t graph the slope is the acceleration and the area is the displacement — mixing these up is the most common motion-graph mistake. Area below the time axis is negative — the object is moving the other way.

🧮 The key equations

For constant acceleration (a straight v–t line) the four suvat equations link the five quantities s, u, v, a and t. All four are given in the data booklet.

v = u + at

No s — use it when displacement is not involved.

$s$: displacement (m)
$u$: initial velocity (m s⁻¹)
$v$: final velocity (m s⁻¹)
$a$: acceleration (m s⁻²)
$t$: time (s)

s = ut + \tfrac{1}{2}at^{2}

No v — use it for displacement from a known acceleration and time.

v^{2} = u^{2} + 2as

No t — use it when time is neither known nor wanted.

s = \frac{u + v}{2}\,t

No a — average velocity × time. This is also the area of the trapezium under a straight v–t line.

$s$: displacement (m)
$u$: initial velocity (m s⁻¹)
$v$: final velocity (m s⁻¹)
$t$: time (s)

List your knowns, mark the one you want, and pick the equation that is missing the quantity you neither know nor need. Always write the formula first, then substitute.

🪂 Free fall & projectiles

Free fall is just constant-acceleration motion with a = g = 9.81 m s⁻² pointing down (independent of mass). A projectile splits into two independent motions — constant horizontal velocity, free-fall vertical — sharing one clock.

R = u_x\,t

Horizontal range = constant horizontal velocity × the time of flight (the time comes from the vertical drop). Not in the booklet — it is just distance = speed × time applied sideways.

$R$: horizontal range (m)
$u_x$: horizontal velocity — stays constant (m s⁻¹)
$t$: time of flight, set by the vertical drop (s)

💨 Fluid resistance & terminal velocity

Real falling objects meet drag (fluid resistance), which acts against the motion and grows with speed. The suvat equations no longer apply (the acceleration is changing), so this part is described, not calculated.

At terminal velocity the forces are not absent — weight and drag are equal and opposite, so they cancel. Zero resultant force means zero acceleration, so the velocity stays constant.

✍️ Worked examples

IB-style question — pick the right suvat equation

A motorbike accelerates uniformly from 6.0 m s⁻¹ to 30 m s⁻¹ over a distance of 90 m. Find its acceleration.

Solution:

Knowns: u = 6.0, v = 30, s = 90; want a. No t, so use the given equation with no t:
$v^{2} = u^{2} + 2as$
Make a the subject — formula first:
$a = \frac{v^{2} - u^{2}}{2s}$
Substitute the values:
$a = \frac{30^{2} - 6.0^{2}}{2 \times 90}$
Work it out — keep the unit:
$a = \frac{900 - 36}{180} = 4.8\ \text{m s}^{-2}$

Final answer:

a = 4.8 m s⁻².

IB-style question — displacement from a v–t graph

A tram's velocity–time graph is a straight line rising from 5.0 m s⁻¹ to 17 m s⁻¹ over 8.0 s. Find the distance it travels.

Solution:

Displacement is the area under the line — a trapezium. Use the given equation first:
$s = \frac{u + v}{2}\,t$
Substitute (u = 5.0, v = 17, t = 8.0):
$s = \frac{5.0 + 17}{2} \times 8.0$
Average velocity, then × time — keep the unit:
$s = 11 \times 8.0 = 88\ \text{m}$

Final answer:

s = 88 m.

IB-style question — free fall

A stone is dropped from rest down a well and takes 1.8 s to reach the water. Find (a) its speed on impact and (b) the depth of the well. Take g = 9.81 m s⁻².

Solution:

(a) Dropped from rest so u = 0, a = g. Want v, have t — use the given equation with no s:
$v = u + at$
Substitute (u = 0, a = 9.81, t = 1.8):
$v = 0 + 9.81 \times 1.8 = 17.7\ \text{m s}^{-1}$
(b) Want s, have u, a, t — use the given equation with no v:
$s = ut + \tfrac{1}{2}at^{2}$
Substitute (u = 0):
$s = 0 + \tfrac{1}{2}(9.81)(1.8)^{2} = 15.9\ \text{m}$

Final answer:

(a) v = 17.7 m s⁻¹ ≈ 18 m s⁻¹; (b) s = 15.9 m ≈ 16 m.

IB-style question — projectile launched horizontally

A ball is thrown horizontally at 7.0 m s⁻¹ from the top of a 31 m cliff. Find (a) the time to land and (b) how far from the base it lands. Take g = 9.8 m s⁻².

Solution:

(a) Vertical motion is free fall from rest (uy = 0). Use the given equation:
$s = u_y t + \tfrac{1}{2}gt^{2}$
Substitute (s = 31, uy = 0, g = 9.8) and rearrange for t:
$t = \sqrt{\frac{2 \times 31}{9.8}} = 2.5\ \text{s}$
(b) Horizontal velocity is constant — use the horizontal relation:
$R = u_x\,t$
Substitute (uₓ = 7.0, t = 2.5 s from part a):
$R = 7.0 \times 2.5 = 17.5\ \text{m}$

Final answer:

(a) t = 2.5 s; (b) R = 17.5 m ≈ 18 m from the base.

✅ Quick self-check

Tap each card to check yourself.

Exam Tips

v–t graph: slope = acceleration, area = displacement. Never swap the two.
The suvat equations apply only when the acceleration is constant (a straight v–t line) — not once drag matters.
Choose a suvat equation by the quantity that is missing: list knowns, mark the unknown, pick the equation without the spare one.
Always write the equation first, then substitute, and keep the unit on every line of working.
Free fall: a = g = 9.81 m s⁻² down, same for every mass. Decide which direction is positive before you start.
Projectiles: treat horizontal (constant uₓ) and vertical (free fall) separately — they share only the time.
Watch the sign: a falling v–t line / a 'deceleration' / area below the axis are all negative.

IB Physics — Kinematics

Key concepts in Kinematics

Key Idea: Kinematics is the maths of motion — describing how things move with displacement, velocity and acceleration, without asking what causes it. It is the foundation of the whole course and is tested in every paper: quick one-step reads in Paper 1A (multiple choice), graph-plotting in Paper 1B, and full determine / show that / sketch questions in Paper 2.

📈 Reading a motion graph

Almost every kinematics graph question comes down to two readings: take the slope or take the area. Which one depends on which graph you are looking at.

On a v–t graph the slope is the acceleration and the area is the displacement — mixing these up is the most common motion-graph mistake. Area below the time axis is negative — the object is moving the other way.

🧮 The key equations

For constant acceleration (a straight v–t line) the four suvat equations link the five quantities s, u, v, a and t. All four are given in the data booklet.

v = u + at

No s — use it when displacement is not involved.

$s$: displacement (m)
$u$: initial velocity (m s⁻¹)
$v$: final velocity (m s⁻¹)
$a$: acceleration (m s⁻²)
$t$: time (s)

s = ut + \tfrac{1}{2}at^{2}

No v — use it for displacement from a known acceleration and time.

v^{2} = u^{2} + 2as

No t — use it when time is neither known nor wanted.

s = \frac{u + v}{2}\,t

No a — average velocity × time. This is also the area of the trapezium under a straight v–t line.

$s$: displacement (m)
$u$: initial velocity (m s⁻¹)
$v$: final velocity (m s⁻¹)
$t$: time (s)

List your knowns, mark the one you want, and pick the equation that is missing the quantity you neither know nor need. Always write the formula first, then substitute.

🪂 Free fall & projectiles

R = u_x\,t

Horizontal range = constant horizontal velocity × the time of flight (the time comes from the vertical drop). Not in the booklet — it is just distance = speed × time applied sideways.

$R$: horizontal range (m)
$u_x$: horizontal velocity — stays constant (m s⁻¹)
$t$: time of flight, set by the vertical drop (s)

💨 Fluid resistance & terminal velocity

At terminal velocity the forces are not absent — weight and drag are equal and opposite, so they cancel. Zero resultant force means zero acceleration, so the velocity stays constant.

✍️ Worked examples

IB-style question — pick the right suvat equation

A motorbike accelerates uniformly from 6.0 m s⁻¹ to 30 m s⁻¹ over a distance of 90 m. Find its acceleration.

Solution:

Knowns: u = 6.0, v = 30, s = 90; want a. No t, so use the given equation with no t:
$v^{2} = u^{2} + 2as$
Make a the subject — formula first:
$a = \frac{v^{2} - u^{2}}{2s}$
Substitute the values:
$a = \frac{30^{2} - 6.0^{2}}{2 \times 90}$
Work it out — keep the unit:
$a = \frac{900 - 36}{180} = 4.8\ \text{m s}^{-2}$

Final answer:

a = 4.8 m s⁻².

IB-style question — displacement from a v–t graph

A tram's velocity–time graph is a straight line rising from 5.0 m s⁻¹ to 17 m s⁻¹ over 8.0 s. Find the distance it travels.

Solution:

Displacement is the area under the line — a trapezium. Use the given equation first:
$s = \frac{u + v}{2}\,t$
Substitute (u = 5.0, v = 17, t = 8.0):
$s = \frac{5.0 + 17}{2} \times 8.0$
Average velocity, then × time — keep the unit:
$s = 11 \times 8.0 = 88\ \text{m}$

Final answer:

s = 88 m.

IB-style question — free fall

A stone is dropped from rest down a well and takes 1.8 s to reach the water. Find (a) its speed on impact and (b) the depth of the well. Take g = 9.81 m s⁻².

Solution:

(a) Dropped from rest so u = 0, a = g. Want v, have t — use the given equation with no s:
$v = u + at$
Substitute (u = 0, a = 9.81, t = 1.8):
$v = 0 + 9.81 \times 1.8 = 17.7\ \text{m s}^{-1}$
(b) Want s, have u, a, t — use the given equation with no v:
$s = ut + \tfrac{1}{2}at^{2}$
Substitute (u = 0):
$s = 0 + \tfrac{1}{2}(9.81)(1.8)^{2} = 15.9\ \text{m}$

Final answer:

(a) v = 17.7 m s⁻¹ ≈ 18 m s⁻¹; (b) s = 15.9 m ≈ 16 m.

IB-style question — projectile launched horizontally

A ball is thrown horizontally at 7.0 m s⁻¹ from the top of a 31 m cliff. Find (a) the time to land and (b) how far from the base it lands. Take g = 9.8 m s⁻².

Solution:

(a) Vertical motion is free fall from rest (uy = 0). Use the given equation:
$s = u_y t + \tfrac{1}{2}gt^{2}$
Substitute (s = 31, uy = 0, g = 9.8) and rearrange for t:
$t = \sqrt{\frac{2 \times 31}{9.8}} = 2.5\ \text{s}$
(b) Horizontal velocity is constant — use the horizontal relation:
$R = u_x\,t$
Substitute (uₓ = 7.0, t = 2.5 s from part a):
$R = 7.0 \times 2.5 = 17.5\ \text{m}$

Final answer:

(a) t = 2.5 s; (b) R = 17.5 m ≈ 18 m from the base.

✅ Quick self-check

Tap each card to check yourself.

Exam Tips

v–t graph: slope = acceleration, area = displacement. Never swap the two.
The suvat equations apply only when the acceleration is constant (a straight v–t line) — not once drag matters.
Choose a suvat equation by the quantity that is missing: list knowns, mark the unknown, pick the equation without the spare one.
Always write the equation first, then substitute, and keep the unit on every line of working.
Free fall: a = g = 9.81 m s⁻² down, same for every mass. Decide which direction is positive before you start.
Projectiles: treat horizontal (constant uₓ) and vertical (free fall) separately — they share only the time.
Watch the sign: a falling v–t line / a 'deceleration' / area below the axis are all negative.

Key concepts in Kinematics

📈 Reading a motion graph

🧮 The key equations

🪂 Free fall & projectiles

💨 Fluid resistance & terminal velocity

✍️ Worked examples

IB-style question — pick the right suvat equation

IB-style question — displacement from a v–t graph

IB-style question — free fall

IB-style question — projectile launched horizontally

✅ Quick self-check

Exam Tips

What you'll learn in Topic 1.1

Study resources — 1.1 Kinematics

Velocity and displacement

Acceleration

Displacement from a velocity–time graph

The suvat equations

Free fall

Projectiles

Fluid resistance and terminal velocity

Ready to study Kinematics?

Ready to practice?

Key concepts in Kinematics

📈 Reading a motion graph

🧮 The key equations

🪂 Free fall & projectiles

💨 Fluid resistance & terminal velocity

✍️ Worked examples

IB-style question — pick the right suvat equation

IB-style question — displacement from a v–t graph

IB-style question — free fall

IB-style question — projectile launched horizontally

✅ Quick self-check

Exam Tips

What you'll learn in Topic 1.1

Study resources — 1.1 Kinematics

Velocity and displacement

Acceleration

Displacement from a velocity–time graph

The suvat equations

Free fall

Projectiles

Fluid resistance and terminal velocity

Ready to study Kinematics?

Ready to practice?