A.1 Kinematics

IB Physics Higher Level· Theme A: A.1 Kinematics· 15 min read

1. Scalar and Vector Quantities★☆☆☆☆⏱ 3 min

📘 Definition

Scalar Quantity

A quantity that has only magnitude (size) and no associated direction.

Example:

Distance, speed, mass, time, energy are all scalars.

📘 Definition

Vector Quantity

A quantity that has both magnitude and direction, and follows the rules of vector addition.

Example:

Displacement, velocity, acceleration, force are all vectors.

For one-dimensional motion, we simplify vector calculations by assigning a positive sign to motion in one direction, and a negative sign to motion in the opposite direction.

📐 Worked Example

A hiker walks 8 km north, then turns around and walks 3 km south. Calculate the total distance travelled and the total displacement.

1
Distance is a scalar, so add all magnitudes:
2
$8 + 3 = 11 \text{ km}$
3
Assign north as positive, so south is negative. Displacement is a vector, so add signed values:
4
$\text{Displacement} = (+8) + (-3) = +5 \text{ km}$
5
Final answer: Total distance = 11 km, total displacement = 5 km north.

Exam tip:

Always check if a question asks for distance or displacement — they are almost always marked as separate answers.

2. Core Kinematic Quantities★★☆☆☆⏱ 4 min

📘 Definition

Average Velocity

v_{\text{avg}}

Total displacement divided by total time taken for the motion

Example:

$v_{avg} = \frac{Δ s}{Δ t}$

📘 Definition

Acceleration

a

Rate of change of velocity, equal to the gradient of a velocity-time graph

Example:

$a = \frac{Δ v}{Δ t}$ for constant acceleration

For motion with constant acceleration, average velocity can also be calculated as the average of the initial and final velocity, regardless of time taken.

📐 Worked Example

A cyclist accelerates from rest to 12 m s⁻¹ over 6 seconds. Calculate the average velocity and acceleration of the cyclist.

1
We know initial velocity $u = 0$ , final velocity $v = 12$ m s⁻¹, time $t = 6$ s.
2
Average velocity for constant acceleration:
3
$v_{\text{avg}} = \frac{u + v}{2} = \frac{0 + 12}{2} = 6 \text{ m s}^{-1}$
4
Acceleration is change in velocity over time:
5
$a = \frac{v - u}{t} = \frac{12 - 0}{6} = 2 \text{ m s}^{-2}$

3. Kinematic Equations (SUVAT)★★☆☆☆⏱ 5 min

For motion with constant acceleration, the four valid kinematic equations are:

v = u + at

s = \frac{(u + v)t}{2}

s = ut + \frac{1}{2} a t^2

v^2 = u^2 + 2as

These equations only work when acceleration is constant. They are used for all one-dimensional constant acceleration problems, including free fall.

📐 Worked Example

A stone is dropped from rest from a 45 m tall cliff. Taking $g = 9.8$ m s⁻², calculate the time taken for the stone to hit the ground.

1
Take downward as positive. List known SUVAT values: $u = 0$ , $s = 45$ m, $a = 9.8$ m s⁻², $t = ?$
2
Select the equation that contains only the known values and the unknown we need to find:
3
$s = ut + \frac{1}{2} a t^2$
4
Substitute values, the $u t$ term cancels out because $u = 0$ :
5
$45 = 0 + \frac{1}{2}(9.8)t^2 \implies t^2 = \frac{90}{9.8} \approx 9.18$
6
Take the positive root (time cannot be negative):
7
$t \approx 3.03 \text{ s}$

Exam tip:

Always pick the equation that directly solves for your unknown, to avoid calculating unnecessary intermediate values and save exam time.

4. Interpreting Motion Graphs★★☆☆☆⏱ 4 min

Motion can be represented graphically, and key kinematic quantities are found from gradients and areas under graphs, regardless of whether acceleration is constant:

Gradient of a displacement-time ( $s$ - $t$ ) graph = instantaneous velocity
Gradient of a velocity-time ( $v$ - $t$ ) graph = instantaneous acceleration
Area under a velocity-time ( $v$ - $t$ ) graph = total displacement
Area under an acceleration-time ( $a$ - $t$ ) graph = total change in velocity

📐 Worked Example

A velocity-time graph for a car is a straight line from $(t = 0, v = 0)$ to $(t = 10 s, v = 20 m s^{- 1})$ . What is the total displacement of the car after 10 seconds?

1
Total displacement equals the area under the velocity-time graph.
2
The graph forms a right triangle with base 10 s and height 20 m s⁻¹:
3
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 20 = 100 \text{ m}$
4
Total displacement = 100 m.

5. Common Pitfalls

Wrong move:

Assuming average speed is always equal to the magnitude of average velocity

Why:

Average speed is total distance over time, while average velocity is total displacement over time. They only match for motion with no direction change.

Correct move:

Always distinguish the two quantities when the object changes direction during motion.

Wrong move:

Using SUVAT equations for motion with non-constant acceleration

Why:

SUVAT equations are derived specifically for constant acceleration, so they will give incorrect results for changing acceleration.

Correct move:

Use gradient and area methods on motion graphs for non-constant acceleration problems.

Wrong move:

Mixing up gradients of displacement-time and velocity-time graphs

Why:

Students often confuse which gradient corresponds to which kinematic quantity in exams.

Correct move:

Memorise: Gradient of $s$ - $t$ = $v$ , gradient of $v$ - $t$ = $a$ .

Wrong move:

Using an inconsistent sign convention for direction

Why:

For example, taking upward as positive but writing acceleration due to gravity as positive leads to wrong sign results.

Correct move:

Always write down your chosen positive direction at the start of any kinematics problem, and stick to it.

Wrong move:

Keeping negative time solutions from quadratic kinematic equations

Why:

Quadratic equations give two solutions, but negative time refers to a moment before motion started, which is unphysical.

Correct move:

Always discard negative time solutions unless the question explicitly asks for time before $t = 0$ .

6. Quick Reference Cheatsheet

Quantity/Relationship	Symbol/Rule
Scalar	Magnitude only, no direction
Vector	Magnitude + direction
Average velocity	$\frac{Δ s}{Δ t}$
Average acceleration	$\frac{Δ v}{Δ t}$
$v =$	$u + a t$
$s =$	$\frac{( u + v ) t}{2}$
$s =$	$u t + \frac{1}{2} a t^{2}$
$v^{2} =$	$u^{2} + 2 a s$
Gradient $s$ - $t$ graph	Velocity
Gradient $v$ - $t$ graph	Acceleration
Area under $v$ - $t$ graph	Displacement

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · Paper 1
Motion graph interpretation
2022 · Paper 2
Constant acceleration problem
2021 · Paper 1
Scalar vs vector distinction

What's Next

A.1 Kinematics is the foundational topic for all mechanics in IB Physics HL. All subsequent motion topics build on the core concepts of displacement, velocity, acceleration and the SUVAT framework introduced here. Next, you will extend these ideas to two-dimensional motion with projectile motion, then use kinematics to understand how forces cause acceleration in Newton's laws of motion. Mastery of this sub-topic is critical for later topics including circular motion, orbital mechanics and simple harmonic motion.