A.1 Kinematics

Lesson 1 — Kinematics (IB Physics SL)

Learning objectives

Distinguish clearly between scalar and vector quantities.
Represent, add, and resolve vectors using components.
Describe motion using displacement, velocity, and acceleration.
Interpret displacement–time, velocity–time, and acceleration–time graphs.
Apply the equations of uniformly accelerated motion (SUVAT).
Analyse projectile motion using independent horizontal and vertical motion.

a) Vectors

Scalars

A scalar is a physical quantity described by magnitude only. Scalars answer “how much?” but not “in which direction?” Examples include
distance, speed, mass, and time.

Vectors

A vector has both magnitude and direction. Vectors answer “how much?” and “which way?” Examples include
displacement, velocity, acceleration, and force.

Representing vectors

Vectors are represented using arrows:
the length represents magnitude and the arrowhead indicates direction.

Fig A.1.2 — Representing a vector with an arrow

Multiplying a vector by a scalar

Multiplying a vector by a scalar changes the magnitude but not the direction, unless the scalar is negative.
A negative scalar reverses the direction of the vector.

Adding vectors

Vectors can be added by:

Tip-to-tail method: place the tail of the next vector at the tip of the previous vector; the resultant joins the start to the final tip.
Component method: resolve each vector into $x$ and $y$ components, add components, then recombine using Pythagoras and trigonometry.

Resolving vectors into components

If a vector of magnitude $F$ makes an angle $\theta$ with the horizontal, then:

$F_x = F\cos\theta, \qquad F_y = F\sin\theta$

Worked example — resolving a force

A force of $10\ \text{N}$ acts at $30^\circ$ above the horizontal. Find $F_x$ and $F_y$ .

$F_x = 10\cos 30^\circ = 8.66\ \text{N}$

$F_y = 10\sin 30^\circ = 5.00\ \text{N}$

Answer: $F_x = 8.66\ \text{N}$ , $F_y = 5.00\ \text{N}$ .

Fig A.1.3 — Resolving a force into components

b) Describing motion

Position

Position describes the location of an object relative to a chosen reference point.

Displacement

Displacement is the change in position and is a vector. It depends only on the starting and ending positions, not the path taken.

Distance

Distance is the total path length travelled and is a scalar. An object can travel a non-zero distance while having zero displacement (for example, returning to its starting point).

Velocity and speed

Velocity is displacement per unit time (a vector), while speed is distance per unit time (a scalar).

$\text{velocity} = \frac{\Delta s}{\Delta t}, \qquad \text{speed} = \frac{\text{distance}}{\Delta t}$

Acceleration

Acceleration is the rate of change of velocity:

$a = \frac{\Delta v}{t}$

Worked example — acceleration

A car increases speed from $10\ \text{m s}^{-1}$ to $20\ \text{m s}^{-1}$ in $5\ \text{s}$ . Find the acceleration.

$a=\frac{\Delta v}{t}=\frac{20-10}{5}=2.0\ \text{m s}^{-2}$

Answer: $a=2.0\ \text{m s}^{-2}$ .

Motion graphs

Displacement–time graph: slope $=$ velocity
Velocity–time graph: slope $=$ acceleration; area $=$ displacement
Acceleration–time graph: area $=$ change in velocity

Displacement-time graphs — Displacement–time graphs

Velocity-time graphs — Velocity–time graphs

Instantaneous vs average

Instantaneous values describe a quantity at a single moment (a “snapshot”).
Average values describe the overall change divided by the total time.

c) Equations of motion (SUVAT)

These equations apply only when acceleration is constant.
Symbols: $s$ displacement (m), $u$ initial velocity (m s $^{-1}$ ), $v$ final velocity (m s $^{-1}$ ), $a$ acceleration (m s $^{-2}$ ), $t$ time (s).

The four SUVAT equations

$v = u + at$
$s = ut + \tfrac{1}{2}at^2$
$v^2 = u^2 + 2as$
$s = \tfrac{(u+v)}{2}t$

Worked examples (full solutions)

1) Accelerating from rest

A car starts from rest and accelerates at $3.0\ \text{m s}^{-2}$ for $6.0\ \text{s}$ . Find the final speed.

$v = u + at = 0 + (3.0)(6.0) = 18\ \text{m s}^{-1}$

Answer: $v = 18\ \text{m s}^{-1}$ .

2) Distance travelled under acceleration

A runner has $u = 2.0\ \text{m s}^{-1}$ and accelerates at $1.5\ \text{m s}^{-2}$ for $4.0\ \text{s}$ . Find the displacement.

$s = ut + \tfrac{1}{2}at^2 = (2.0)(4.0) + \tfrac{1}{2}(1.5)(4.0)^2 = 8.0 + 12.0 = 20\ \text{m}$

Answer: $s = 20\ \text{m}$ .

3) Braking distance

A car travels at $25\ \text{m s}^{-1}$ and brakes uniformly to rest with $a = -5.0\ \text{m s}^{-2}$ . Find the stopping distance.

$v^2 = u^2 + 2as \Rightarrow 0 = 25^2 + 2(-5.0)s \Rightarrow 0 = 625 - 10s \Rightarrow s = 62.5\ \text{m}$

Answer: $s = 62.5\ \text{m}$ .

4) Distance from average velocity

A train speeds up uniformly from $10\ \text{m s}^{-1}$ to $22\ \text{m s}^{-1}$ in $6.0\ \text{s}$ . Find the displacement.

$s = \tfrac{(u+v)}{2}t = \tfrac{10+22}{2}\times 6.0 = 16\times 6.0 = 96\ \text{m}$

Answer: $s = 96\ \text{m}$ .

Quick exam tips

Choose an equation that does not contain the unknown you are solving for.
Use a consistent sign convention (e.g., upward positive, downward negative).
Check units: $s$ in m, $u,v$ in m s $^{-1}$ , $a$ in m s $^{-2}$ , $t$ in s.

d) Projectile motion

Projectile motion diagram — Projectile motion (ideal model)

Definition: Projectile motion occurs when an object is launched with initial speed $u$ at an angle $\theta$ above the horizontal and then moves under the influence of gravity alone.
Horizontal and vertical motions are independent (air resistance neglected).

Horizontal: constant velocity $u\cos\theta$
Vertical: constant acceleration $-g$ , where $g = 9.81\ \text{m s}^{-2}$

Position as a function of time

Taking the launch point as $(0,0)$ :

$x(t) = u\cos\theta \, t, \qquad y(t) = u\sin\theta \, t - \tfrac{1}{2}gt^2$

Velocities:

$v_x = u\cos\theta \quad(\text{constant}), \qquad v_y = u\sin\theta - gt$

Key results (same launch and landing height)

$T = \frac{2u\sin\theta}{g}, \qquad H_{\max} = \frac{u^2\sin^2\theta}{2g}, \qquad R = \frac{u^2\sin 2\theta}{g}$

Note: In the ideal model (no air resistance), the range is maximised when $\theta = 45^\circ$ .

Worked example — projectile motion

A ball is launched from ground level with $u = 20.0\ \text{m s}^{-1}$ at $\theta = 40^\circ$ . Take $g = 9.81\ \text{m s}^{-2}$ .
Find (i) time of flight $T$ , (ii) maximum height $H_{\max}$ , (iii) range $R$ , and (iv) the position $(x,y)$ at $t=1.0\ \text{s}$ .

(i) Time of flight

$T=\frac{2u\sin\theta}{g} =\frac{2(20.0)\sin 40^\circ}{9.81} \approx 2.62\ \text{s}$

(ii) Maximum height

$H_{\max}=\frac{u^2\sin^2\theta}{2g} =\frac{(20.0)^2\sin^2 40^\circ}{2(9.81)} \approx 8.42\ \text{m}$

(iii) Range

$R=\frac{u^2\sin 2\theta}{g} =\frac{(20.0)^2\sin 80^\circ}{9.81} \approx 40.2\ \text{m}$

(iv) Position at $t=1.0\ \text{s}$

$x(1.0)=u\cos\theta(1.0)=20.0\cos 40^\circ \approx 15.32\ \text{m}$

$y(1.0)=u\sin\theta(1.0)-\tfrac12 g(1.0)^2 =20.0\sin 40^\circ-4.905 \approx 7.95\ \text{m}$

Answers: $T\approx 2.62\ \text{s}$ , $H_{\max}\approx 8.42\ \text{m}$ , $R\approx 40.2\ \text{m}$ , $(x,y)\approx (15.32,\ 7.95)\ \text{m}$ .

Model assumptions

Air resistance is neglected.
$g$ is constant and acts vertically downward.
“Same height” means the launch and landing points have the same vertical coordinate.
If launch and landing heights differ, use $x(t)$ and $y(t)$ and solve for the required time first.

Lesson 1 — Kinematics (IB Physics SL)

Learning objectives

a) Vectors

Scalars

Vectors

Representing vectors

Multiplying a vector by a scalar

Adding vectors

Resolving vectors into components

Worked example — resolving a force

b) Describing motion

Position

Displacement

Distance

Velocity and speed

Acceleration

Worked example — acceleration

Motion graphs

Instantaneous vs average

c) Equations of motion (SUVAT)

The four SUVAT equations

Worked examples (full solutions)

Quick exam tips

d) Projectile motion

Position as a function of time

Key results (same launch and landing height)

Worked example — projectile motion

Model assumptions

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IB Physics SL 2025; Papers 1 & 2 Practice

A.1 Kinematics

Lesson 1 — Kinematics (IB Physics SL)

Learning objectives

a) Vectors

Scalars

Vectors

Representing vectors

Multiplying a vector by a scalar

Adding vectors

Resolving vectors into components

Worked example — resolving a force

b) Describing motion

Position

Displacement

Distance

Velocity and speed

Acceleration

Worked example — acceleration

Motion graphs

Instantaneous vs average

c) Equations of motion (SUVAT)

The four SUVAT equations

Worked examples (full solutions)

Quick exam tips

d) Projectile motion

Position as a function of time

Key results (same launch and landing height)

Worked example — projectile motion

Model assumptions

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