1. Kinematics

🏃‍♂️ Kinematics – IB Physics Notes

a) Vectors

Scalars

Magnitude only (e.g., speed, distance)

Scalars are physical quantities that are described only by their magnitude, meaning they tell us “how much” of something but not “in which direction.” For example, distance, speed, mass, and time are all scalars. If a person walks 5 kilometers, the number alone gives the complete description of the distance traveled, without needing to specify a direction.

Fig A.1.1

Vectors

Magnitude & direction (e.g., velocity, force)

Vectors, on the other hand, have both magnitude and direction. They tell us not only “how much” but also “which way.” Common examples of vectors include displacement, velocity, acceleration, and force. For instance, saying a car is moving at 20 meters per second eastward gives a vector quantity, because it combines the speed of the motion with its direction.

Representing Vectors

Use arrows; direction = direction, length = magnitude

Representing vectors is typically done using arrows. The length of the arrow corresponds to the magnitude of the vector, while the direction in which the arrow points shows the direction of the vector. For example, a force of 10 newtons to the right would be represented by an arrow of proportional length pointing rightward.

Fig A.1.2

Multiplying by a Scalar

Changes magnitude, not direction

When a vector is multiplied by a scalar, its magnitude changes, but its direction stays the same. If the scalar is greater than one, the vector becomes longer, while if it is between zero and one, the vector becomes shorter. A negative scalar reverses the direction of the vector. For example, doubling a velocity vector of 5 meters per second east produces a velocity of 10 meters per second east, while multiplying it by –1 would give 5 meters per second west.

Adding Vectors

Use tip-to-tail or components

There are two main ways of adding vectors. The first is the tip-to-tail method. In this approach, the tail of the second vector is placed at the tip of the first, and the resultant vector is drawn from the tail of the first vector to the tip of the last. The second method is the component method, in which each vector is broken into horizontal (x) and vertical (y) parts. The x-components are added together, as are the y-components, and then the resultant is found by recombining them using Pythagoras’ theorem and trigonometry.

Resolving Vectors

Fx = F cos(θ), Fy = F sin(θ)

Example: 10 N at 30° → Fx = 8.66 N, Fy = 5.00 N

Fig A.1.3

Resolving vectors means breaking a single vector into its horizontal and vertical components. If a vector of magnitude F makes an angle θ with the horizontal, its horizontal component is given by Fx = F cos θ and its vertical component is Fy = F sin θ. For example, a 10 newton force at an angle of 30° to the horizontal can be split into two components: a horizontal force of approximately 8.66 newtons and a vertical force of 5 newtons. This technique is particularly useful in analyzing forces in mechanics problems.

b) Describing Motion

Position

Location from a reference point

In physics, we often need to describe where something is. This is called its position. Position tells us the location of an object compared to a chosen reference point. For example, if you say your classroom is 20 meters north of the school gate, you are giving its position relative to the gate.

Displacement

Change in position (vector)

When an object moves, its position changes. The change in position is called displacement. Displacement is a vector, which means it has both size and direction. If you walk 50 meters east from your starting point, your displacement is 50 meters east. Displacement only cares about where you started and where you ended, not about the exact path you took.

Distance

Total path length (scalar)

By contrast, distance measures the total path length you actually traveled, and it is a scalar. For example, if you walk around a park in a big circle and end up back at your starting point, your displacement is zero (since you ended where you began), but the distance you traveled is the full length of the circle.

Velocity

Displacement ÷ time (vector)

To describe how fast displacement changes, we use velocity. Velocity is displacement divided by time, and since displacement is a vector, velocity is also a vector. If you walk 100 meters east in 20 seconds, your velocity is 5 meters per second east. Notice that velocity includes direction, which makes it different from speed.

Speed

Distance ÷ time (scalar)

Speed is distance divided by time. Since distance is a scalar, speed is also a scalar. If you ran 400 meters around a track in 80 seconds, your speed would be 5 meters per second. Unlike velocity, speed does not include direction, it only tells you “how fast.”

Acceleration

a = Δv / t

Finally, when velocity changes, we get acceleration. Acceleration measures how quickly the velocity of an object increases or decreases, or how quickly its direction changes. The formula is a = Δv / t, where Δv is the change in velocity and t is the time taken. For example, if a car speeds up from 10 m/s to 20 m/s in 5 seconds, its acceleration is (20 − 10) ÷ 5 = 2 m/s². Acceleration can be positive (speeding up) or negative (slowing down), and it can also happen if an object simply changes direction while keeping the same speed.

Graphs
- Displacement–time: slope = velocity
- Velocity–time: slope = acceleration; area = displacement
- Acceleration–time: area = change in velocity

In physics, graphs are a powerful way to describe motion. They allow us to see how quantities like displacement, velocity, and acceleration change with time. There are three main types of motion graphs: displacement–time, velocity–time, and acceleration–time.

The displacement-time graph shows how an object’s position changes over time. This graph’s slope (the steepness of the line) tells us the object’s velocity. The object moves with constant velocity if the graph is a straight line sloping upward. A steeper slope means a higher velocity. If the graph is flat (horizontally), the object is not moving — its displacement is constant.

The velocity–time graph shows how velocity changes with time. The slope of this graph tells us the acceleration of the object. A straight, sloping line means constant acceleration, while a flat line means constant velocity (no acceleration). Another important thing about the velocity–time graph is that the area under the graph gives the displacement. For example, if the graph is a straight horizontal line at 10 m/s for 5 seconds, the area under the line is 10 × 5 = 50 meters, which is the displacement of the object.

The acceleration–time graph shows how acceleration changes over time. The area under this graph gives the change in velocity. For example, if an object has a constant acceleration of 2 m/s² for 5 seconds, the area under the line is 2 × 5 = 10 m/s, which means the velocity increased by 10 m/s during that time.

Instantaneous

value at one moment

An instantaneous value is the value of a quantity at a single moment in time. It is like taking a snapshot of motion at one exact instant. For example, when you glance at a car’s speedometer, it shows you the instantaneous speed — how fast the car is moving right now, not how fast it was moving a few seconds ago or will be moving later.

Average

total ÷ total time

On the other hand, an average value looks at the total change over a period of time and divides it by the total time taken. For example, if you drive 120 kilometers in 2 hours, your average speed is 60 kilometers per hour, even though your actual speed may have been faster on the highway and slower in the city. In physics, we use averages to simplify motion into something easier to calculate, while instantaneous values help us describe motion at precise moments.

Example: 0 → 20 m/s in 5s → a = 4 m/s²

c) Equations of Motion

Free fall: g = 9.81 m/s² downward

Example: Rock dropped for 2s → s = 19.6 m

Non-uniform motion: Use graphs or numerical methods

SUVAT: The Four Equations of Uniformly Accelerated Motion

Symbols used: s = displacement (m), u = initial velocity (m/s), v = final velocity (m/s), a = acceleration (m/s²), t = time (s). These equations apply only when a is constant.

The Four SUVAT Equations

Here are the four standard equations (often used together and called the SUVAT equations):

\( \displaystyle v = u + a t \)
\( \displaystyle s = u t + \tfrac{1}{2} a t^2 \)
\( \displaystyle v^2 = u^2 + 2 a s \)
\( \displaystyle s = \tfrac{(u + v)}{2}\,t \) (distance = average velocity × time)

Worked Examples

1) Using \(v = u + at\) — “Speeding up from rest“

Problem. A car starts from rest and accelerates at \(a = 3.0\ \text{m/s}^2\) for \(t = 6.0\ \text{s}\). What is its final speed?

Given: \(u = 0\), \(a = 3.0\ \text{m/s}^2\), \(t = 6.0\ \text{s}\).

Solution.

\( v = u + at = 0 + (3.0)(6.0) = 18\ \text{m/s} \).

Answer: \(v = 18\ \text{m/s}\).

2) Using \(s = ut + \tfrac12 a t^2\) — “How far during a push?”

Problem. A runner is already moving at \(u = 2.0\ \text{m/s}\) and accelerates at \(a = 1.5\ \text{m/s}^2\) for \(t = 4.0\ \text{s}\). How far does the runner travel?

Solution.

\( s = ut + \tfrac12 a t^2 = (2.0)(4.0) + \tfrac12(1.5)(4.0)^2 = 8.0 + 0.5\times1.5\times16 = 8.0 + 12.0 = 20\ \text{m} \).

Answer: \(s = 20\ \text{m}\).

3) Using \(v^2 = u^2 + 2as\) — “Braking distance”

Problem. A car traveling at \(u = 25\ \text{m/s}\) brakes uniformly to a stop with acceleration \(a = -5.0\ \text{m/s}^2\). What stopping distance \(s\) is required?

Given: \(v = 0\).

Solution.

\(0^2 = (25)^2 + 2(-5.0)s \Rightarrow 0 = 625 – 10s \Rightarrow 10s = 625 \Rightarrow s = 62.5\ \text{m}.\)

Answer: \(s = 62.5\ \text{m}\).

4) Using \(s = \tfrac{(u+v)}{2}t\) — “Distance while speeding up”

Problem. A train speeds up uniformly from \(u = 10\ \text{m/s}\) to \(v = 22\ \text{m/s}\) in \(t = 6.0\ \text{s}\). How far does it travel?

Solution.

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

Answer: \(s = 96\ \text{m}\).

Quick Tips for Students

1. Choose the equation that doesn’t include the unknown you are solving for. For example, if time is not given, try \(v^2 = u^2 + 2as\).

2. Use a sign convention (e.g. upward/forward positive, downward/backward negative) and be consistent with it when plugging in values for acceleration or displacement.

3. Units: check that \(s\) is in metres (m), \(u\) and \(v\) in metres per second (m/s), \(a\) in metres per second squared (m/s²), and \(t\) in seconds (s).

d) Projectile Motion

Definition. Projectile motion: an object launched with initial speed \(u\) at angle \(\theta\) above the horizontal moves under constant gravitational acceleration \(g\) (no air resistance, flat ground). Horizontal and vertical motions are independent.

Horizontal: constant velocity \(u\cos\theta\).
Vertical: uniformly accelerated with acceleration \(-g\).

Position at any time \(t\)

Taking the launch point as \((x_0,y_0)\):

\[
x(t) = x_0 + u\cos\theta \, t,
\qquad
y(t) = y_0 + u\sin\theta \, t – \frac{1}{2} g t^2.
\]

Velocities:
\[
v_x = u\cos\theta \quad(\text{constant}), \qquad
v_y = u\sin\theta – gt.
\]

Time of flight \(T\) (same launch and landing height)

Set \(y(T)=y_0\) and solve for \(T\):

\[
T = \frac{2u\sin\theta}{g}.
\]

Maximum height \(H_{\max}\)

At the top \(v_y=0\Rightarrow t_{\text{top}}=u\sin\theta/g\). Substituting:

\[
H_{\max} = \frac{u^2 \sin^2\theta}{2g}.
\]

Horizontal range \(R\) (same launch and landing height)

Using \(R=(u\cos\theta)T\) and \(\sin 2\theta = 2\sin\theta\cos\theta\):

\[
R = \frac{u^2 \sin 2\theta}{g}.
\]

Note: \(R\) is maximized at \(\theta=45^\circ\) (neglecting air resistance).

Worked Example

Problem. 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 \(T\), \(H_{\max}\), \(R\), and \((x,y)\) at \(t=1.0\ \text{s}\).

Solution

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

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

\[
R=\frac{u^2\sin 2\theta}{g}
=\frac{(20.0)^2\sin 80^\circ}{9.81}
\approx 40.16\ \text{m}.
\]

\[
x(1.0)=u\cos\theta\,(1.0)=20.0\cos 40^\circ \approx 15.32\ \text{m},
\]
\[
y(1.0)=u\sin\theta\,(1.0)-\tfrac12 g (1.0)^2
=20.0\sin 40^\circ-4.905
\approx 7.95\ \text{m}.
\]

Assumptions & Notes

No air resistance; \(g\) constant.
Flat ground; “same height” means launch and landing at the same vertical level.
For different heights, use the general \(x(t)\), \(y(t)\) and solve for the required quantity.

When an object is projected with velocity v at an angle of θ to the horizontal and is then assumed to move freely under gravity in a vertical plane, its motion can be modelled by the following equations:

Horizontal motion: constant velocity → x = vₓt
Vertical motion: affected by gravity → y = vᵧt + ½(-g)t²
Ignore air resistance: trajectory = symmetrical parabola
With fluid resistance: trajectory is flattened; range and height reduced

Example: 20 m/s at 30° → analyze vx = 17.32 m/s, vy = 10 m/s separately

1. Kinematics

🏃‍♂️ Kinematics – IB Physics Notes

a) Vectors

Scalars

Vectors

Representing Vectors

Multiplying by a Scalar

Adding Vectors

Resolving Vectors

b) Describing Motion

Position

Displacement

Distance

Velocity

Speed

Acceleration

Graphs

Instantaneous

Average

c) Equations of Motion

The Four SUVAT Equations

Worked Examples

1) Using \(v = u + at\) — “Speeding up from rest“

2) Using \(s = ut + \tfrac12 a t^2\) — “How far during a push?”

3) Using \(v^2 = u^2 + 2as\) — “Braking distance”

4) Using \(s = \tfrac{(u+v)}{2}t\) — “Distance while speeding up”

Quick Tips for Students

d) Projectile Motion window.MathJax = { tex: { inlineMath: [['\\(','\\)'], ['$', '$']], displayMath: [['\\[','\\]']], processEscapes: true }, options: { skipHtmlTags: ['script','noscript','style','textarea','pre','code'] } };

Position at any time \(t\)

Time of flight \(T\) (same launch and landing height)

Maximum height \(H_{\max}\)

Horizontal range \(R\) (same launch and landing height)

Worked Example

Solution

Assumptions & Notes

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d) Projectile Motion