A.1 Kinematics

UNIT SYLLABUS

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Kinematics is the language physics uses to describe motion — where something is, how fast it moves, and how its velocity changes. Everything rests on three ideas: position, velocity (the rate of change of position) and acceleration (the rate of change of velocity). Master the four 'suvat' equations for uniform acceleration and you can predict where a body will be at any instant. The unit's showpiece is projectile motion: by splitting a projectile's velocity into independent horizontal and vertical components, a curved flight becomes two simple one-dimensional problems solved simultaneously.

Guiding Questions

? How can the motion of a body be described both quantitatively and qualitatively?
? How can the position of a body in space and time be predicted?
? How does the analysis of motion in one and two dimensions solve real-life problems?

Projectile motion: the horizontal velocity component stays constant while the vertical component changes uniformly under gravity, producing a parabolic path.

✓ What the IB expects you to master

Describe motion using position, displacement, velocity and acceleration — and state which are vectors.
Distinguish distance from displacement, and speed from velocity.
Distinguish instantaneous from average values of velocity, speed and acceleration, and determine each from data or graphs.
Select and apply the four equations of uniformly accelerated motion ( $v = u + at$ , $s = ut + \frac{1}{2}at^2$ , $v^2 = u^2 + 2as$ , $s = \frac{(u+v)}{2}t$ ).
Interpret and sketch displacement–time, velocity–time and acceleration–time graphs, using gradients and areas.
Analyse projectile motion without fluid resistance by resolving into horizontal (constant velocity) and vertical (constant acceleration) components.
Describe qualitatively how fluid resistance changes a projectile's time of flight, trajectory, velocity, acceleration, range and terminal speed.

1 Key Formulas

Final velocity

v = u + at

Displacement

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

Velocity–displacement

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

Displacement (average velocity)

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

2 Exam Preparation & Topic Explanations

Mastering projectile motion

Projectile questions are guaranteed marks once the routine is automatic. Resolve the initial velocity into $u_x = u\cos\theta$ and $u_y = u\sin\theta$ ; treat the horizontal direction as constant velocity and the vertical as uniform acceleration $-g$ ; and remember the two directions share only one quantity — time.

Almost every projectile problem is solved by finding the time from the vertical motion, then substituting it into the horizontal motion (or the reverse).

Pro Exam Strategy

At maximum height $v_y = 0$ — but $v_x$ and the acceleration $g$ are unchanged.
Pick the suvat equation that omits the variable you neither know nor need.
Keep a strict sign convention: choose up as positive and stick to it for $u$ , $a$ and $s$ .
Graph questions: gradient of $s$ – $t$ is velocity; gradient of $v$ – $t$ is acceleration; area under $v$ – $t$ is displacement.

3 MCQ Practice

Q1. A ball is thrown vertically upwards. At the highest point of its motion, which statement is correct?

Velocity and acceleration are both zero
Velocity is zero; acceleration is $9.81\ \text{m s}^{-2}$ downwards
Velocity is zero; acceleration is $9.81\ \text{m s}^{-2}$ upwards
Velocity and acceleration are both non-zero

Q2. A car accelerates uniformly from rest to $24\ \text{m s}^{-1}$ over a distance of $144\ \text{m}$ . What is its acceleration?

$1.0\ \text{m s}^{-2}$
$2.0\ \text{m s}^{-2}$
$3.0\ \text{m s}^{-2}$
$4.0\ \text{m s}^{-2}$

Q3. Two identical balls are launched horizontally from the same height, one at $5\ \text{m s}^{-1}$ and one at $10\ \text{m s}^{-1}$ . Ignoring air resistance, which lands first?

The slower ball
The faster ball
They land at the same time
It depends on their masses

4 Short Answer Questions

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A.2 Forces and momentum

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