Kinematics Study Notes, Formulas & Summary

1. Introduction to Kinematics

Kinematics is the branch of classical mechanics that describes the motion of objects without reference to the forces that cause that motion. It provides a mathematical framework for quantifying how objects move, using the fundamental concepts of displacement, velocity, and acceleration as functions of time.

Distance (scalar) is the total path length travelled by an object.
Displacement ($s$, vector) is the straight-line distance from the initial position to the final position, with direction specified. It can be positive or negative depending on direction.
Speed (scalar) is the rate of change of distance.
Velocity ($v$, vector) is the rate of change of displacement with time: $v = \frac{\Delta s}{\Delta t}$
Acceleration ($a$, vector) is the rate of change of velocity with time: $a = \frac{\Delta v}{\Delta t}$. An object accelerates if its speed changes, direction changes, or both.

In everyday life, a car braking is decelerating (negative acceleration), while a ball thrown upward decelerates as it rises and accelerates as it falls.

2. Equations of Uniformly Accelerated Motion (SUVAT)

When acceleration is constant (uniform), five kinematic variables are interrelated: displacement ($s$), initial velocity ($u$), final velocity ($v$), acceleration ($a$), and time ($t$). The SUVAT equations connect any three of these to find the remaining two.

Equation	Connects
$v = u + at$	v, u, a, t
$s = ut + \frac{1}{2}at^2$	s, u, a, t
$v^2 = u^2 + 2as$	v, u, a, s
$s = \frac{(u+v)}{2}t$	s, u, v, t
$s = vt - \frac{1}{2}at^2$	s, v, a, t

Strategy for solving SUVAT problems: 1. List all known quantities with their signs. 2. Identify which quantity is unknown. 3. Choose the equation that contains only known quantities and the unknown. 4. Always define a positive direction at the start — typically up or right.

graph TD
    A[Kinematics Problem] --> B{Is acceleration constant?}
    B -- Yes --> C[Use SUVAT Equations]
    C --> D[Identify 3 known variables]
    D --> E[Choose correct equation]
    B -- No --> F[Use Calculus]
    F --> G[v = ds/dt]
    F --> H[a = dv/dt]

3. Motion Graphs

Interpreting motion graphs is one of the most important skills in A-Level kinematics.

Displacement–Time (s–t) Graph

The gradient at any point equals the instantaneous velocity.
A straight line indicates constant velocity (zero acceleration).
A curve indicates changing velocity (acceleration or deceleration).
A horizontal line (zero gradient) means the object is stationary.
A negative gradient means the object is moving in the negative direction.

Velocity–Time (v–t) Graph

The gradient equals the instantaneous acceleration.
The area under the graph equals the displacement. This applies whether the motion is uniform or not.
A horizontal line indicates constant velocity (zero acceleration).
A straight line with positive gradient indicates uniform acceleration.
When the line crosses the time axis (velocity = 0), the object has momentarily stopped.

Acceleration–Time (a–t) Graph

The area under the graph equals the change in velocity ($\Delta v$).
A horizontal line at $a = 0$ means constant velocity.

4. Free Fall and Air Resistance

When an object falls freely under gravity only (no air resistance), it accelerates uniformly at $g = 9.81 \text{ m s}^{-2}$ downward. Using SUVAT with $a = g$, the speed and distance fallen can be calculated.

In reality, air resistance acts upward on a falling object. As the object speeds up, air resistance increases, reducing the net downward force. Eventually, the upward drag equals the downward weight, and the object falls at terminal velocity — a constant speed with zero acceleration.

The sequence of events for a skydiver: 1. Initial free fall: large acceleration (~9.81 m s⁻²) 2. Acceleration decreases as drag increases 3. Terminal velocity reached (zero acceleration, constant speed ~55 m s⁻¹) 4. Parachute opens: drag suddenly exceeds weight → deceleration 5. New, much lower terminal velocity reached (~5–7 m s⁻¹)

5. Projectile Motion

A projectile is any object that is launched into the air and moves solely under the influence of gravity (ignoring air resistance). The key insight is that the horizontal and vertical components of motion are completely independent of each other.

Horizontal Component

No horizontal force acts (ignoring air resistance), so horizontal acceleration = 0.
Horizontal velocity $v_x = u\cos\theta$ remains constant throughout the flight.
Horizontal displacement: $s_x = u\cos\theta \cdot t$

Vertical Component

Only gravity acts vertically, so vertical acceleration = $g = 9.81 \text{ m s}^{-2}$ downward.
Vertical velocity: $v_y = u\sin\theta - gt$
Vertical displacement: $s_y = u\sin\theta \cdot t - \frac{1}{2}gt^2$

Key Results for Projectile Launched at Angle θ with Speed u

Time of flight: $T = \frac{2u\sin\theta}{g}$
Maximum height: $H = \frac{u^2\sin^2\theta}{2g}$
Maximum range: $R = \frac{u^2\sin 2\theta}{g}$, maximised at $\theta = 45°$

At maximum height, the vertical velocity is zero, but horizontal velocity is unchanged. The path followed is a parabola. This is confirmed by eliminating $t$ from the two displacement equations.

6. Relative Velocity

The velocity of object A relative to object B is: $\vec{v}_{A \text{ rel } B} = \vec{v}_A - \vec{v}_B$

This concept is important when dealing with boats crossing rivers, aircraft in wind, and collisions. For example, a boat with velocity 3 m s⁻¹ east crossing a river with current 4 m s⁻¹ north has a resultant velocity found by vector addition: magnitude $\sqrt{3^2+4^2} = 5$ m s⁻¹ at angle $\arctan(4/3) = 53°$ from east.

Kinematics — Study Notes