Welcome to Kinematics!
Welcome to Kinematics, one of the most exciting and practical topics in your CCEA GCSE Further Mathematics Mechanics course! Have you ever wondered how engineers calculate the braking distance of a roller coaster, or how sports scientists track a sprinter accelerating off the blocks? Kinematics is the branch of mechanics that describes the motion of objects without worrying about the forces that cause the motion.
Don't worry if mechanics feels a bit intimidating at first. By breaking motion down into simple diagrams, graphs, and a set of five powerful formulas (often called the SUVAT equations), you will quickly build the confidence to solve any kinematics question on your exam!
1. The Language of Motion
Before we calculate anything, we need to know the fundamental quantities used in mechanics. In physics and further maths, quantities are divided into scalars (which have only size/magnitude) and vectors (which have both size and direction).
Key Terms and Units
• Displacement (\(s\)): The straight-line distance from the starting point to the final position, including direction. Measured in metres (\(\text{m}\)). Unlike distance, displacement can be positive, negative, or zero.
• Initial Velocity (\(u\)): The speed of an object in a given direction at the very start of the time period being measured. Measured in metres per second (\(\text{m s}^{-1}\) or \(\text{m/s}\)).
• Final Velocity (\(v\)): The speed and direction of an object at the end of the time period. Measured in \(\text{m s}^{-1}\).
• Acceleration (\(a\)): The rate at which velocity changes over time. Measured in metres per second squared (\(\text{m s}^{-2}\) or \(\text{m/s}^2\)). A positive acceleration means velocity is increasing in the chosen direction, while a negative acceleration (deceleration or retardation) means it is slowing down.
• Time (\(t\)): The time taken for the motion to occur. Measured in seconds (\(\text{s}\)).
Analogy: Distance vs. Displacement
Imagine walking \(50\text{ m}\) to the end of a corridor and walking \(50\text{ m}\) right back to where you started. Your total distance travelled is \(100\text{ m}\), but your overall displacement is \(0\text{ m}\) because you ended up exactly where you began!
Did You Know?
The term "kinematics" comes from the Greek word kinema, meaning movement. It is the exact same root word that gives us the word cinema (moving pictures)!
Key Takeaway: Always check your units before starting calculations! Distances must be in metres (\(\text{m}\)), speeds/velocities in \(\text{m s}^{-1}\), accelerations in \(\text{m s}^{-2}\), and time in seconds (\(\text{s}\)).
2. Travel Graphs: Seeing Motion in Action
Graphs are a wonderful way to see what an object is doing over time. In CCEA GCSE Further Mathematics, you need to understand two main types of graphs: Displacement-Time Graphs and Velocity-Time Graphs.
Displacement-Time (\(s\)-\(t\)) Graphs
On a displacement-time graph, displacement (\(s\)) is on the vertical axis and time (\(t\)) is on the horizontal axis.
• Gradient (Slope) = Velocity: \(\text{Gradient} = \frac{\text{Change in } s}{\text{Change in } t} = v\)
• Horizontal flat line: The object is stationary (velocity = \(0\)).
• Straight sloped line: The object is moving at a constant velocity.
• Curved line: The velocity is changing (the object is accelerating or decelerating).
Velocity-Time (\(v\)-\(t\)) Graphs
Velocity-time graphs are the most common graph type in mechanics exam questions. They give us two crucial pieces of information:
1. Gradient = Acceleration:
\(\text{Acceleration } (a) = \frac{\text{Change in velocity}}{\text{Time taken}} = \frac{v - u}{t}\)
• A straight upward line means constant acceleration.
• A horizontal line means constant speed / zero acceleration.
• A straight downward line means constant deceleration (negative acceleration).
2. Area Under the Graph = Distance / Displacement:
To find the total distance travelled from a velocity-time graph, divide the area under the line into simple geometric shapes: triangles, rectangles, and trapezia.
• \(\text{Area of a rectangle} = \text{base} \times \text{height}\)
• \(\text{Area of a triangle} = \frac{1}{2} \times \text{base} \times \text{height}\)
• \(\text{Area of a trapezium} = \frac{1}{2}(a + b)h\), where \(a\) and \(b\) are the parallel sides and \(h\) is the perpendicular height.
Worked Example: Analysing a \(v\)-\(t\) Graph
A car accelerates uniformly from rest to a speed of \(12\text{ m s}^{-1}\) in \(6\text{ s}\). It then travels at this constant speed for \(14\text{ s}\), before decelerating uniformly to rest in \(4\text{ s}\).
Step 1: Find the acceleration during the first \(6\text{ s}\).
\(\text{Acceleration} = \text{Gradient} = \frac{12 - 0}{6} = 2\text{ m s}^{-2}\)
Step 2: Find the deceleration in the final \(4\text{ s}\).
\(\text{Gradient} = \frac{0 - 12}{4} = -3\text{ m s}^{-2}\)
So the deceleration (retardation) is \(3\text{ m s}^{-2}\).
Step 3: Calculate the total distance travelled.
The shape under the graph is a single trapezium with parallel sides of length \(24\text{ s}\) (total time: \(6 + 14 + 4\)) and \(14\text{ s}\) (constant speed stage), and a height of \(12\text{ m s}^{-1}\).
\(\text{Total Distance} = \frac{1}{2}(a + b)h = \frac{1}{2}(24 + 14) \times 12 = \frac{1}{2}(38) \times 12 = 19 \times 12 = 228\text{ m}\)
Common Mistake to Avoid: Don't forget that if the graph goes below the time axis (negative velocity), the object is travelling in the opposite direction. Total displacement considers negative area, while total distance treats all area as positive.
Key Takeaway: For velocity-time graphs, always remember: Gradient gives Acceleration and Area gives Distance.
3. Constant Acceleration Equations (The SUVAT Equations)
When an object moves in a straight line with constant (uniform) acceleration, we can use a set of five connected formulas. These are affectionately known as the SUVAT equations.
The Five SUVAT Variables
• \(s\) = displacement (\(\text{m}\))
• \(u\) = initial velocity (\(\text{m s}^{-1}\))
• \(v\) = final velocity (\(\text{m s}^{-1}\))
• \(a\) = acceleration (\(\text{m s}^{-2}\))
• \(t\) = time (\(\text{s}\))
The SUVAT Formulae
1. \(v = u + at\) (relates \(u, v, a, t\); does not need \(s\))
2. \(s = \frac{1}{2}(u + v)t\) (relates \(s, u, v, t\); does not need \(a\))
3. \(s = ut + \frac{1}{2}at^2\) (relates \(s, u, a, t\); does not need \(v\))
4. \(v^2 = u^2 + 2as\) (relates \(s, u, v, a\); does not need \(t\))
5. \(s = vt - \frac{1}{2}at^2\) (relates \(s, v, a, t\); does not need \(u\))
The 4-Step Method to Solve Any SUVAT Problem
Step 1 (List): Write down "\(s =\)", "\(u =\)", "\(v =\)", "\(a =\)", "\(t =\)" down the side of your page and fill in the values you know from the question.
Step 2 (Identify): Identify the value you need to calculate, and the one variable you don't care about.
Step 3 (Choose): Pick the formula containing your knowns and your target variable.
Step 4 (Substitute & Solve): Plug in your numbers and solve carefully for the unknown.
Worked Example: Horizontal Motion
A train starts from rest and accelerates uniformly at \(0.8\text{ m s}^{-2}\) along a straight track for a distance of \(250\text{ m}\). Find the final speed of the train.
Step 1 (List):
• \(s = 250\text{ m}\)
• \(u = 0\text{ m s}^{-1}\) (from the clue "starts from rest")
• \(v = ?\)
• \(a = 0.8\text{ m s}^{-2}\)
• \(t = \text{not given and not needed}\)
Step 2 & 3 (Choose formula): We have \(s\), \(u\), \(a\) and want \(v\). The formula without \(t\) is:
\(v^2 = u^2 + 2as\)
Step 4 (Substitute & Solve):
\(v^2 = (0)^2 + 2(0.8)(250)\)
\(v^2 = 0 + 400\)
\(v = \sqrt{400} = 20\text{ m s}^{-1}\)
Quick Review: Look out for hidden clues in words!
• "Starts from rest" \(\implies u = 0\)
• "Comes to rest" or "brakes to a halt" \(\implies v = 0\)
• "Constant speed" \(\implies a = 0\)
Key Takeaway: SUVAT equations can only be used when acceleration is constant. Always list your 5 variables first!
4. Vertical Motion Under Gravity
When an object is thrown upwards or dropped downwards and moves freely through the air (ignoring air resistance), it accelerates downwards due to the pull of gravity.
Key Rules for Gravity Problems
• Acceleration due to gravity (\(g\)): Near the Earth's surface, \(g = 9.8\text{ m s}^{-2}\) (unless a question specifies \(g = 10\text{ m s}^{-2}\) or \(g = 9.81\text{ m s}^{-2}\)). Gravity ALWAYS acts downwards toward the centre of the Earth.
• Sign Conventions: Pick one direction as positive at the start of your calculation and stick to it! If you choose upwards as positive (+), then the acceleration due to gravity is negative: \(a = -9.8\text{ m s}^{-2}\).
• Highest Point: At the maximum height of its flight, an object momentarily stops before falling back down. That means \(v = 0\text{ m s}^{-1}\) at the top!
Worked Example: Throwing a Ball Upwards
A ball is projected vertically upwards from ground level with an initial speed of \(24.5\text{ m s}^{-1}\). Taking \(g = 9.8\text{ m s}^{-2}\) and ignoring air resistance:
(a) Find the maximum height reached.
(b) Find the total time the ball is in the air before hitting the ground.
Part (a): Maximum Height
Let upwards be the positive direction.
• \(s = ?\)
• \(u = +24.5\text{ m s}^{-1}\)
• \(v = 0\text{ m s}^{-1}\) (at maximum height)
• \(a = -9.8\text{ m s}^{-2}\)
• \(t = \text{not needed}\)
Choose \(v^2 = u^2 + 2as\):
\(0^2 = (24.5)^2 + 2(-9.8)s\)
\(0 = 600.25 - 19.6s\)
\(19.6s = 600.25\)
\(s = \frac{600.25}{19.6} = 30.625\text{ m}\)
The maximum height reached is \(30.625\text{ m}\) (or \(30.6\text{ m}\) to 3 s.f.).
Part (b): Total Time in the Flight
When the ball returns to the ground, its displacement from the start is \(s = 0\).
• \(s = 0\text{ m}\)
• \(u = +24.5\text{ m s}^{-1}\)
• \(a = -9.8\text{ m s}^{-2}\)
• \(t = ?\)
Choose \(s = ut + \frac{1}{2}at^2\):
\(0 = 24.5t + \frac{1}{2}(-9.8)t^2\)
\(0 = 24.5t - 4.9t^2\)
Factorise out \(t\):
\(0 = t(24.5 - 4.9t)\)
This gives two solutions:
\(t = 0\text{ s}\) (the moment of launch)
\(24.5 - 4.9t = 0 \implies 4.9t = 24.5 \implies t = \frac{24.5}{4.9} = 5\text{ s}\)
The total time in the air is \(5\text{ s}\).
Symmetry Trick for Projectiles:
Due to symmetry, the time taken to reach the top is exactly equal to the time taken to fall back down to the ground. In the example above, it took \(2.5\text{ s}\) to reach the peak and \(2.5\text{ s}\) to fall back down, giving a total of \(5\text{ s}\)!
Key Takeaway: For vertical motion questions, clearly state your positive direction at the very start. If UP is positive, then \(a = -g = -9.8\text{ m s}^{-2}\).
5. Top Exam Tips & Checklist
1. Look out for unit mismatches: If speed is given in \(\text{km/h}\), convert it to \(\text{m s}^{-1}\) by dividing by \(3.6\). If time is given in minutes, multiply by \(60\) to get seconds.
2. Draw a clear diagram: For graph questions and multi-stage journeys, sketch a quick diagram showing the starting point, positive direction, and key stages.
3. Check positive/negative signs: If a car is decelerating, remember that \(a\) must have the opposite sign to velocity \(u\).
4. Multi-stage journeys: If a car accelerates and then travels at constant speed, treat this as two separate stages. The final velocity \(v\) of Stage 1 becomes the initial velocity \(u\) of Stage 2!
Quick Formula Reference Card
• \(v = u + at\)
• \(s = \frac{1}{2}(u + v)t\)
• \(s = ut + \frac{1}{2}at^2\)
• \(v^2 = u^2 + 2as\)
• \(s = vt - \frac{1}{2}at^2\)
You are now fully equipped to tackle any Kinematics problem in CCEA GCSE Further Mathematics! Practice drawing the graphs, listing your SUVAT variables, and watch your confidence grow.