Welcome to Linear Motion!
Welcome to one of the most fundamental and exciting chapters in CCEA AS Physics: Linear Motion! Whether you are tracking a 100-metre sprinter, analysing a falling raindrop, or calculating the stopping distance of a car, linear motion gives you the exact mathematical toolkit to describe how things move in straight lines.
Don't worry if physics equations have felt overwhelming in the past! We will break down every single concept into bite-sized, digestible steps with clear analogies, real-world examples, and step-by-step problem-solving guides.
1. The Foundations: Scalars vs. Vectors in Motion
Before we look at moving objects, we need to distinguish between two types of physical quantities: scalars and vectors.
• Scalar: A quantity that has magnitude (size) only. Direction does not matter.
• Vector: A quantity that has both magnitude and a specific direction.
Distance vs. Displacement
• Distance (\(d\)): A scalar quantity representing the total ground covered during motion, measured in metres (\(\text{m}\)).
• Displacement (\(s\)): A vector quantity representing the shortest straight-line distance from the start point to the finish point, along with the direction.
Everyday Analogy: Imagine running exactly one full lap around a \(400\text{ m}\) athletics track. Your distance travelled is \(400\text{ m}\), but because you ended up right back where you started, your total displacement is \(0\text{ m}\)!
Speed vs. Velocity
• Speed: The rate of change of distance (scalar). Calculated as \(\text{Speed} = \frac{\text{Distance}}{\text{Time}}\), unit: \(\text{m s}^{-1}\).
• Velocity (\(v\)): The rate of change of displacement (vector). Calculated as \(\text{Velocity} = \frac{\Delta s}{\Delta t}\), unit: \(\text{m s}^{-1}\).
Did you know? A car driving around a circular roundabout at a steady \(30\text{ mph}\) has a constant speed, but its velocity is constantly changing because its direction is constantly changing!
Acceleration
Acceleration (\(a\)) is defined as the rate of change of velocity. It is a vector quantity measured in metres per second squared (\(\text{m s}^{-2}\)).
\(a = \frac{v - u}{t} = \frac{\Delta v}{\Delta t}\)
Where:
• \(u\) = initial velocity (\(\text{m s}^{-1}\))
• \(v\) = final velocity (\(\text{m s}^{-1}\))
• \(t\) = time taken (\(\text{s}\))
Important Note: If an object slows down, its acceleration is in the opposite direction to its velocity. We often call this deceleration or negative acceleration.
Key Takeaway: Always check if direction matters! Vectors (displacement, velocity, acceleration) must include or account for direction (+ or -), whereas scalars (distance, speed) do not.
2. Motion Graphs: Reading the Story of Motion
Graphs are one of the most powerful tools in Physics. In CCEA AS 1, you must be able to interpret and calculate values from three main types of motion graphs.
Displacement-Time (\(s\)-\(t\)) Graphs
• Gradient (Slope): Represents velocity (\(\text{Gradient} = \frac{\Delta s}{\Delta t} = v\)).
• Flat horizontal line: Object is stationary (\(v = 0\text{ m s}^{-1}\)).
• Straight sloping line: Object is moving at a constant velocity.
• Curved line: Velocity is changing, meaning the object is accelerating.
• Instantaneous velocity: To find velocity on a curved \(s\)-\(t\) graph at a specific instant, draw a tangent to the curve at that point and find its gradient.
Velocity-Time (\(v\)-\(t\)) Graphs
Velocity-time graphs contain the most information of all motion graphs:
1. Gradient: Represents acceleration (\(\text{Gradient} = \frac{\Delta v}{\Delta t} = a\)).
2. Area under the graph: Represents the displacement (\(s\)) or total distance travelled.
• Flat horizontal line: Constant velocity (\(a = 0\text{ m s}^{-2}\)).
• Straight upward slope: Uniform (constant) acceleration.
• Straight downward slope: Uniform deceleration.
• Section below the time axis: Object has reversed direction (negative velocity).
Acceleration-Time (\(a\)-\(t\)) Graphs
• Area under the graph: Represents the change in velocity (\(\Delta v\)).
• A horizontal line above the time axis indicates constant, uniform acceleration.
Memory Trick for Graphs:
Going from \(s \rightarrow v \rightarrow a\)? Take the Gradient (differentiate).
Going from \(a \rightarrow v \rightarrow s\)? Find the Area under the graph (integrate).
Key Takeaway: When faced with a graph question, ask yourself: "Do I need the gradient, the area, or just a direct coordinate read-off?"
3. The SUVAT Equations (Uniform Acceleration)
When an object moves with constant (uniform) acceleration in a straight line, we can use the famous SUVAT equations.
Meet the SUVAT Variables:
• \(s\) = displacement (\(\text{m}\))
• \(u\) = initial velocity (\(\text{m s}^{-1}\))
• \(v\) = final velocity (\(\text{m s}^{-1}\))
• \(a\) = constant acceleration (\(\text{m s}^{-2}\))
• \(t\) = time taken (\(\text{s}\))
The Core Equations:
1. \(v = u + at\)
2. \(s = \frac{(u + v)}{2}t\)
3. \(s = ut + \frac{1}{2}at^2\)
4. \(v^2 = u^2 + 2as\)
5. \(s = vt - \frac{1}{2}at^2\)
Step-by-Step Strategy for Solving SUVAT Problems:
Step 1: Define a positive direction (e.g., choose 'upwards' or 'to the right' as positive).
Step 2: List the five letters: \(s\), \(u\), \(v\), \(a\), \(t\).
Step 3: Fill in the 3 known values from the question and identify the 1 unknown you want to find.
Step 4: Select the equation that contains your 3 knowns and your 1 target unknown.
Step 5: Substitute the values and solve carefully.
Worked Example:
A sports car accelerates uniformly from rest to a speed of \(28\text{ m s}^{-1}\) over a distance of \(140\text{ m}\). Calculate its acceleration.
• \(s = 140\text{ m}\)
• \(u = 0\text{ m s}^{-1}\) (since it starts "from rest")
• \(v = 28\text{ m s}^{-1}\)
• \(a = ?\)
• \(t\) = not needed
Select the equation without \(t\): \(v^2 = u^2 + 2as\)
Substitute: \((28)^2 = (0)^2 + 2(a)(140)\)
\(784 = 280a\)
\(a = \frac{784}{280} = 2.8\text{ m s}^{-2}\)
Common Mistake to Avoid: Forgetting that words give hidden values! "Starts from rest" means \(u = 0\), and "comes to a stop" means \(v = 0\).
4. Motion Under Gravity & Free Fall
When an object falls solely under the influence of gravity (ignoring air resistance), it accelerates downward at a constant rate known as the acceleration due to gravity, denoted by \(g\).
On Earth, the standard value used in CCEA AS Physics is:
\(g = 9.81\text{ m s}^{-2}\) (downwards)
Key Rules for Vertical Motion Under Gravity:
• Always choose a positive direction at the start (e.g., upwards = positive).
• If upwards is positive, then acceleration is negative: \(a = -9.81\text{ m s}^{-2}\).
• At the maximum height of a vertical throw, the vertical velocity is momentarily zero (\(v = 0\text{ m s}^{-1}\)).
• The path is symmetrical: the time taken to rise to maximum height equals the time taken to fall back to the release level.
Required Practical: Experimental Determination of \(g\)
A classic CCEA practical involves determining \(g\) by free fall using an electromagnet and trapdoor or light gates.
Apparatus & Method:
1. A small steel ball is held by an electromagnet at a measured height \(h\) above a trapdoor or light gate.
2. When the switch is opened, the electromagnet releases the ball and simultaneously starts an electronic digital timer.
3. When the ball strikes the trapdoor, the circuit breaks, stopping the timer and recording the time of fall \(t\).
4. Repeat the drop for several different heights \(h\), taking repeated time measurements at each height to calculate an average.
Graphical Analysis:
From SUVAT: \(s = ut + \frac{1}{2}at^2\)
Since the ball drops from rest (\(u = 0\)) and \(s = h\), \(a = g\):
\(h = \frac{1}{2}gt^2\)
Comparing this to the equation of a straight line (\(y = mx + c\)):
• Plot \(h\) on the y-axis against \(t^2\) on the x-axis.
• The graph is a straight line through the origin with \(\text{Gradient} = \frac{1}{2}g\).
• Therefore: \(g = 2 \times \text{Gradient}\).
Sources of Experimental Error:
• Systematic error: Residual magnetism in the electromagnet can cause a slight delay in releasing the ball, making \(t\) slightly too long.
• Random error: Parallax error when reading the ruler for height \(h\).
Key Takeaway: By linearising the equation \(h = \frac{1}{2}gt^2\) into a plot of \(h\) against \(t^2\), the gradient directly gives \(\frac{1}{2}g\).
5. Projectile Motion
A projectile is an object given an initial velocity and left to move freely under the influence of gravity alone.
The Golden Rule of Projectiles:
Horizontal motion and vertical motion are completely independent of each other!
• Horizontal Motion: There is no horizontal resultant force (ignoring air resistance), so horizontal acceleration is zero (\(a_x = 0\)). The horizontal velocity (\(u_x\)) remains constant throughout the entire flight.
\(\text{Horizontal distance (Range)} = u_x \times t\)
• Vertical Motion: The object experiences a constant downward acceleration due to gravity (\(a_y = -9.81\text{ m s}^{-2}\)). Use the SUVAT equations for vertical motion!
• Connecting Variable: Time (\(t\)) is the single quantity shared between horizontal and vertical calculations.
Resolving Initial Velocity at an Angle \(\theta\) to the Horizontal:
If a projectile is launched with velocity \(u\) at an angle \(\theta\) above the horizontal:
• Horizontal component: \(u_x = u \cos\theta\)
• Vertical component: \(u_y = u \sin\theta\)
Standard Problem-Solving Steps for Projectiles:
1. Split the motion into two distinct columns: Horizontal and Vertical.
2. Resolve the initial launch velocity into \(u_x = u \cos\theta\) and \(u_y = u \sin\theta\).
3. Use the vertical information with SUVAT to find the total time of flight \(t\).
4. Use the time \(t\) with \(\text{Distance} = u_x \times t\) to find the horizontal range.
Key Takeaway: Never mix horizontal and vertical values in the same equation—except for time \(t\), which links both dimensions together.
6. Road Safety & Stopping Distances
A vital real-world application of linear motion in the AS 1 specification is vehicle stopping distance.
Total Stopping Distance = Thinking Distance + Braking Distance
1. Thinking Distance
The distance travelled by the vehicle during the driver's reaction time (the time between seeing the hazard and applying the brakes).
Because speed is constant during reaction time:
\(\text{Thinking Distance} = \text{Speed} \times \text{Reaction Time}\)
Factors Increasing Thinking Distance:
• Higher vehicle speed.
• Driver fatigue / tiredness.
• Influence of alcohol or drugs.
• Distractions (e.g., mobile phone usage).
2. Braking Distance
The distance travelled by the vehicle while the brakes are being applied and the car is decelerating to a stop.
From SUVAT (\(v^2 = u^2 + 2as\) where \(v = 0\)):
\(s = \frac{u^2}{2a}\)
Notice that braking distance is proportional to the square of initial speed (\(s \propto u^2\))! If you double your speed, your braking distance quadruples!
Factors Increasing Braking Distance:
• Higher vehicle speed.
• Poor road conditions (wet, icy, gravel).
• Worn tyres or worn brakes (reduced friction).
• Greater vehicle mass (requires a larger braking force for the same deceleration).
Key Takeaway: Thinking distance depends purely on the driver's reaction time and speed; braking distance depends on vehicle physics, tyre grip, road surface, and speed squared.
Chapter Quick Review & Exam Checklist
Before sitting your exam, make sure you can confidently:
• State the difference between scalars and vectors, giving motion examples for each.
• Calculate velocity from the gradient of an \(s\)-\(t\) graph and acceleration from the gradient of a \(v\)-\(t\) graph.
• Calculate displacement by determining the area under a \(v\)-\(t\) graph.
• Select and apply the correct SUVAT equation for uniform acceleration.
• Explain the experimental method to determine \(g\) and how to extract \(g\) from a straight-line graph.
• Solve projectile problems by strictly separating horizontal (\(a = 0\)) and vertical (\(a = -9.81\text{ m s}^{-2}\)) components.
• Define thinking, braking, and stopping distances and identify the physical factors affecting each.