Year 11 Physics Module 2: Dynamics 45 min Lesson 5 of 15

Acceleration and Graphical Analysis

A graph is not just a picture of data — it is a proof. The gradient of a force vs acceleration graph is a measurement of mass, and from that single line Newton's Second Law emerges.

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Think First

A student pushes a trolley with a constant force and measures how fast it accelerates. She repeats the experiment with the trolley carrying more and more mass. She plots force on the y-axis and acceleration on the x-axis. What shape will the graph be — and what physical quantity does the gradient represent?

Type your prediction below — you will revisit it at the end.

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Formula Reference — This Lesson

$F_{net} = ma$ | gradient $= \Delta y/\Delta x$

F vs a gradient = mass (kg) | y-intercept = 0 (through origin) v-t gradient = acceleration (m/s²) | area = displacement (m) a = Δv / Δt | v = u + at

Find m from $m = \Delta F/\Delta a$ (gradient of F vs a) | Find a from v-t $a = (v_2 - v_1)/(t_2 - t_1)$

F=ma
Formula Reference — Newton's Second Law and Graphical Analysis

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$F_{net} = ma$

Newton's Second Law

F_net = net force (N) | m = mass (kg) | a = acceleration (m/s²)

Graph form: y = mx — where F is y, a is x, and the gradient m is the mass. A straight line through the origin on an F vs a graph confirms direct proportionality.

Common trap: The gradient of an F vs a graph is mass — not acceleration. Students often confuse the variable m (gradient) with the quantity a. Always check units: N ÷ (m/s²) = kg.

$\text{gradient} = \Delta y/\Delta x$

Gradient of a straight line graph

Always calculated using two points on the line of best fit — never using data points directly.

For F vs a: gradient = ΔF / Δa = mass (kg)

Common trap: Using data points instead of points on the line of best fit gives the wrong gradient. Always draw the line first, then pick two widely-spaced points on the line.

$v = u + at$

Velocity-time relationship (uniform acceleration)

v = final velocity (m/s) | u = initial velocity (m/s) | a = acceleration (m/s²) | t = time (s)

Graph form: On a v-t graph, gradient = acceleration, area under line = displacement.

Common trap: Confusing gradient of v-t graph (acceleration) with gradient of s-t graph (velocity). Different graphs — different physical meanings for the gradient.

$a = \Delta v/\Delta t$

Acceleration from velocity-time graph

a = gradient of v-t graph (m/s²)

Applies when: acceleration is constant — straight line on v-t graph

Common trap: For non-uniform acceleration, draw a tangent at the point of interest — the gradient of the tangent gives instantaneous acceleration.

Know

What a v-t graph gradient and area represent
What an F vs a graph gradient represents
F_net = ma as derived from graphical data
What uniform acceleration looks like on graphs

Understand

Why gradient of F vs a graph = mass
Why a straight line through origin proves direct proportionality
How experimental data derives F_net = ma rather than assumes it
Why line of best fit is used — not data points — for gradient

Can Do

Draw and label a v-t graph for uniform acceleration
Calculate gradient of F vs a graph and interpret in physics terms
Describe what a straight line through origin means physically
Write a NESA-standard gradient interpretation sentence

Misconceptions to Fix

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Wrong: Vectors and scalars are just different ways of writing the same thing.

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Right: Vectors have magnitude and direction; scalars have magnitude only. They follow different mathematical rules.

📚 Core Content

Key Terms

graphnot just a picture of data — it is a proof

force vs acceleration grapha measurement of mass, and from that single line Newton's Second Law emerges

accelerationconstant — straight line on v-t graph

line of best fitused — not data points — for gradient

Vectors and scalarsjust different ways of writing the same thing

change of velocityconstant; and doubling the force doubles the rate of acceleration

01Qualitative Description of Accelerated Motion

Qualitative Description of Accelerated Motion

Acceleration is a change in velocity — if an object is speeding up, slowing down, or changing direction, it is accelerating.

A student in a trolley experiment applies a constant net force and observes the trolley accelerate uniformly. Qualitatively, we can describe this as: the velocity increases steadily over time; the rate of change of velocity is constant; and doubling the force doubles the rate of acceleration.

Observation	Qualitative description	What it tells us
Trolley speeds up steadily	Velocity increases uniformly with time	Acceleration is constant — uniform acceleration
Heavier trolley speeds up more slowly	Same force, larger mass → smaller acceleration	a inversely proportional to m (at constant F)
Harder push → faster acceleration	Larger force, same mass → larger acceleration	a directly proportional to F (at constant m)

Real-World Anchor A shopping trolley at Woolworths accelerates slowly when full and quickly when empty — same push, different mass. This is Newton's Second Law experienced daily. The key insight is that both relationships (a ∝ F and a ∝ 1/m) combine into a single equation: F_net = ma.

02Graphs and Vectors — Representing Accelerated Motion

Graphs and Vectors — Representing Accelerated Motion

A velocity-time graph turns invisible motion into a visible shape — and the shape contains more information than the numbers alone.

Velocity-Time Graph for Uniform Acceleration

For an object under constant net force (uniform acceleration), the v-t graph is a straight line. The gradient of this line equals the acceleration, and the area under the line equals displacement.

Vector Representation

Acceleration is a vector — it has both magnitude and direction. For an object accelerating in the positive direction, the acceleration vector points in the same direction as the net force. For a decelerating object, the acceleration vector points opposite to the velocity vector.

Vector Protocol — graphical problems

Step 1 — Define positive direction before reading any values from the graph

Step 2 — Identify gradient (direction and magnitude) from line of best fit

Step 3 — Interpret gradient in physics terms using NESA language before calculating

03Deriving F = ma

Deriving F_net = ma from Graphical Data

Newton didn't write F = ma on a whiteboard. Scientists measured it from data — and you can too.

In the trolley experiment, a constant force F is applied to a trolley of mass m. The acceleration a is measured for several different values of F (keeping m constant). The results are plotted as F on the y-axis and a on the x-axis.

The Four-Step Graphing Scaffold

Graph-to-Physics Interpretation — use every time

Draw labelled axes with variable name, symbol, and unit in brackets — e.g. "Net Force, F (N)"

Plot all data points and draw a line of best fit (straight or curved as appropriate)

Calculate gradient — use two widely-spaced points ON the line, not data points. Show rise/run working.

4
Interpret the gradient in physics terms using NESA language: "The gradient of the F vs a graph represents the mass of the object, as Fnet = ma can be rearranged to F = ma, which has the form y = mx where the gradient equals m."

Step 4 is where marks are lost. Calculating the gradient correctly earns 1 mark. Interpreting it in physics terms earns the second mark. A gradient of 10 kg/m·s⁻² is not an answer — "the gradient represents the mass of the trolley, which is 10 kg" is the answer.

Common Misconceptions

The gradient of a v-t graph is displacement.

The gradient of a v-t graph is acceleration (Δv/Δt). The area under a v-t graph is displacement. These are frequently confused — always state which property of the graph you are using.

You read the gradient from the data table, not the graph.

The gradient must be calculated from two points on the line of best fit — not from data points. Data points have experimental error; the line of best fit averages these out. Using data points gives an unreliable gradient.

A line of best fit must pass through the origin.

A line of best fit passes through the region of highest data density — it does not automatically pass through the origin unless theory predicts direct proportionality. If theory says F ∝ a, a line through the origin is expected — but it still has to be drawn to fit the data.

Interactive: Motion Graph Animator

Interactive: Kinematics and Acceleration (3D)

Interactive: v–t Graph

✏️ Worked Examples

Worked Example 1 Type 2 — Kinetic: Single Object

Problem Setup

Problem type: Type 2 — Kinetic Single Object. Constant net force, finding acceleration from F = ma.

Scenario: A 4 kg trolley is pushed with a constant net force of 12 N on a frictionless surface. Find the acceleration and describe the motion qualitatively and using vectors.

m = 4 kg
F_net = 12 N (forward)
Surface frictionless — no opposing force
Positive direction: forward

Solution

Positive direction: forward = positive. F_net = +12 N

Vector Protocol Step 1. Sign of F_net is positive because force acts in the positive direction.

a = F_net / m = 12 / 4 = 3 m/s² (forward)

Newton's Second Law rearranged. Positive acceleration confirms it acts in the positive (forward) direction.

Qualitative: velocity increases at a constant rate of 3 m/s every second

Uniform acceleration means equal velocity increases in equal time intervals. v-t graph is a straight line with gradient = 3.

Vector: acceleration vector → (forward, 3 m/s²), same direction as F_net

Newton's Second Law: direction of a always equals direction of F_net. Draw arrow pointing forward with label "a = 3 m/s²".

What would change if...

The surface were not frictionless — instead μ_k = 0.2. What would the new acceleration be? Draw the new FBD and recalculate F_net before applying F = ma.

Worked Example 2 Type 2 — Kinetic: Graphical Analysis

Problem Setup

Problem type: Type 2 — Graphical Analysis. Derive F = ma from an F vs a graph.

Scenario: An F vs a graph for a trolley produces a straight line through the origin. Two points on the line are (1.0 m/s², 8 N) and (4.0 m/s², 32 N). Find the mass of the trolley and write the relationship in words.

Point 1 on line: a = 1.0 m/s², F = 8 N
Point 2 on line: a = 4.0 m/s², F = 32 N
Graph passes through origin — direct proportionality

Solution

gradient = ΔF / Δa = (32 − 8) / (4.0 − 1.0) = 24 / 3.0 = 8 kg

Use the two points ON the line of best fit — not data points. Show full rise/run working. Units: N ÷ m/s² = kg.

mass of trolley = 8 kg

The gradient of an F vs a graph equals mass. This is the physical interpretation — not just a number.

Relationship (NESA language): The gradient of the F vs a graph represents the mass of the trolley. F_net = ma can be written as F = (8)a, confirming mass = 8 kg.

Step 4 of the graphing scaffold — always write the gradient interpretation in full. This earns the interpretation mark.

Straight line through origin confirms F ∝ a (direct proportionality at constant mass)

The shape of the graph is evidence for Newton's Second Law — not just an illustration of it.

What would change if...

A second trolley of mass 16 kg were used in the same experiment. Sketch what the new F vs a graph would look like alongside the first — steeper or shallower? At a = 2 m/s², what F would be required for each trolley?

Visual Break

Copy into your books

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v-t Graph Interpretation

Gradient of v-t graph = acceleration (m/s²)
Area under v-t graph = displacement (m)
Straight line = uniform (constant) acceleration
Horizontal line = constant velocity (a = 0)

F vs a Graph

Straight line through origin = F ∝ a (direct proportionality)
Gradient = ΔF / Δa = mass (kg)
Units check: N ÷ m/s² = kg
Steeper line = greater mass

Graphing Protocol

Step 1: Labelled axes (variable, symbol, unit)
Step 2: Plot points, draw line of best fit
Step 3: Gradient = rise/run from line (not data points)
Step 4: Interpret gradient in physics terms — NESA language

NESA Gradient Statement Template

"The gradient of the [F] vs [a] graph represents [mass]..."
"...because F = ma can be rearranged to F = m × a..."
"...which has the form y = mx, where gradient = m (mass)"
Always include units in the interpretation

🏃 Activities

Activity 01 — Pattern B

Data Table and Graph Analysis

Complete the table, plot the graph, calculate the gradient, and write a NESA-standard interpretation.

A student applies different net forces to a trolley and measures the resulting acceleration. Results are shown below.

Trial	Net Force F_net (N)	Acceleration a (m/s²)
1	4.0	0.80
2	8.0	1.62
3	12.0	2.40
4	16.0	3.18
5	20.0	4.05

Complete the F/a ratio column. What do you notice about these values?
Plot F (y-axis) vs a (x-axis) in your book. Draw a line of best fit.
Calculate the gradient using two points on your line of best fit. Show rise/run working.
Write a NESA-standard sentence interpreting the gradient in physics terms.
Does the line pass through the origin? What does this indicate about the relationship between F and a?

Type your gradient calculation and interpretation below. Draw the graph in your book.

Complete the table, draw the graph, and write your answers in your book.

Complete the table, graph, and analysis in your book

Saved

Activity 02 — Pattern B (Error-spot)

Find and Fix the Errors

A student's graphical analysis contains four deliberate errors. Identify each, explain why it is wrong, and write the correct version.

Student's working:

"I plotted my F vs a data and drew a line of best fit. To find the gradient I used the first and last data points: gradient = (20 − 4) / (4.05 − 0.80) = 16 / 3.25 = 4.9.

The gradient of the graph is 4.9 m/s² — this represents the acceleration of the trolley.

Since the graph is a straight line, this proves F = ma is correct.

The area under the F vs a graph equals the displacement of the trolley."

For each error: (a) state what is wrong, (b) explain why using physics reasoning, (c) write the correct version.

Type your error analysis below — aim to find all four.

Write your error analysis in your book — aim to find all four errors.

Write your error analysis in your book

Saved

✅ Check Your Understanding

Revisit Your Thinking

Earlier you were asked: What shape will the F vs a graph be — and what physical quantity does the gradient represent?

The full answer: the graph is a straight line through the origin, reflecting the direct proportionality between net force and acceleration when mass is constant. The gradient equals mass — specifically, gradient = ΔF/Δa with units of N ÷ m/s² = kg.

The deeper point: this is not just a calculation. The straight line through the origin is experimental evidence for Newton's Second Law. When students measure this in a trolley experiment, they are not verifying a formula — they are deriving it. The law emerges from the data.

Look back at your initial prediction. What did you get right? What is clearer now?

Annotate your prediction in your book with what you now understand differently.

Annotate your prediction in your book

Saved

Revisit Your Initial Thinking

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

Multiple Choice

5 random questions from a replayable lesson bank — feedback shown immediately

UnderstandBand 3

7. A v-t graph for a moving object shows a straight line with a positive gradient starting from the origin. Describe the motion of the object and state what the gradient and area under the line each represent. 3 MARKS

Answer in your book

Saved

AnalyseBand 5

8. An F vs a graph for a 5 kg trolley is expected to be a straight line through the origin. However the student's line has a y-intercept of +2 N. Suggest two possible reasons for this and explain what each would mean physically. 3 MARKS

Answer in your book

Saved

EvaluateBand 6

9. A student investigates Newton's Second Law by applying different net forces to a trolley and measuring the resulting acceleration. The F vs a graph produces a straight line through the origin with a gradient of 2.4 kg. (a) State the mass of the trolley. (b) The student repeats the experiment with an additional 1.2 kg mass placed on the trolley. Predict the gradient of the new F vs a graph. (c) On the same set of axes, sketch both lines and label them. (d) Write a conclusion in NESA-standard language that includes the relationship between gradient, mass, and Newton's Second Law. 4 MARKS

Answer in your book — include your sketch

Saved

Comprehensive Answers

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Activity 01 — Data Table and Graph

F/a ratios: Trial 1: 4.0/0.80 = 5.0 | Trial 2: 8.0/1.62 = 4.94 | Trial 3: 12.0/2.40 = 5.0 | Trial 4: 16.0/3.18 = 5.03 | Trial 5: 20.0/4.05 = 4.94. All approximately 5 — this is the mass of the trolley (5 kg).

Gradient: Using points on the line of best fit at a = 1.0 and a = 4.0: gradient ≈ (20 − 4) / (4.0 − 0.8) ≈ 5.0 kg.

NESA interpretation: The gradient of the F vs a graph represents the mass of the trolley (5 kg), as F_net = ma can be rearranged to F = m × a, which has the form y = mx where the gradient m equals mass.

Line through origin: Yes — confirming that F_net is directly proportional to a when mass is constant. This is consistent with Newton's Second Law.

Activity 02 — Four Errors

Error 1 — Used data points instead of line of best fit: The gradient must be calculated from two points on the line of best fit, not from data points. Data points include experimental error; the line of best fit minimises this error. Using data points gives an unreliable gradient.

Error 2 — Wrong units and interpretation (gradient called "m/s²"): The gradient of an F vs a graph has units of N ÷ m/s² = kg. It represents mass, not acceleration. The student stated "4.9 m/s²" — this is both wrong units and wrong physical interpretation.

Error 3 — Incorrect claim that straight line "proves" F = ma: A straight line through the origin is consistent with F ∝ a — it supports Newton's Second Law but does not prove it. Scientific language requires "consistent with" or "supports" rather than "proves".

Error 4 — Area under F vs a graph equals displacement: The area under a v-t graph equals displacement. The area under an F vs a graph has units of N × m/s² which equals W (watts, power per unit time) — it has no simple kinematic meaning. This is a confusion between graph types.

Multiple Choice

1. B — Gradient of v-t graph = Δv/Δt = acceleration. Area under v-t graph = displacement (common confusion).

2. C — Straight line through origin means y ∝ x, i.e. F ∝ a — Newton's Second Law at constant mass.

3. A — 5 kg. Gradient = ΔF/Δa = (30 − 10)/(6 − 2) = 20/4 = 5 kg.

4. D — Data points have random experimental error. Using them instead of the line of best fit introduces that error into the gradient. The error could be in either direction depending on which data points are high or low.

5. B — 9 N. a = Δv/Δt = (14 − 2)/4 = 3 m/s². F_net = ma = 3 × 3 = 9 N.

6. C — The 6 kg trolley requires more force to achieve the same acceleration (F = ma — larger m, same a → larger F). On an F vs a graph, this means a steeper line (larger gradient = larger mass).

Short Answer — Model Answers

Q7 (3 marks): The object is moving in the positive direction with uniform (constant) acceleration from rest. The velocity increases linearly with time — equal increases in velocity for equal time intervals. The gradient of the v-t graph represents the acceleration of the object (a = Δv/Δt, units m/s²). The area under the v-t graph represents the displacement of the object (units m).

Q8 (3 marks): Reason 1 — A constant friction force is present that was not subtracted from the applied force. If a constant friction force f acts on the trolley, then F_applied = F_net + f, shifting the line upward by f and producing a positive y-intercept equal to the friction force. Reason 2 — A systematic measurement error in the force sensor (zero offset). If the force sensor was not zeroed before the experiment, all force readings would be consistently too high by a fixed amount, producing a y-intercept equal to the offset error. In this case the gradient (mass) would still be correct, but the y-intercept would not be zero.

Q9 (4 marks):

(a) Mass of trolley = gradient = 2.4 kg

(b) New total mass = 2.4 + 1.2 = 3.6 kg. New gradient = 3.6 kg (steeper line)

(c) Two straight lines through the origin on the same axes: the 3.6 kg line is steeper than the 2.4 kg line. Both pass through the origin. Label each with its mass.

(d) The gradient of the F vs a graph represents the mass of the trolley, as F_net = ma can be rearranged to F = m × a, which has the form y = mx where the gradient equals m (mass, in kg). A straight line through the origin confirms that net force is directly proportional to acceleration when mass is constant, consistent with Newton's Second Law of Motion. The steeper gradient for the heavier trolley confirms that greater force is required to produce the same acceleration when mass is increased.

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