Year 11 PhysicsModule 2⏱ ~40 min5 MC · 3 Short AnswerLesson 5 of 15
Acceleration and Graphical Analysis
At the 2023 Kitzbuehel Hahnenkamm downhill race, French skier Cyprien Sarrazin reached 162 km/h (45 m/s). From his GPS v-t trace, analysts calculated his peak acceleration at 8.6 m/s² by reading the gradient of the steepest segment, a direct application of $a = \Delta v/\Delta t$ from a v-t graph. This is precisely the graphical analysis method NESA requires in your investigation reports.
Today's hook: At Kitzbuehel 2023, Cyprien Sarrazin hit 162 km/h. His GPS data produced a v-t graph. Analysts drew a best-fit line through the steepest section and calculated the gradient. What physical quantity does that gradient represent, and why use a graph instead of just using $F = ma$ directly?
0/5TASKS
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Warm up first
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
Worksheets
Practise this lesson
Graphical analysis and F=ma investigation worksheets.
A student pushes a trolley with a constant force and measures the acceleration. She then doubles the force and measures again. 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?
On a velocity-time graph of an object under constant net force, the gradient equals:
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Learning Intentions
Learning Intentions
goals
Know
Gradient of a v-t graph = acceleration
Area under v-t graph = displacement
Gradient of F vs a graph = mass
$F_{net} = ma$ derivable from the F vs a graph
What uniform acceleration looks like on a v-t graph
Understand
Why the gradient of F vs a graph gives mass
Why the graph passes through the origin
The difference between best-fit line and connecting dots
How to derive $F = ma$ experimentally from a graph
Can Do
Draw and label a v-t graph for constant acceleration
Calculate the gradient of an F vs a graph
Describe the graph: straight line through origin, gradient = mass
Write a NESA-standard result statement for the F vs a investigation
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Key Terms
Key Terms
vocab
Gradient (slope)the rise over run on a graph; for a v-t graph: gradient = $\Delta v/\Delta t = a$; for F vs a: gradient = $F/a = m$
Line of best fita straight line drawn through scattered data points that minimises the total distance from all points to the line; does not necessarily pass through any data point
Uniform accelerationconstant acceleration; on a v-t graph this gives a straight line with non-zero gradient
F vs a grapha graph where net force is on the y-axis and acceleration is on the x-axis; for a fixed mass, gives a straight line through the origin with gradient = mass
Misconceptions to fix
The gradient of a v-t graph equals displacement.The gradient of a v-t graph equals acceleration ($\Delta v/\Delta t$). The area under a v-t graph equals displacement.
The gradient of an F vs a graph equals acceleration.Since $F = ma$, rearranging gives $F/a = m$. The gradient of F vs a is mass, not acceleration.
Cross-lesson links: L05 is the investigative capstone of Phase 1. The v-t gradient you read here is the acceleration that drives $F_{net} = ma$ (L04); the area under the curve gives displacement used in $W = Fs$ (L06). The F vs a graphical method reappears in senior investigations across all modules. Mastering graph construction and gradient interpretation here pays dividends through Year 12.
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Velocity-Time Graphs
Core Content
01
Velocity-Time Graphs
+5 XP
A car accelerates from rest at a traffic light and reaches 60 km/h after 8 seconds. Plotted as velocity vs time, this is a straight upward line from (0, 0) to (8, 16.7). You can see two things directly from the graph: the steepness of the line shows how quickly it accelerated, and the area of the triangle underneath shows how far it travelled.
$$a = \text{gradient of v-t graph} = \frac{\Delta v}{\Delta t}$$$$\text{displacement} = \text{area under v-t graph}$$
Under constant net force, velocity changes uniformly. On a v-t graph this produces a straight line. The steeper the line, the greater the acceleration. A horizontal line means zero acceleration (constant velocity).
On a velocity-time graph: gradient = acceleration ($a = \Delta v / \Delta t$, units m s$^{-2}$); area under graph = displacement (m); a straight line indicates uniform (constant) acceleration; a horizontal line indicates zero acceleration (constant velocity).
Pause, copy the highlighted graph rules into your book before moving on.
True or false: On a velocity-time graph, a steeper positive gradient means a greater net force (assuming constant mass).
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Force vs Acceleration Graph, Deriving $F = ma$
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Force vs Acceleration Graph, Deriving $F = ma$
+5 XP
We just saw that v-t gradient gives acceleration. That raises a question: how can you experimentally confirm the relationship between force and acceleration, and what does the graph look like? This card answers it → plotting F vs a for a fixed mass gives a straight line through the origin with gradient = mass.
When you plot net force against acceleration for a fixed mass, you always get a straight line through the origin. The gradient of that line is the mass, this is the experimental derivation of Newton's Second Law.
$$F = ma \quad \Rightarrow \quad \text{gradient of F vs a} = m$$
The graph passes through the origin because: when the net force is zero, the acceleration is zero (Newton's First Law). No offset, no constant.
A graph that does NOT pass through the origin signals a systematic error in the experiment, perhaps unmeasured friction or a miscalibrated force sensor.
Data Analysis ExampleF vs a Investigation
A 5 kg trolley was pulled with varying forces. Data collected:
Net Force (N)
Acceleration (m/s²)
F/a ratio
4.0
0.80
5.0 kg
8.0
1.61
5.0 kg
12.0
2.38
5.0 kg
16.0
3.19
5.0 kg
20.0
4.05
4.9 kg
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Plot F (y-axis) vs a (x-axis). Draw a line of best fit through the origin.
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Gradient = $\Delta F / \Delta a = (20.0 - 0) / (4.05 - 0) \approx 4.9\text{ kg}$
Use the full extent of the best-fit line, not individual data points, to calculate gradient.
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Conclusion: gradient $\approx 5.0\text{ kg}$, consistent with the known mass of the trolley.
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NESA statement: "The force applied to the trolley is directly proportional to its acceleration. The gradient of the F vs a graph equals the mass of the trolley (5.0 kg), consistent with Newton's Second Law $F_{net} = ma$."
Read axes before calculating: on $v$ against $t$, divide velocity change by time change for acceleration and use area for displacement. On $F$ against $a$, divide force change by acceleration change; the units reduce to kilograms, so the gradient is mass.
F vs a graph for fixed mass: straight line through origin with gradient = mass ($m = \Delta F/\Delta a$, units kg); a non-zero y-intercept indicates a systematic error (e.g. unmeasured friction or sensor offset); NESA requires: "F is directly proportional to a; gradient equals mass."
Pause, copy the highlighted graph interpretation and NESA statement into your book before moving on.
A student plots F (y-axis) vs a (x-axis) for a trolley and finds a gradient of 3.5 kg. She then plots the same data with F on the x-axis and a on the y-axis. The gradient of this second graph is:
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The Four-Step Graphing Scaffold
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The Four-Step Graphing Scaffold
+5 XP
We just saw that a well-constructed F vs a graph reveals the mass as its gradient. That raises a question: what specific steps does NESA reward in graph construction to maximise marks? This card answers it → axes with labels and units, all points plotted, best-fit straight line, gradient calculated from the line.
NESA rewards systematic graph construction. Follow these four steps every time.
Set up axes: Label each axis with quantity AND unit. Choose a scale that uses at least 2/3 of the grid space. Mark evenly spaced values.
Plot data points: Use a clear mark (cross or dot with circle). Plot ALL data points, don't skip outliers.
Draw best-fit line: A single straight line (not a curve) with roughly equal numbers of points above and below. Does NOT need to pass through any data point. Extend to the axes if appropriate.
Calculate gradient: Use two widely-spaced points on the line (not data points). Show working: $m = \Delta y/\Delta x$. Include units.
Four-step graphing scaffold: (1) label axes with quantity and unit, use 2/3 of grid; (2) plot ALL data points including outliers; (3) draw a single best-fit straight line with roughly equal points above and below; (4) calculate gradient using two widely-spaced points ON the line, not data points.
Pause, copy the four steps into your book before moving on.
When calculating the gradient of a best-fit line, you should use ______ and NOT individual ______.
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Deriving the Kinematic (SUVAT) Equations from the Graph
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Deriving the Kinematic (SUVAT) Equations from the Graph
+5 XP
We just saw the four-step graphing scaffold for constructing a NESA-standard F vs a graph. That raises a question: the kinematic (SUVAT) equations, $v = u + at$, $s = ut + \frac{1}{2}at^2$, and $v^2 = u^2 + 2as$, are usually handed to you as ready-made formulae, but where do they actually come from? This card answers it → every SUVAT equation follows directly from the gradient and the area of a v-t graph for uniform acceleration.
Consider an object with uniform acceleration $a$, starting at velocity $u$ and reaching velocity $v$ after time $t$. On a v-t graph this is a single straight line from the point $(0, u)$ to the point $(t, v)$. Every SUVAT equation is hidden inside this one line, once from its gradient, once from its area.
DerivationFrom Gradient and Area
Straight-line v-t graph from $(0, u)$ to $(t, v)$, uniform acceleration $a$.
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From the gradient: $a = \dfrac{\Delta v}{\Delta t} = \dfrac{v - u}{t - 0} = \dfrac{v-u}{t}$
This is exactly the gradient rule from Card 01, rise over run on the v-t graph.
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Rearranging: $at = v - u \;\;\Rightarrow\;\; v = u + at$
The first SUVAT equation, derived purely from the gradient definition.
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From the area: split the region under the line into a rectangle (height $u$, width $t$) plus a triangle on top (base $t$, height $v-u = at$).
Displacement = area under a v-t graph, established in Card 01.
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$s = \underbrace{ut}_{\text{rectangle}} + \underbrace{\tfrac{1}{2}t(at)}_{\text{triangle}} = ut + \tfrac{1}{2}at^2$
The second SUVAT equation, derived purely from the area of the graph.
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Eliminate $t$ algebraically: from Step 2, $t = (v-u)/a$. Substitute into Step 4: $s = u\left(\dfrac{v-u}{a}\right) + \tfrac{1}{2}a\left(\dfrac{v-u}{a}\right)^2$, which simplifies to $v^2 = u^2 + 2as$.
The third SUVAT equation needs no further graph reading, it is a purely algebraic combination of the first two.
All three SUVAT equations come from the same v-t graph: $v = u + at$ from the gradient; $s = ut + \frac{1}{2}at^2$ from the area (rectangle + triangle); $v^2 = u^2 + 2as$ from algebraically eliminating $t$ between the first two. None of the three requires any information beyond the straight-line v-t graph for uniform acceleration.
Pause, copy the highlighted derivation summary into your book before moving on.
The kinematic equation $s = ut + \frac{1}{2}at^2$ comes from which feature of a v-t graph?
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Graph Construction and Analysis
Activities
Graph Construction and Analysis
ApplyBand 3
Using the data table from the Worked Example above, construct a fully labelled F vs a graph. Draw a best-fit line and calculate the gradient. Show all working.
Error Spot, Common Graph Mistakes
EvaluateBand 6
A student's F vs a graph has a y-intercept of 2.5 N (not through the origin) and a gradient of 5.2 kg. Identify (a) what the non-zero intercept suggests about the experiment, and (b) whether the mass reading is reliable despite the error.
✓
Show what you have learned
Quick recall, Graphical Analysis
+5 XP
Pick your answer, then rate your confidence.
Short Answer, 10 marks
+5 XP
UnderstandBand 3(3 marks) 1. A student plots a v-t graph for a trolley under constant force. The line starts at the origin and has a positive gradient. Describe what the gradient and the area under the graph represent, including units.
ApplyBand 4(3 marks) 2. A student investigates Newton's Second Law using a 4 kg trolley. She plots F (N) on the y-axis and a (m/s²) on the x-axis. The best-fit line passes through (0, 0) and (8.0, 2.0). Calculate the gradient and explain what it represents.
EvaluateBand 5(4 marks) 3. A student's F vs a graph has a gradient of 5.1 kg but a y-intercept of +1.8 N. The student claims: "My result proves $F = ma$ because the gradient equals the mass." Evaluate this claim and suggest what the non-zero y-intercept indicates about the experiment's validity.
Show all answers
Short Answer, Model Answers
Q1 (3 marks): The gradient of the v-t graph represents acceleration, the rate of change of velocity, measured in m/s². A steeper gradient means greater acceleration. The area under the v-t graph represents displacement, how far the trolley has moved, measured in metres. For a uniform (constant) acceleration, the area is a triangle: $\text{displacement} = \frac{1}{2} \times t \times v$.
Q2 (3 marks): Gradient $= \Delta F/\Delta a = (8.0 - 0)/(2.0 - 0) = 4.0\text{ kg}$. The gradient represents the mass of the trolley. From $F_{net} = ma$, rearranging gives $m = F/a$. This is exactly what the gradient calculates, it equals 4.0 kg, consistent with the known mass of the trolley.
Q3 (4 marks): The student's claim is partially correct: the gradient of 5.1 kg is consistent with Newton's Second Law and does represent mass. However, $F = ma$ requires the graph to pass through the origin, when net force is zero, acceleration must also be zero. A y-intercept of +1.8 N suggests a systematic error: the force sensor may have a non-zero offset (e.g., it reads 1.8 N even with no force), or there is unmeasured friction in the system acting as a constant force. This systematic error means the investigation's validity is compromised, the student has not controlled all variables.
✓
Retrieve, reflect and finish
Check what actually stuck
Boss Battle, Checkpoint Quiz 1
boss
You have completed Phase 1 (Forces). This checkpoint covers L01-L05. Beat it to unlock Phase 2.
At Kitzbuehel 2023, Cyprien Sarrazin's peak acceleration of 8.6 m/s² was calculated from the gradient of the steepest v-t segment, not by measuring force directly. The F vs a approach works identically: plot experimental data, draw a best-fit line, and read the gradient to extract mass. This is Newton's Second Law derived from experimental data rather than assumed from theory, exactly what NESA rewards in investigation reports.
