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.
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:
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
Core Content
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.
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).
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.
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.
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 |
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:
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 ______.
Activities
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.
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.
Pick your answer, then rate your confidence.
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.
You have completed Phase 1 (Forces). This checkpoint covers L01-L05. Beat it to unlock Phase 2.
⚔ Enter Checkpoint 1At 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.