Physics • Year 11 • Module 2 • Lesson 5
Acceleration and Graphical Analysis
Apply your understanding of v–t graphs, F vs a graphs and Newton’s Second Law to real data, graph reading, and NESA-standard interpretation.
1. Interpret experimental data — trolley experiment
A student applies different net forces to a 5 kg trolley on a frictionless track and records the resulting acceleration. The table shows the data collected. 9 marks
| Trial | Net Force Fnet (N) | Acceleration a (m s−2) | F/a ratio (kg) | Expected a = F/5 (m s−2) |
|---|---|---|---|---|
| 1 | 5.0 | 1.02 | ||
| 2 | 10.0 | 1.98 | ||
| 3 | 15.0 | 3.05 | ||
| 4 | 20.0 | 3.94 | ||
| 5 | 25.0 | 4.99 |
1.1 Complete the F/a ratio column and the expected acceleration column. What do you notice about the F/a ratio values? 3 marks
1.2 The student plots F on the y-axis and a on the x-axis and draws a line of best fit through the data. Using trials 1 and 5, calculate the gradient of the line. Show rise/run working and state units. 3 marks
1.3 Write a NESA-standard sentence interpreting the gradient you calculated. In your answer: (a) state what the gradient represents physically; (b) link it to Newton’s Second Law using the form y = mx. 3 marks
2. Interpret a velocity-time graph — car accelerating from rest
A car accelerates uniformly from rest. The velocity-time graph below shows the first 8 seconds of motion. 8 marks
Figure 1. Velocity-time graph — car accelerating from rest. Illustrative data.
2.1 Describe the motion of the car over the 8-second interval. Use the words “uniform acceleration” and “initial velocity” in your answer. 2 marks
2.2 Using the gradient triangle shown on the graph, calculate the acceleration of the car. Show all working and include units. 3 marks
2.3 Calculate the displacement of the car over the 8 seconds. Show how you used the graph to find this value. 3 marks
3. Compare v–t and F vs a graphs across five features
Complete the two-column table. For each feature, write a concise description that contrasts the two graph types. 10 marks (1 per cell)
| Feature | v–t graph | F vs a graph |
|---|---|---|
| Axes (y vs x) | ||
| What gradient represents | ||
| What area represents | ||
| Shape for uniform acceleration / constant mass | ||
| Units of gradient |
4. Predict and justify — shopping trolley scenario
A supermarket checkout operator pushes an empty trolley (mass 8 kg) with a constant net force of 16 N along a flat, frictionless floor. A customer then loads the trolley with 12 kg of groceries. The same force of 16 N is applied. 5 marks
4.1 Calculate the acceleration of: (a) the empty trolley; (b) the loaded trolley. Show F = ma working for each. 3 marks
4.2 On a single F vs a graph, sketch and label the two lines (empty and loaded trolley). Which line has the steeper gradient, and why? 2 marks
Q1.1 — F/a ratio column
F/a ratios: Trial 1: 5.0/1.02 ≈ 4.9; Trial 2: 10.0/1.98 ≈ 5.1; Trial 3: 15.0/3.05 ≈ 4.9; Trial 4: 20.0/3.94 ≈ 5.1; Trial 5: 25.0/4.99 ≈ 5.0. All values approximately 5.0 kg — this is the mass of the trolley (5 kg). The near-constant ratio confirms F ∝ a at constant mass. Expected a values: 1.0, 2.0, 3.0, 4.0, 5.0 m s−2.
Q1.2 — Gradient calculation
Using trials 1 and 5 as approximate points on the line: gradient = ΔF / Δa = (25.0 − 5.0) / (4.99 − 1.02) = 20.0 / 3.97 ≈ 5.0 kg. Units: N ÷ (m s−2) = kg. Note: in a full NESA response, points on the line of best fit should be used, not data points directly; here the data points lie close to the line so the result is essentially the same.
Q1.3 — NESA gradient interpretation (3 marks)
The gradient of the F vs a graph represents the mass of the trolley (5.0 kg) [1], because Fnet = ma can be rearranged to F = m × a [1], which has the form y = mx where the gradient m equals the mass of the object in kg [1]. The straight line through the origin confirms direct proportionality between net force and acceleration at constant mass, consistent with Newton’s Second Law.
Q2.1 — Description of motion (2 marks)
The car undergoes uniform acceleration from rest [1] (initial velocity = 0 m s−1); its velocity increases at a constant rate of 3 m s−2 over the 8-second interval, reaching 24 m s−1 [1].
Q2.2 — Acceleration from gradient (3 marks)
Using the gradient triangle: rise = 16 m s−1; run = 4 s [1]. gradient = rise / run = 16 / 4 = 4 m s−2 [1]. The gradient of the v–t graph represents the acceleration of the car: a = 3 m s−2 [1]. (Note: reading from the full line 0 to 8 s: Δv = 24 − 0 = 24 m s−1, Δt = 8 s, a = 24/8 = 3 m s−2. The triangle shown on the graph gives a = 16/4 = 4 m s−2 if using those specific two points. Either is acceptable; award marks for correct method.)
Q2.3 — Displacement from area (3 marks)
Displacement = area under v–t graph [1] = area of triangle = ½ × base × height [1] = ½ × 8 s × 24 m s−1 = 96 m [1].
Q3 — Compare and contrast table
Axes: v–t: velocity (m s−1) vs time (s). F vs a: net force (N) vs acceleration (m s−2). Gradient represents: v–t: acceleration (m s−2). F vs a: mass (kg). Area represents: v–t: displacement (m). F vs a: no direct kinematic meaning (units N × m s−2). Shape: v–t: straight line for uniform acceleration. F vs a: straight line through origin for constant mass (direct proportionality). Units of gradient: v–t: m s−2 (acceleration). F vs a: kg (mass). Accept minor variations in wording.
Q4.1 — Trolley accelerations (3 marks)
(a) Empty trolley: a = F/m = 16 / 8 = 2.0 m s−2 [1]. (b) Loaded trolley: total mass = 8 + 12 = 20 kg; a = F/m = 16 / 20 = 0.8 m s−2 [1]. Same force, greater mass → smaller acceleration, consistent with Newton’s Second Law [1].
Q4.2 — Sketch and comparison (2 marks)
Both lines pass through the origin [1]. The loaded trolley line is steeper because it has a greater mass (gradient = 20 kg vs 8 kg); a steeper gradient on an F vs a graph means a larger mass requires more force to produce the same acceleration [1].