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].