A-Level · Physics · AQA · Mark scheme decoded
AQA A-Level Physics: Displacement, Speed, Velocity, and Acceleration — mark scheme explained
The short answer
In this section, we will explore the fundamental concepts of displacement, speed, velocity, and acceleration. These concepts are crucial for understanding motion in a straight line and form the basis for more advanced topics in mechanics.
The question
A car accelerates uniformly from rest to a speed of 20 m/s in 5 seconds. Calculate the acceleration and the distance traveled during this time.
[Paraphrased for study — not reproduced from any exam paper.]
Mark scheme, decoded
How the examiner actually awards the marks on this topic.
Gradora's own decode of the marking approach — not the exam board's published mark scheme.
How marks are awarded
For calculation questions, marks are typically awarded for correct substitution of values into the appropriate formula, showing working, and providing the final answer with units. For conceptual questions, marks are given for clear and accurate explanations.
What the command words demand
- Calculate
- Perform a mathematical operation to find a numerical answer.
- Determine
- Find a value or quantity using given data and appropriate formulas.
- Explain
- Provide a clear and detailed account of how something works or why something happens.
- Identify
- Recognize and name specific quantities, errors, or methods.
- Sketch
- Draw a graph or diagram to represent motion or relationships between variables.
Model answer
A full-mark response to the question above, worked through step by step.
Timing: Allocate approximately 1 minute per mark. For a 6-mark question, spend about 6 minutes to ensure you have enough time to show all necessary steps and provide a well-structured answer.
- 1. Identify the given values: initial velocity (u) = 0 m/s, final velocity (v) = 20 m/s, time (t) = 5 s.0 marks
- 2. Use the formula for acceleration: a = (v - u) / t1 mark
- 3. Substitute the values: a = (20 - 0) / 5 = 4 m/s 22 marks
- 4. Use the formula for displacement: s = ut + (1/2)at 21 mark
- 5. Substitute the values: s = 0 × 5 + (1/2) × 4 × 5 2 = 50 m2 marks
Final answer: The acceleration is 4 m/s 2 and the distance traveled is 50 m.
Work through every step correctly and you earn all 6 marks.
Another worked example
A ball is thrown vertically upwards with an initial velocity of 15 m/s. Calculate the maximum height it reaches and the time taken to reach this height.
- 1. Identify the given values: initial velocity (u) = 15 m/s, final velocity at maximum height (v) = 0 m/s, acceleration due to gravity (a) = -9.81 m/s 2 (negative because it acts downwards).1 mark
- 2. Use the formula for final velocity: v 2 = u 2 + 2as0 marks
- 3. Substitute the values: 0 2 = 15 2 + 2 × (-9.81) × s1 mark
- 4. Solve for s: 0 = 225 - 19.62s → 19.62s = 225 → s = 225 / 19.62 ≈ 11.47 m1 mark
- 5. Use the formula for time: v = u + at0 marks
- 6. Substitute the values: 0 = 15 - 9.81t1 mark
- 7. Solve for t: 9.81t = 15 → t = 15 / 9.81 ≈ 1.53 s2 marks
Final answer: The maximum height is approximately 11.47 m and the time taken to reach this height is approximately 1.53 s.
Work through every step correctly and you earn all 6 marks.
Common mistakes
Confusing displacement with distance
Why it happens: Students often mix up the scalar quantity (distance) with the vector quantity (displacement). Distance is the total length of the path traveled, while displacement is the shortest straight-line distance from the initial to the final position.
Fix: Always check if the question asks for distance or displacement. Displacement can be positive, negative, or zero, depending on the direction of movement relative to a reference point.
Using average velocity instead of instantaneous velocity
Why it happens: Students sometimes use the formula for average velocity (v avg = s / t) when they should be using the formula for instantaneous velocity (v = ds / dt).
Fix: Read the question carefully to determine if it asks for average or instantaneous velocity. Instantaneous velocity is the velocity at a specific moment in time, while average velocity is over an interval.
Forgetting to include units in calculations
Why it happens: Students often forget to include units in their final answers, which can lead to marks being deducted for incorrect or missing units.
Fix: Always write down the units for each quantity used in your calculations and ensure that the final answer has the correct units. For example, velocity should be in m/s, acceleration in m/s 2 , and displacement in meters.
Incorrectly interpreting the area under a velocity-time graph
Why it happens: Students sometimes misinterpret the area under a velocity-time graph as representing acceleration instead of displacement.
Fix: Remember that the area under a velocity-time graph represents the displacement. The gradient (slope) of the graph gives the acceleration.
Using the wrong formula for uniform acceleration
Why it happens: Students may use the incorrect equation from the set of equations for uniform acceleration, leading to incorrect answers.
Fix: Memorize and understand the four main equations for uniform acceleration: v = u + at, s = (u + v)t / 2, s = ut + (1/2)at 2 , and v 2 = u 2 + 2as. Choose the appropriate equation based on the given information in the question.
Neglecting the direction of acceleration
Why it happens: Students often forget that acceleration is a vector quantity and can be positive or negative, depending on the direction of motion.
Fix: Always consider the direction of acceleration. For example, in free fall, the acceleration due to gravity (g) acts downwards and is therefore negative when using upward as the positive direction.
Incorrectly identifying random and systematic errors in practical experiments
Why it happens: Students may not clearly distinguish between random and systematic errors, leading to incorrect suggestions for improving the experiment.
Fix: Understand that random errors can be reduced by taking multiple measurements and calculating an average, while systematic errors require calibration of equipment or a consistent method. Identify potential sources of error in the experiment and suggest specific ways to minimize them.
Where the marks go
The question types you’ll meet on this topic and the marks each one carries — so you know what to expect and where to focus.
| Question type | What you’re asked to do | Marks |
|---|---|---|
| Calculate Acceleration And Distance | Find the car's acceleration and distance travelled using the kinematics equations. | 6 |
| Kinematics Calculation | Use suvat equations to find maximum height and time for a vertically thrown ball. | 6 |
| Calculate Deceleration And Distance | Use kinematics equations to find deceleration and braking distance from given speed and time. | 6 |
| Kinematics Calculation | Use equations of motion to find fall time and impact velocity of a dropped ball. | 6 |
| Total across these question types | 24 | |
Question types and mark tariffs are Gradora’s guidance based on how this topic is typically examined — not the board’s official paper structure.