AQA AS Paper 1 Specimen — Question 15 5 marks

Exam BoardAQA
ModuleAS Paper 1 (AS Paper 1)
SessionSpecimen
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeMulti-stage motion with velocity-time graph given
DifficultyModerate -0.8 Part (a) is a direct gradient calculation from a graph (1 mark). Part (b) requires calculating area under a speed-time graph to find distance, then solving for T—a standard kinematics application with straightforward arithmetic across 4 marks. Both parts are routine AS-level mechanics with no problem-solving insight required.
Spec3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae

The graph shows how the speed of a cyclist varies during a timed section of length 120 metres along a straight track. \includegraphics{figure_15}
  1. Find the acceleration of the cyclist during the first 10 seconds. [1 mark]
  2. After the first 15 seconds, the cyclist travels at a constant speed of 5 m s⁻¹ for a further \(T\) seconds to complete the 120-metre section. Calculate the value of \(T\). [4 marks]

Question 15:

AnswerMarks Guidance
15(a)Finds correct acceleration AO1.1b
(b)Identifies 5T as the distance
travelled after the first 15 secondsAO3.4 B1
Distance in first 15 secs =
1 1
× (3 + 8) × 10 + × (8 + 5) × 5
2 2
= 55 + 32.5 = 87.5
5T + 87.5 = 120
So T = 6.5
Uses the information given to form
an equation to find T
(award mark for either trapezium
expression separate, totalled or
AnswerMarks Guidance
implied)AO3.1b M1
Correctly calculates the distance
AnswerMarks Guidance
for the first 15 secsAO1.1b A1
Deduces the values of T from the
AnswerMarks Guidance
mathematical models appliedAO2.2a A1
Total5
QMarking Instructions AO
Question 15:
--- 15(a) ---
15(a) | Finds correct acceleration | AO1.1b | B1 | 0.5 m s2
(b) | Identifies 5T as the distance
travelled after the first 15 seconds | AO3.4 | B1 | Distance at constant speed = 5T
Distance in first 15 secs =
1 1
× (3 + 8) × 10 + × (8 + 5) × 5
2 2
= 55 + 32.5 = 87.5
5T + 87.5 = 120
So T = 6.5
Uses the information given to form
an equation to find T
(award mark for either trapezium
expression separate, totalled or
implied) | AO3.1b | M1
Correctly calculates the distance
for the first 15 secs | AO1.1b | A1
Deduces the values of T from the
mathematical models applied | AO2.2a | A1
Total | 5
Q | Marking Instructions | AO | Marks | Typical Solution
The graph shows how the speed of a cyclist varies during a timed section of length 120 metres along a straight track.

\includegraphics{figure_15}

\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of the cyclist during the first 10 seconds. [1 mark]

\item After the first 15 seconds, the cyclist travels at a constant speed of 5 m s⁻¹ for a further $T$ seconds to complete the 120-metre section.

Calculate the value of $T$. [4 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA AS Paper 1  Q15 [5]}}