AQA AS Paper 1 2021 June — Question 13 5 marks

Exam BoardAQA
ModuleAS Paper 1 (AS Paper 1)
Year2021
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeModel refinement or criticism
DifficultyEasy -1.2 Part (a) is a straightforward area-under-graph calculation requiring trapezium/triangle areas from a velocity-time graph—standard AS mechanics with no conceptual challenge. Part (b) tests basic understanding that real acceleration changes smoothly, requiring only a sketch with rounded corners rather than sharp vertices. Both parts are routine applications well below average A-level difficulty.
Spec3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area

A car, initially at rest, is driven along a straight horizontal road. The graph below is a simple model of how the car's velocity, \(v\) metres per second, changes with respect to time, \(t\) seconds. \includegraphics{figure_13}
  1. Find the displacement of the car when \(t = 45\) [3 marks]
  2. Shona says: "This model is too simple. It is unrealistic to assume that the car will instantaneously change its acceleration." On the axes below sketch a graph, for the first 10 seconds of the journey, which would represent a more realistic model. [2 marks] \includegraphics{figure_13b}

Question 13:

AnswerMarks Guidance
13(a)Finds any area either above or
below the axis3.1b M1
1
= 252
�2�(12)(10+32)
Area below = = 52
1
�2�(13)(8)
= 252 – 52 = 200 m
𝑟𝑟
Obtains correct values of area
above or area below the axis
AnswerMarks Guidance
PI1.1b A1
Obtains correct displacement
AnswerMarks Guidance
condone missing units1.1b A1
Subtotal3
QMarking instructions AO

AnswerMarks
13(b)Sketches a curve at the end =
0 correctly
AnswerMarks Guidance
𝑡𝑡3.5c B1
Sketches a curve at the end =
10 correctly
AnswerMarks Guidance
𝑡𝑡3.5c B1
Subtotal2
Question Total5
QMarking instructions AO
Question 13:
--- 13(a) ---
13(a) | Finds any area either above or
below the axis | 3.1b | M1 | Area above =
1
= 252
�2�(12)(10+32)
Area below = = 52
1
�2�(13)(8)
= 252 – 52 = 200 m
𝑟𝑟
Obtains correct values of area
above or area below the axis
PI | 1.1b | A1
Obtains correct displacement
condone missing units | 1.1b | A1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 13(b) ---
13(b) | Sketches a curve at the end =
0 correctly
𝑡𝑡 | 3.5c | B1
Sketches a curve at the end =
10 correctly
𝑡𝑡 | 3.5c | B1
Subtotal | 2
Question Total | 5
Q | Marking instructions | AO | Marks | Typical solution
A car, initially at rest, is driven along a straight horizontal road.

The graph below is a simple model of how the car's velocity, $v$ metres per second, changes with respect to time, $t$ seconds.

\includegraphics{figure_13}

\begin{enumerate}[label=(\alph*)]
\item Find the displacement of the car when $t = 45$
[3 marks]

\item Shona says:
"This model is too simple. It is unrealistic to assume that the car will instantaneously change its acceleration."

On the axes below sketch a graph, for the first 10 seconds of the journey, which would represent a more realistic model.
[2 marks]

\includegraphics{figure_13b}
\end{enumerate}

\hfill \mbox{\textit{AQA AS Paper 1 2021 Q13 [5]}}