AQA AS Paper 1 2023 June — Question 15 4 marks

Exam BoardAQA
ModuleAS Paper 1 (AS Paper 1)
Year2023
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConstant acceleration (SUVAT)
TypeRead and interpret velocity-time graph
DifficultyEasy -1.2 This is a straightforward mechanics question requiring only basic graph interpretation: (a) calculating acceleration as gradient (change in velocity ÷ time) and (b) finding displacement as area under a velocity-time graph using simple geometric shapes (trapezium/triangle). Both parts are 'show that' questions with answers given, requiring minimal problem-solving and testing only routine application of standard AS-level mechanics formulas.
Spec3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area
A particle is moving in a straight line such that its velocity, \(v \text{ m s}^{-1}\), changes with respect to time, \(t\) seconds, as shown in the graph below. \includegraphics{figure_15}
  1. Show that the acceleration of the particle over the first 4 seconds is \(3.5 \text{ m s}^{-2}\) [1 mark]
  2. The particle is initially at a fixed point \(P\) Show that the displacement of the particle from \(P\), when \(t = 9\), is 62 metres. [3 marks]
Question 15:
AnswerMarks
15(a)Uses gradient for 0  t  4 to
show given acceleration value.
AnswerMarks Guidance
AG1.1b B1
a = = 3.5
4
AnswerMarks Guidance
Subtotal1
QMarking instructions AO
AnswerMarks
15(b)Selects a suitable method for
calculating the displacement by
working out an appropriate area.
or
Uses a constant acceleration
equation and substitutes
AnswerMarks Guidance
appropriate values.3.1a M1
s = 12 m
Since displacement = area
When t = 9
s = 12 + (9 – 4)10 = 62 m
Obtains a displacement of 12
when t = 4
or
− 1 6
Obtains a displacement of
7
8
when t =
7
16
Allow area of triangle =
7
or
100
Obtains a displacement of
7
8
from t = to t = 4
7
or
4 5 0
Obtains a displacement of
7
8
from t = to t = 9
7
AnswerMarks Guidance
OE1.1b A1
Completes reasoned argument
with fully correct working to
show the given displacement.
Do not award if decimal values
are repeatedly used throughout
to only one decimal place.
AnswerMarks Guidance
AG2.1 R1
Subtotal3
Question 15 Total4
QMarking instructions AO 
Question 15:
--- 15(a) ---
15(a) | Uses gradient for 0  t  4 to
show given acceleration value.
AG | 1.1b | B1 | 1 0 − − 4
a = = 3.5
4
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 15(b) ---
15(b) | Selects a suitable method for
calculating the displacement by
working out an appropriate area.
or
Uses a constant acceleration
equation and substitutes
appropriate values. | 3.1a | M1 | 100 = 16 + 7s
s = 12 m
Since displacement = area
When t = 9
s = 12 + (9 – 4)10 = 62 m
Obtains a displacement of 12
when t = 4
or
− 1 6
Obtains a displacement of
7
8
when t =
7
16
Allow area of triangle =
7
or
100
Obtains a displacement of
7
8
from t = to t = 4
7
or
4 5 0
Obtains a displacement of
7
8
from t = to t = 9
7
OE | 1.1b | A1
Completes reasoned argument
with fully correct working to
show the given displacement.
Do not award if decimal values
are repeatedly used throughout
to only one decimal place.
AG | 2.1 | R1
Subtotal | 3
Question 15 Total | 4
Q | Marking instructions | AO | Marks | Typical solution
A particle is moving in a straight line such that its velocity, $v \text{ m s}^{-1}$, changes with respect to time, $t$ seconds, as shown in the graph below.

\includegraphics{figure_15}

\begin{enumerate}[label=(\alph*)]
\item Show that the acceleration of the particle over the first 4 seconds is $3.5 \text{ m s}^{-2}$
[1 mark]

\item The particle is initially at a fixed point $P$

Show that the displacement of the particle from $P$, when $t = 9$, is 62 metres.
[3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA AS Paper 1 2023 Q15 [4]}}
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