AQA Paper 2 Specimen — Question 14 7 marks

Exam BoardAQA
ModulePaper 2 (Paper 2)
SessionSpecimen
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeDistance from velocity-time graph
DifficultyModerate -0.3 This is a standard mechanics question testing basic kinematics concepts: distance as area under velocity-time graph, displacement vs distance, and F=ma. Part (a) requires calculating areas of simple shapes, (b) accounts for direction, (c) applies Newton's second law using gradient for acceleration, and (d) asks for a modeling critique. All parts are routine applications with no novel problem-solving required, making it slightly easier than average.
Spec3.01b Derived quantities and units3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.03c Newton's second law: F=ma one dimension
The graph below models the velocity of a small train as it moves on a straight track for 20 seconds. The front of the train is at the point \(A\) when \(t = 0\) The mass of the train is 800kg. \includegraphics{figure_14}
  1. Find the total distance travelled in the 20 seconds. [3 marks]
  2. Find the distance of the front of the train from the point \(A\) at the end of the 20 seconds. [1 mark]
  3. Find the maximum magnitude of the resultant force acting on the train. [2 marks]
  4. Explain why, in reality, the graph may not be an accurate model of the motion of the train. [1 mark]
Question 14:
AnswerMarks Guidance
14(a)Calculates two (or four)
appropriate distancesAO3.1b M1
s  (610)864 m
1
2
1
s  10210 m
2
2
s s 641074 m
1 2
OR
S = 6  8 = 48 m
1
1
S = 4816 m
2
2
1
S = 424 m
3
2
1
S = 626 m
4
2
S + S + S + S = 74 m
1 2 3 4
AnswerMarks Guidance
Obtains correct distancesAO1.1b A1
Obtains correct sum of ‘their’
AnswerMarks Guidance
distancesAO1.1b A1F
(b)Finds difference of ‘their’
distances from part (a)AO2.2a B1F
(c)Calculates magnitude of
accelerationAO3.1b M1
a  2
max
4
F 80021600 N
max
AnswerMarks Guidance
Obtains correct resultant forceAO1.1b A1
(d)Explains that abrupt changes
and straight lines in the graph
AnswerMarks Guidance
are unlikely in realityAO3.5b E1
result in abrupt changes I would
expect to see curves on the
graph
AnswerMarks Guidance
Total7
QMarking Instructions AO 
Question 14:
--- 14(a) ---
14(a) | Calculates two (or four)
appropriate distances | AO3.1b | M1 | 1
s  (610)864 m
1
2
1
s  10210 m
2
2
s s 641074 m
1 2
OR
S = 6  8 = 48 m
1
1
S = 4816 m
2
2
1
S = 424 m
3
2
1
S = 626 m
4
2
S + S + S + S = 74 m
1 2 3 4
Obtains correct distances | AO1.1b | A1
Obtains correct sum of ‘their’
distances | AO1.1b | A1F
(b) | Finds difference of ‘their’
distances from part (a) | AO2.2a | B1F | 64 – 10 = 54 m
(c) | Calculates magnitude of
acceleration | AO3.1b | M1 | 8
a  2
max
4
F 80021600 N
max
Obtains correct resultant force | AO1.1b | A1
(d) | Explains that abrupt changes
and straight lines in the graph
are unlikely in reality | AO3.5b | E1 | Change of velocity is unlikely to
result in abrupt changes I would
expect to see curves on the
graph
Total | 7
Q | Marking Instructions | AO | Marks | Typical Solution
The graph below models the velocity of a small train as it moves on a straight track for 20 seconds.

The front of the train is at the point $A$ when $t = 0$

The mass of the train is 800kg.

\includegraphics{figure_14}

\begin{enumerate}[label=(\alph*)]
\item Find the total distance travelled in the 20 seconds.
[3 marks]

\item Find the distance of the front of the train from the point $A$ at the end of the 20 seconds.
[1 mark]

\item Find the maximum magnitude of the resultant force acting on the train.
[2 marks]

\item Explain why, in reality, the graph may not be an accurate model of the motion of the train.
[1 mark]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 2  Q14 [7]}}
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