Question 7 - A-Level Maths

Edexcel AS Paper 2 2018 June — Question 7 7 marks

Exam Board	Edexcel
Module	AS Paper 2 (AS Paper 2)
Year	2018
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Constant acceleration (SUVAT)
Type	Multi-phase journey: find unknown speed or time
Difficulty	Moderate -0.3 This is a standard three-stage SUVAT problem requiring straightforward application of kinematic equations and interpretation of a velocity-time graph. While it involves multiple parts and algebraic manipulation to find the constant velocity time, all steps follow routine procedures with no novel problem-solving required, making it slightly easier than average.
Spec	3.02c Interpret kinematic graphs: gradient and area 3.02d Constant acceleration: SUVAT formulae

A train travels along a straight horizontal track between two stations, \(A\) and \(B\).

In a model of the motion, the train starts from rest at \(A\) and moves with constant acceleration \(0.3 \mathrm {~m} \mathrm {~s} ^ { - 2 }\) for 80 s .
The train then moves at constant velocity before it moves with a constant deceleration of \(0.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }\), coming to rest at \(B\).

For this model of the motion of the train between \(A\) and \(B\),
1. state the value of the constant velocity of the train,
2. state the time for which the train is decelerating,
3. sketch a velocity-time graph. The total distance between the two stations is 4800 m .
Using the model, find the total time taken by the train to travel from \(A\) to \(B\).
Suggest one improvement that could be made to the model of the motion of the train from \(A\) to \(B\) in order to make the model more realistic.

Show mark scheme Show mark scheme source

Question 7:

Part (a)(i):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
\(24 \text{ (m s}^{-1}\text{)}\)	B1	Must be stated, not just inserted on graph

Part (a)(ii):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
\(48 \text{ (s)}\)	B1	Allow \(-48\) changed to 48; must be stated, not just on graph

Part (a)(iii):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
Trapezium shape	B1	A trapezium starting at origin and ending on \(t\)-axis

Part (b):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
Equating area under graph to 4800 to give equation in one unknown	M1	Complete method using trapezium rule with correct structure or two triangles and rectangle
\(\frac{1}{2}(T+T+80+48)\times 24 = 4800\) OR \((\frac{1}{2}\times80\times24)+24T+(\frac{1}{2}\times48\times24)=4800\)	A1ft	Correct equation ft on their 24 and 48 (must be positive times); N.B. \(\frac{1}{2}(T+80+48)\times24=4800\) is M0
\(T=136\) so total time is \(264\text{ (s)}\)	A1	For 264 (s)

Part (c):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
Either: smooth change from acceleration to constant velocity or from constant velocity to deceleration. Or: train accelerating and/or decelerating at a variable rate	B1	Do not accept: air resistance, resistive forces, straightness/horizontality of track, friction, length/mass of train, not having constant velocity. B0 if incorrect extra included. Variable acceleration due to variable air resistance is B1

# Question 7:

## Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $24 \text{ (m s}^{-1}\text{)}$ | B1 | Must be stated, not just inserted on graph |

## Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $48 \text{ (s)}$ | B1 | Allow $-48$ changed to 48; must be stated, not just on graph |

## Part (a)(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Trapezium shape | B1 | A trapezium starting at origin and ending on $t$-axis |

## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Equating area under graph to 4800 to give equation in one unknown | M1 | Complete method using trapezium rule with correct structure or two triangles and rectangle |
| $\frac{1}{2}(T+T+80+48)\times 24 = 4800$ **OR** $(\frac{1}{2}\times80\times24)+24T+(\frac{1}{2}\times48\times24)=4800$ | A1ft | Correct equation ft on their 24 and 48 (must be positive times); N.B. $\frac{1}{2}(T+80+48)\times24=4800$ is M0 |
| $T=136$ so total time is $264\text{ (s)}$ | A1 | For 264 (s) |

## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Either: smooth change from acceleration to constant velocity or from constant velocity to deceleration. Or: train accelerating and/or decelerating at a variable rate | B1 | Do not accept: air resistance, resistive forces, straightness/horizontality of track, friction, length/mass of train, not having constant velocity. B0 if incorrect extra included. Variable acceleration due to **variable** air resistance is B1 |

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Show LaTeX source

\begin{enumerate}
  \item A train travels along a straight horizontal track between two stations, $A$ and $B$.
\end{enumerate}

In a model of the motion, the train starts from rest at $A$ and moves with constant acceleration $0.3 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ for 80 s .\\
The train then moves at constant velocity before it moves with a constant deceleration of $0.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$, coming to rest at $B$.\\
(a) For this model of the motion of the train between $A$ and $B$,\\
(i) state the value of the constant velocity of the train,\\
(ii) state the time for which the train is decelerating,\\
(iii) sketch a velocity-time graph.

The total distance between the two stations is 4800 m .\\
(b) Using the model, find the total time taken by the train to travel from $A$ to $B$.\\
(c) Suggest one improvement that could be made to the model of the motion of the train from $A$ to $B$ in order to make the model more realistic.

\hfill \mbox{\textit{Edexcel AS Paper 2 2018 Q7 [7]}}

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