OCR MEI M1 — Question 1 6 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeDistance from velocity-time graph
DifficultyModerate -0.8 This is a straightforward interpretation question requiring students to read values from a speed-time graph and recall basic relationships (area = distance, gradient = acceleration). All parts involve direct reading or simple calculations with no problem-solving or novel insight required, making it easier than average.
Spec3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area
1 Fig. 1 shows the speed-time graph of a runner during part of his training. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{569e7c0e-7c33-47c9-b986-8587ea239f0a-1_1068_1586_319_273} \captionsetup{labelformat=empty} \caption{Fig. 1}
\end{figure} For each of the following statements, say whether it is true or false. If it is false give a brief explanation.
(A) The graph shows that the runner finishes where he started.
(B) The runner's maximum speed is \(8 \mathrm {~ms} ^ { - 1 }\).
(C) At time 58 seconds, the runner is slowing down at a rate of \(1.6 \mathrm {~ms} ^ { - 2 }\).
(D) The runner travels 400 m altogether.
Question 1:
(i)
When \(t = 4\), \(s = \frac{1}{2} \times 4 \times 10\)
\(s = 20\)
B1
When \(t = 18\), \(s = \frac{1}{2} \times (18 + 12) \times 10\)
\(s = 150\)
M1
A1
[3]
Finding the area of the triangle or equivalent.
A complete method of finding the area of the trapezium or equivalent.
CAO
(ii)
Graph joining \((0,0)\), \((4,20)\) and \((18, 150)\)
B1
Allow FT for their \((4,20)\) and \((18, 150)\)
The graph goes through \((16, 140)\)
B1
Condone extension to \((20, 150)\) with a horizontal line.
Allow SC1 for the first two marks if there is a consistent displacement from a correct scale, eg plotting \((18,150)\) at \((19, 150)\)
Curves at both ends
B1
The sections from \(t = 0\) to \(t = 4\) and from \(t = 16\) to \(t = 18\) are both curves
[3]
Question 1:

(i)

When $t = 4$, $s = \frac{1}{2} \times 4 \times 10$

$s = 20$

B1

When $t = 18$, $s = \frac{1}{2} \times (18 + 12) \times 10$

$s = 150$

M1

A1

[3]

Finding the area of the triangle or equivalent.

A complete method of finding the area of the trapezium or equivalent.

CAO

(ii)

Graph joining $(0,0)$, $(4,20)$ and $(18, 150)$

B1

Allow FT for their $(4,20)$ and $(18, 150)$

The graph goes through $(16, 140)$

B1

Condone extension to $(20, 150)$ with a horizontal line.

Allow SC1 for the first two marks if there is a consistent displacement from a correct scale, eg plotting $(18,150)$ at $(19, 150)$

Curves at both ends

B1

The sections from $t = 0$ to $t = 4$ and from $t = 16$ to $t = 18$ are both curves

[3]
1 Fig. 1 shows the speed-time graph of a runner during part of his training.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{569e7c0e-7c33-47c9-b986-8587ea239f0a-1_1068_1586_319_273}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}

For each of the following statements, say whether it is true or false. If it is false give a brief explanation.\\
(A) The graph shows that the runner finishes where he started.\\
(B) The runner's maximum speed is $8 \mathrm {~ms} ^ { - 1 }$.\\
(C) At time 58 seconds, the runner is slowing down at a rate of $1.6 \mathrm {~ms} ^ { - 2 }$.\\
(D) The runner travels 400 m altogether.

\hfill \mbox{\textit{OCR MEI M1  Q1 [6]}}
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