OCR MEI M1 2011 January — Question 1 8 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Year2011
SessionJanuary
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSUVAT in 2D & Gravity
TypeVertical motion: velocity-time graph
DifficultyModerate -0.8 This is a straightforward velocity-time graph question requiring basic SUVAT skills: completing a graph from given information, calculating acceleration from gradient (change in velocity over time), and finding displacement from area under the graph. All steps are routine applications of standard mechanics techniques with no problem-solving insight required.
Spec3.02a Kinematics language: position, displacement, velocity, acceleration3.02b Kinematic graphs: displacement-time and velocity-time

1 An object C is moving along a vertical straight line. Fig. 1 shows the velocity-time graph for part of its motion. Initially C is moving upwards at \(14 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and after 10 s it is moving downwards at \(6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e36ef805-beff-4125-b332-439ccb0d91c4-2_878_933_479_607} \captionsetup{labelformat=empty} \caption{Fig. 1}
\end{figure} C then moves as follows.
  • In the interval \(10 \leqslant t \leqslant 15\), the velocity of C is constant at \(6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) downwards.
  • In the interval \(15 \leqslant t \leqslant 20\), the velocity of C increases uniformly so that C has zero velocity at \(t = 20\).
    1. Complete the velocity-time graph for the motion of C in the time interval \(0 \leqslant t \leqslant 20\).
    2. Calculate the acceleration of C in the time interval \(0 < t < 10\).
    3. Calculate the displacement of C from \(t = 0\) to \(t = 20\).

1 An object C is moving along a vertical straight line. Fig. 1 shows the velocity-time graph for part of its motion. Initially C is moving upwards at $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and after 10 s it is moving downwards at $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{e36ef805-beff-4125-b332-439ccb0d91c4-2_878_933_479_607}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}

C then moves as follows.

\begin{itemize}
  \item In the interval $10 \leqslant t \leqslant 15$, the velocity of C is constant at $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ downwards.
  \item In the interval $15 \leqslant t \leqslant 20$, the velocity of C increases uniformly so that C has zero velocity at $t = 20$.\\
(i) Complete the velocity-time graph for the motion of C in the time interval $0 \leqslant t \leqslant 20$.\\
(ii) Calculate the acceleration of C in the time interval $0 < t < 10$.\\
(iii) Calculate the displacement of C from $t = 0$ to $t = 20$.
\end{itemize}

\hfill \mbox{\textit{OCR MEI M1 2011 Q1 [8]}}