OCR MEI M1 — Question 2

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
TopicSUVAT & Travel Graphs

2 The speed of a 100 metre runner in \(\mathrm { m } \mathrm { s } ^ { - 1 }\) is measured electronically every 4 seconds.
The measurements are plotted as points on the speed-time graph in Fig. 6. The vertical dotted line is drawn through the runner's finishing time. Fig. 6 also illustrates Model P in which the points are joined by straight lines. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{4f80ea36-001f-4a00-849f-542f5072516b-2_1022_1503_524_290} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure}
  1. Use Model P to estimate
    (A) the distance the runner has gone at the end of 12 seconds,
    (B) how long the runner took to complete 100 m . A mathematician proposes Model Q in which the runner's speed, \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at time \(t \mathrm {~s}\), is given by $$v = \frac { 5 } { 2 } t - \frac { 1 } { 8 } t ^ { 2 } .$$
  2. Verify that Model Q gives the correct speed for \(t = 8\).
  3. Use Model Q to estimate the distance the runner has gone at the end of 12 seconds.
  4. The runner was timed at 11.35 seconds for the 100 m . Which model places the runner closer to the finishing line at this time? In this question take \(g\) as \(10 \mathrm {~m} \mathrm {~s} ^ { - 2 }\).
    A small ball is released from rest. It falls for 2 seconds and is then brought to rest over the next 5 seconds. This motion is modelled in the speed-time graph Fig. 6. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{4f80ea36-001f-4a00-849f-542f5072516b-3_658_1101_281_503} \captionsetup{labelformat=empty} \caption{Fig. 6}
    \end{figure} For this model,
  5. calculate the distance fallen from \(t = 0\) to \(t = 7\),
  6. find the acceleration of the ball from \(t = 2\) to \(t = 6\), specifying the direction,
  7. obtain an expression in terms of \(t\) for the downward speed of the ball from \(t = 2\) to \(t = 6\),
  8. state the assumption that has been made about the resistance to motion from \(t = 0\) to \(t = 2\). The part of the motion from \(t = 2\) to \(t = 7\) is now modelled by \(v = - \frac { 3 } { 2 } t ^ { 2 } + \frac { 19 } { 2 } t + 7\).
  9. Verify that \(v\) agrees with the values given in Fig, 6 at \(t = 2 , t = 6\) and \(t = 7\).
  10. Calculate the distance fallen from \(t = 2\) to \(t = 7\) according to this model.
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