| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Variable acceleration with initial conditions |
| Difficulty | Moderate -0.3 This is a straightforward variable acceleration question requiring two integrations with initial conditions (parts i-ii) and interpretation of displacement-time graphs (part iii). The calculus is routine (integrating polynomial functions), and the graph interpretation tests basic understanding of concavity. Slightly easier than average due to the mechanical nature of the integration and standard graph analysis. |
| Spec | 3.02c Interpret kinematic graphs: gradient and area3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(V = \int 0.8t \, dt\) | M1* | Attempt at integration |
| \(v = 0.8t^2/2 \, (+c)\) | A1 | Award if \(c\) omitted |
| \(t=0, v=13 \Rightarrow (c=13)\) | M1 | |
| \(v = 0.4x \, 6^2 \, (+c)\) | D*M1 | |
| \(v = 27.4 \text{ ms}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(s = \int 0.4t^2 \, (+c) \, dt\) | M1* | Attempt at integration of \(v(t)\) |
| \(s = 0.4t^3/3 + 13t \, (+k)\) | A1ft | ft \(cv(v(t)\) in (i)) |
| \(t=0, s=0 \Rightarrow (k=0)\) | M1 | |
| \(s = 0.4 \times 6^3/3 + 13 \times 6\) | D*M1 | |
| \(s = 106.8 \text{ m}\) | A1 | Allow if \(k=0\) assumed. Accept 107 m. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Fig. 2 | B1 | |
| Fig. 1 has zero initial velocity/gradient | B1 | |
| Fig. 3 does not have an increasing velocity/gradient | B1 |
# Question 5:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $V = \int 0.8t \, dt$ | M1* | Attempt at integration |
| $v = 0.8t^2/2 \, (+c)$ | A1 | Award if $c$ omitted |
| $t=0, v=13 \Rightarrow (c=13)$ | M1 | |
| $v = 0.4x \, 6^2 \, (+c)$ | D*M1 | |
| $v = 27.4 \text{ ms}^{-1}$ | A1 | |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $s = \int 0.4t^2 \, (+c) \, dt$ | M1* | Attempt at integration of $v(t)$ |
| $s = 0.4t^3/3 + 13t \, (+k)$ | A1ft | ft $cv(v(t)$ in (i)) |
| $t=0, s=0 \Rightarrow (k=0)$ | M1 | |
| $s = 0.4 \times 6^3/3 + 13 \times 6$ | D*M1 | |
| $s = 106.8 \text{ m}$ | A1 | Allow if $k=0$ assumed. Accept 107 m. |
## Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Fig. 2 | B1 | |
| Fig. 1 has zero initial velocity/gradient | B1 | |
| Fig. 3 does not have an increasing velocity/gradient | B1 | |
---5 A car is travelling at $13 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ along a straight road when it passes a point $A$ at time $t = 0$, where $t$ is in seconds. For $0 \leqslant t \leqslant 6$, the car accelerates at $0.8 t \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(i) Calculate the speed of the car when $t = 6$.\\
(ii) Calculate the displacement of the car from $A$ when $t = 6$.\\
(iii) Three $( t , x )$ graphs are shown below, for $0 \leqslant t \leqslant 6$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{470e70de-66ba-4dcc-a205-0c92f29471b1-3_382_458_1366_340}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{470e70de-66ba-4dcc-a205-0c92f29471b1-3_382_460_1366_881}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{470e70de-66ba-4dcc-a205-0c92f29471b1-3_384_461_1366_1420}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item State which of these three graphs is most appropriate to represent the motion of the car.
\item For each of the two other graphs give a reason why it is not appropriate to represent the motion of the car.
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2009 Q5 [13]}}