| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Sketching velocity-time graphs |
| Difficulty | Moderate -0.3 This is a straightforward mechanics question requiring basic integration, trapezium rule application, and differentiation. All techniques are standard M1 content with clear step-by-step procedures and no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02c Interpret kinematic graphs: gradient and area3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Distance travelled \(=\) Area under the graph | M1 | Attempt to find area |
| \(\frac{1}{2}\times4\times8 + \frac{1}{2}\times4\times(8+12) + 4\times12\) | M1 | Splitting into suitable parts |
| \(104\text{ m}\) | A1 | Cao; Allow all 3 marks for 104 without any working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Either: Working backwards from distance when \(t=12\): \(12 - \frac{(104-100)}{12}\) | M1, M1 | Allow this mark for \(0.33...\); Follow through from their total distance |
| \(11.67\text{ s}\) | A1 | Cao |
| Or: Working forwards from when \(t=8\): \(8 + \frac{(100-56)}{12}\) | M1, M1 | Allow this mark for \(3.67...\); Follow through from their distance at time \(8\text{ s}\) |
| \(11.67\text{ s}\) | A1 | Cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substituting \(t=8\) gives \(v = \frac{5}{2}\times8 - \frac{1}{8}\times8^2 = 12\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Distance \(= \int_0^{12}\left(\frac{5t}{2} - \frac{t^2}{8}\right)dt\) | M1 | Integrating \(v\). Condone no limits. |
| \(\left[\frac{5t^2}{4} - \frac{t^3}{24}\right]_0^{12}\) | A1 | Condone no limits |
| \(\left[180 - 72\right] - \left(-\left[0\right]\right)\) | M1 | Substituting \(t = 12\) |
| \(108\) m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Model P: distance at \(t = 11.35\) is \(96.2\) | B1 | Cao |
| Model Q: distance at \(t = 11.35\) is \(\left[\frac{5t^2}{4} - \frac{t^3}{24}\right]_0^{11.35} = 100.1\) | M1 | Substituting \(11.35\) in their expression from part (iii) |
| Model Q places the runner closer | E1 | Cao from correct previous working for both models |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Model P: Greatest acceleration \(\frac{8}{4} = 2\) m s\(^{-2}\) | B1 | |
| Model Q: \(a = \frac{dv}{dt} = \frac{5}{2} - \frac{t}{4}\) | M1 | Differentiating \(v\) |
| A1 | ||
| Model Q: Greatest acceleration is \(2.5\) m s\(^{-2}\) | B1 | Award if correct answer seen |
## Question 6:
**Part (i)(A)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Distance travelled $=$ Area under the graph | M1 | Attempt to find area |
| $\frac{1}{2}\times4\times8 + \frac{1}{2}\times4\times(8+12) + 4\times12$ | M1 | Splitting into suitable parts |
| $104\text{ m}$ | A1 | Cao; **Allow all 3 marks for 104 without any working** |
**Part (i)(B)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| **Either:** Working backwards from distance when $t=12$: $12 - \frac{(104-100)}{12}$ | M1, M1 | Allow this mark for $0.33...$; Follow through from their total distance |
| $11.67\text{ s}$ | A1 | Cao |
| **Or:** Working forwards from when $t=8$: $8 + \frac{(100-56)}{12}$ | M1, M1 | Allow this mark for $3.67...$; Follow through from their distance at time $8\text{ s}$ |
| $11.67\text{ s}$ | A1 | Cao |
**Part (ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substituting $t=8$ gives $v = \frac{5}{2}\times8 - \frac{1}{8}\times8^2 = 12$ | B1 | |
## Question 6:
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Distance $= \int_0^{12}\left(\frac{5t}{2} - \frac{t^2}{8}\right)dt$ | M1 | Integrating $v$. Condone no limits. |
| $\left[\frac{5t^2}{4} - \frac{t^3}{24}\right]_0^{12}$ | A1 | Condone no limits |
| $\left[180 - 72\right] - \left(-\left[0\right]\right)$ | M1 | Substituting $t = 12$ |
| $108$ m | A1 | |
### Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Model P: distance at $t = 11.35$ is $96.2$ | B1 | Cao |
| Model Q: distance at $t = 11.35$ is $\left[\frac{5t^2}{4} - \frac{t^3}{24}\right]_0^{11.35} = 100.1$ | M1 | Substituting $11.35$ in their expression from part (iii) |
| Model Q places the runner closer | E1 | Cao from correct previous working for both models |
### Part (v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Model P: Greatest acceleration $\frac{8}{4} = 2$ m s$^{-2}$ | B1 | |
| Model Q: $a = \frac{dv}{dt} = \frac{5}{2} - \frac{t}{4}$ | M1 | Differentiating $v$ |
| | A1 | |
| Model Q: Greatest acceleration is $2.5$ m s$^{-2}$ | B1 | Award if correct answer seen |
---6 The speed of a 100 metre runner in $\mathrm { ms } ^ { - 1 }$ 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]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{13f555cc-d506-48e5-a0e4-225cae4251dc-6_1025_1504_641_260}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item 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, $v \mathrm {~ms} ^ { - 1 }$ at time $t \mathrm {~s}$, is given by
$$v = \frac { 5 } { 2 } t - \frac { 1 } { 8 } t ^ { 2 }$$
\item Verify that Model Q gives the correct speed for $t = 8$.
\item Use Model Q to estimate the distance the runner has gone at the end of 12 seconds.
\item 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?
\item Find the greatest acceleration of the runner according to each model.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M1 2013 Q6 [18]}}