| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Displacement-time graph interpretation or sketching |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question requiring basic gradient calculations from a displacement-time graph and understanding that gradient equals velocity. All parts involve direct application of standard techniques with no problem-solving or novel insight required, making it easier than average. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | 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 |
| \([180 - 72] \quad (-[0])\) | M1 | Substituting \(t = 12\) |
| \(108\) m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Model P: distance at \(t = 11.35\) is \(96.2\) | B1 | Ca |
| 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 | Mark | 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 1:
## Part (iii)
| Answer | Mark | 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 |
| $[180 - 72] \quad (-[0])$ | M1 | Substituting $t = 12$ |
| $108$ m | A1 | |
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| Model P: distance at $t = 11.35$ is $96.2$ | B1 | Ca |
| 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 | Mark | 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 |
---1 A ring is moving up and down a vertical pole. The displacement, $s \mathrm {~m}$, of the ring above a mark on the pole is modelled by the displacement-time graph shown in Fig. 1. The three sections of the graph are straight lines.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bdbebc7f-0cb1-4203-8058-7614ba291508-1_763_1057_439_580}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Calculate the velocity of the ring in the interval $0 < t < 2$ and in the interval $2 < t < 3.5$.
\item Sketch a velocity-time graph for the motion of the ring during the 4 seconds.
\item State the direction of motion of the ring when\\
(A) $t = 1$,\\
(B) $t = 2.75$,\\
(C) $t = 3.25$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M1 Q1 [5]}}