| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Read and interpret velocity-time graph |
| Difficulty | Moderate -0.3 This is a multi-part mechanics question covering standard SUVAT and kinematics concepts. Part (i) requires basic trapezium rule area calculation from a graph, parts (ii)-(iv) involve straightforward integration of a given velocity function (v = 5t/2 - t²/8) to find distance, and comparing models. The second section on the falling ball uses standard constant acceleration formulas with g=10. All techniques are routine M1 content with no novel problem-solving required, though the multi-part nature and need to interpret graphs adds slight complexity beyond pure recall. |
| Spec | 1.02z Models in context: use functions in modelling1.08d Evaluate definite integrals: between limits3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Distance travelled = Area under the graph | M1 | Attempt to find area |
| \(\frac{1}{2} \times 4 \times 8 + \frac{1}{2} \times 4 \times (8+12) + 4 \times 12\) | M1 | Splitting into suitable parts |
| 104 m | A1 | cao. Allow all 3 marks for 104 without any working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Either: Working backwards from distance when \(t = 12\) | M1 | |
| \(12 - \frac{(104-100)}{12}\) | M1 | Allow this mark for 0.33… Follow through from their total distance |
| 11.67 s | A1 | cao |
| Or: Working forwards from when \(t = 8\) | M1 | |
| \(8 + \frac{(100-56)}{12}\) | M1 | Allow this mark for 3.67… Follow through from their distance at time 8s |
| 11.67 s | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substituting \(t = 8\) gives \(v = \frac{5}{2} \times 8 - \frac{1}{8} \times 8^2 = 12\) | B1 |
| 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 |
| \([180 - 72]\ (-[0])\) | M1 | Substituting \(t = 12\) |
| 108 m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Model P: distance at \(t = 11.35\) is 96.2 | B1 | 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 | Mark | Guidance |
| Model P: Greatest acceleration \(\frac{8}{4} = 2 \text{ m s}^{-2}\) | B1 | |
| Model Q: \(a = \frac{dv}{dt} = \frac{5}{2} - \frac{t}{4}\) | M1 | Differentiating \(v\) |
| A1 | ||
| Model Q: Greatest acceleration is \(2.5 \text{ m s}^{-2}\) | B1 | Award if correct answer seen |
## Question 2:
### Part (i)(A):
| Answer | Mark | Guidance |
|--------|------|----------|
| Distance travelled = Area under the graph | M1 | Attempt to find area |
| $\frac{1}{2} \times 4 \times 8 + \frac{1}{2} \times 4 \times (8+12) + 4 \times 12$ | M1 | Splitting into suitable parts |
| 104 m | A1 | cao. **Allow all 3 marks for 104 without any working** |
### Part (i)(B):
| Answer | Mark | Guidance |
|--------|------|----------|
| **Either:** Working backwards from distance when $t = 12$ | M1 | |
| $12 - \frac{(104-100)}{12}$ | M1 | Allow this mark for 0.33… Follow through from their total distance |
| 11.67 s | A1 | cao |
| **Or:** Working forwards from when $t = 8$ | M1 | |
| $8 + \frac{(100-56)}{12}$ | M1 | Allow this mark for 3.67… Follow through from their distance at time 8s |
| 11.67 s | A1 | cao |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Substituting $t = 8$ gives $v = \frac{5}{2} \times 8 - \frac{1}{8} \times 8^2 = 12$ | B1 | |
### 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 |
| $[180 - 72]\ (-[0])$ | M1 | Substituting $t = 12$ |
| 108 m | A1 | |
### Part (iv):
| Answer | Mark | 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 | Mark | Guidance |
|--------|------|----------|
| Model P: Greatest acceleration $\frac{8}{4} = 2 \text{ m s}^{-2}$ | B1 | |
| Model Q: $a = \frac{dv}{dt} = \frac{5}{2} - \frac{t}{4}$ | M1 | Differentiating $v$ |
| | A1 | |
| Model Q: Greatest acceleration is $2.5 \text{ m s}^{-2}$ | B1 | Award if correct answer seen |
---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]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4f80ea36-001f-4a00-849f-542f5072516b-2_1022_1503_524_290}
\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, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ 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?
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]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4f80ea36-001f-4a00-849f-542f5072516b-3_658_1101_281_503}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
For this model,
\item calculate the distance fallen from $t = 0$ to $t = 7$,
\item find the acceleration of the ball from $t = 2$ to $t = 6$, specifying the direction,
\item obtain an expression in terms of $t$ for the downward speed of the ball from $t = 2$ to $t = 6$,
\item 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 $v = - \frac { 3 } { 2 } t ^ { 2 } + \frac { 19 } { 2 } t + 7$.
\item Verify that $v$ agrees with the values given in Fig, 6 at $t = 2 , t = 6$ and $t = 7$.
\item Calculate the distance fallen from $t = 2$ to $t = 7$ according to this model.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M1 Q2 [18]}}