| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Velocity from acceleration by integration |
| Difficulty | Moderate -0.3 This is a straightforward mechanics question requiring students to find velocity from an acceleration-time graph by calculating areas (integration). Part (i) involves finding area under the graph and adding to initial velocity. Part (ii) requires setting up and solving a simple equation. While it tests understanding of the relationship between acceleration and velocity, it's a standard textbook exercise with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.08d Evaluate definite integrals: between limits3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Section from \(t = 10\) to \(t = 15\) | B1 | |
| Section from \(t = 15\) to \(t = 20\) | B1 | FT connecting from their point when \(t = 15\). Ignore graph outside \(0 \leq t \leq 20\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{-6 - 14}{10} = -2\), so \(-2\) m s\(^{-2}\) | M1 | Attempt at \(\frac{\Delta v}{\Delta t}\) |
| A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Displacement is \(\frac{14}{2} \times 7 - \frac{13+5}{2} \times 6\) | M1 | Attempt at whole area. Condone 'overlap' but not 'gaps'. |
| or \(\frac{14}{2} \times 7 - \frac{3 \times 6}{2} - 5 \times 6 - \frac{5 \times 6}{2}\) | B1 | 'Positive' area expression correct. Condone sign error. |
| B1 | 'Negative' area expression correct. Condone overall sign error. | |
| \(= -5\) so \(5\) m downwards | A1 | Accept \(-5\) m cao |
| or Displacement is \(14 \times 10 + \frac{1}{2} \times (-2) \times 10^2 - 5 \times 6 + \frac{-6+0}{2} \times 5\) | M1 | Using suvat from \(0\) to \(10\) or \(15\) to \(20\). Condone 'overlap' but not 'gaps' |
| \(= 140 - 100 - 30 - 15 = -5\) so \(5\) m downwards | A1, B1, A1 | Subtracting \(30\) or \(15\) or \(45\). Accept \(-5\) m cao |
# Question 2:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Section from $t = 10$ to $t = 15$ | B1 | |
| Section from $t = 15$ to $t = 20$ | B1 | FT connecting from their point when $t = 15$. Ignore graph outside $0 \leq t \leq 20$ |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{-6 - 14}{10} = -2$, so $-2$ m s$^{-2}$ | M1 | Attempt at $\frac{\Delta v}{\Delta t}$ |
| | A1 | |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Displacement is $\frac{14}{2} \times 7 - \frac{13+5}{2} \times 6$ | M1 | Attempt at whole area. Condone 'overlap' but not 'gaps'. |
| or $\frac{14}{2} \times 7 - \frac{3 \times 6}{2} - 5 \times 6 - \frac{5 \times 6}{2}$ | B1 | 'Positive' area expression correct. Condone sign error. |
| | B1 | 'Negative' area expression correct. Condone overall sign error. |
| $= -5$ so $5$ m downwards | A1 | Accept $-5$ m cao |
| **or** Displacement is $14 \times 10 + \frac{1}{2} \times (-2) \times 10^2 - 5 \times 6 + \frac{-6+0}{2} \times 5$ | M1 | Using suvat from $0$ to $10$ or $15$ to $20$. Condone 'overlap' but not 'gaps' |
| $= 140 - 100 - 30 - 15 = -5$ so $5$ m downwards | A1, B1, A1 | Subtracting $30$ or $15$ or $45$. Accept $-5$ m cao |
---2 Fig. 2 shows an acceleration-time graph modelling the motion of a particle.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bdbebc7f-0cb1-4203-8058-7614ba291508-2_684_1068_408_586}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
At $t = 0$ the particle has a velocity of $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the positive direction.\\
(i) Find the velocity of the particle when $t = 2$.\\
(ii) At what time is the particle travelling in the negative direction with a speed of $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ ?
\hfill \mbox{\textit{OCR MEI M1 Q2 [4]}}