| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Maximum or minimum velocity |
| Difficulty | Standard +0.3 This is a straightforward mechanics question requiring students to (1) find when acceleration is zero by differentiating the velocity function, (2) integrate velocity to find displacement, and (3) add the initial position. All steps are routine applications of standard techniques with no conceptual challenges, making it slightly easier than average. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = 12 - 6t\) | M1 | Differentiation, at least one term correct |
| \(a = 12 - 6t\) (correct) | A1 | |
| \(a = 0\) gives \(t = 2\) | F1 | Follow their \(a\) |
| \(x = \int(2 + 12t - 3t^2)\,dt\) | M1 | Integration indefinite or definite, at least one term correct |
| \(2t + 6t^2 - t^3 + C\) | A1 | Correct. Need not be simplified. Allow as definite integral. Ignore \(C\) or limits |
| \(x = 3\) when \(t = 0\) | M1 | Allow \(x = \pm 3\) or argue it is \(\int_0^2\) from A, then \(\pm 3\) |
| so \(3 = C\) and \(x = 2t + 6t^2 - t^3 + 3\) | A1 | Award if seen WWW or \(x = 2t + 6t^2 - t^3\) seen with \(+3\) added later |
| \(x(2) = 4 + 24 - 8 + 3 = 23\) m | B1 | FT their \(t\) and their \(x\) if obtained by integration but not if \(-3\) obtained instead of \(+3\). [If 20 m seen WWW award SC6] [Award SC1 for position if constant acceleration used for displacement and then \(+3\) applied] |
## Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = 12 - 6t$ | M1 | Differentiation, at least one term correct |
| $a = 12 - 6t$ (correct) | A1 | |
| $a = 0$ gives $t = 2$ | F1 | Follow **their** $a$ |
| $x = \int(2 + 12t - 3t^2)\,dt$ | M1 | Integration indefinite or definite, at least one term correct |
| $2t + 6t^2 - t^3 + C$ | A1 | Correct. Need not be simplified. Allow as definite integral. Ignore $C$ or limits |
| $x = 3$ when $t = 0$ | M1 | Allow $x = \pm 3$ or argue it is $\int_0^2$ from A, then $\pm 3$ |
| so $3 = C$ and $x = 2t + 6t^2 - t^3 + 3$ | A1 | Award if seen WWW or $x = 2t + 6t^2 - t^3$ seen with $+3$ added later |
| $x(2) = 4 + 24 - 8 + 3 = 23$ m | B1 | FT **their** $t$ and **their** $x$ if obtained by integration but not if $-3$ obtained instead of $+3$. [If 20 m seen WWW award SC6] [Award SC1 for position if constant acceleration used for displacement and then $+3$ applied] |
**Total: 8 marks**
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{34e4ce80-21b0-48f5-865c-de4dd837f7c5-1_98_836_1073_718}
\captionsetup{labelformat=empty}
\caption{Fig. 5}
\end{center}
\end{figure}
A toy car is moving along the straight line $\mathrm { O } x$, where O is the origin. The time $t$ is in seconds. At time $t = 0$ the car is at $\mathrm { A } , 3 \mathrm {~m}$ from O as shown in Fig. 5. The velocity of the car, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, is given by
$$v = 2 + 12 t - 3 t ^ { 2 }$$
Calculate the distance of the car from O when its acceleration is zero.
\hfill \mbox{\textit{OCR MEI M1 Q2 [8]}}