| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | SUVAT simultaneous equations: find u and a |
| Difficulty | Standard +0.3 This is a straightforward SUVAT problem with standard multi-part structure. Part (i) uses basic kinematic equations with given values, part (ii) requires checking if a displacement is reached (routine discriminant check), and part (iii) involves differentiating a polynomial and applying initial conditions. All techniques are standard A-level mechanics with no novel insight required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| \(18 = \left(\frac{8 + u}{2}\right)(9)\) | M1 | AO3.4 |
| \(u = -4\) therefore the speed of \(P\) is \(4\ (\text{ms}^{-1})\) | A1 | AO1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| eg \(8 = -4 + 9a\) | M1 | AO3.4 |
| \(a = \frac{4}{3}\ (\text{ms}^{-2})\) | A1 | AO1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0 = -4 + \frac{4}{3}t\) | M1 | Use of \(v = u + at\) with \(v = 0\) and their \(a\) and \(u\) |
| \(t = 3\) | A1 | 1.1 |
| \(-s_{\max} = -4t + \frac{1}{2}\left(\frac{4}{3}\right)t^2\) | M1 | Use of \(s = ut + \frac{1}{2}at^2\) with their \(a\), \(u\) & \(t\) |
| \(s_{\max} = 6 < 10\) so \(P\) is never at \(B\) | A1 [4] | Compare with 10 or suitable comment |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(-10 = -4t + \frac{1}{2}\left(\frac{4}{3}\right)t^2\) | M1, A1 | Use of \(s = ut + \frac{1}{2}at^2\) with their \(u\) and \(a\) and suitable \(s\) |
| e.g. \(\det = -24\) therefore not possible | M1, A1 | Consider \(b^2 - 4ac\) or attempt to solve three term quadratic in \(t\); Or \(36 - 60 < 0\) therefore not possible |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0 = (\pm 4)^2 + 2\left(\frac{4}{3}\right)s\) or \(v^2 = (\pm 4)^2 + 2\left(\frac{4}{3}\right)(-10)\) | M2 | Use of \(v^2 = u^2 + 2as\) with their \(a\) and \(u\) and either \(v = 0\) or \(s = \pm 10\) |
| \(s = -6\) or \(v^2 = -\frac{32}{3}\) | A1 | |
| Suitable conclusion | A1 | Dependent on previous A mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = at^3 + bt^2 + ct\) | ||
| \(\dot{x} = 3at^2 + 2bt + c\) | M1 | 1.1 — Attempt to differentiate once; Two terms differentiated correctly |
| \(\ddot{x} = 6at + 2b\) | M1 | 2.1 — Attempt to differentiate again and substitute \(t = 0\) into both equations and substitute their acceleration in their second derivative and their \(u\) in their first derivative; Two terms differentiated correctly following through from their first derivative |
| \(c = -4\) and \(b = \frac{2}{3}\) | A1ft | 1.1 — \(b = 0.5 \times\) their accel. and \(c = \pm 4\); Allow \(b = 0.665\) from accel. \(= 1.33\) |
| \(18 = a(6)^3 + \frac{2}{3}(6)^2 - 4(6) \Rightarrow a = \frac{1}{12}\) | A1ft | 1.1 — Allow \(a = -\frac{5}{36}, -\frac{7}{108}, \frac{1}{108}\) which come from \(u = 4\) |
| Velocity of \(Q = \left(\frac{1}{4}(6)^2 + \frac{4}{3}(6) - 4\right) = 13\ (\text{ms}^{-1})\) | A1 [5] | 1.1 — cao |
## Question 11:
### Part (i)(a):
$18 = \left(\frac{8 + u}{2}\right)(9)$ | **M1** | AO3.4 | Use of $s = \left(\frac{u+v}{2}\right)t$
$u = -4$ therefore the speed of $P$ is $4\ (\text{ms}^{-1})$ | **A1** | AO1.1 | AG
### Part (i)(b):
eg $8 = -4 + 9a$ | **M1** | AO3.4 | Use of $v = u + at$ with their $u$ **or** $s = vt - \frac{1}{2}at^2$ **or** $v^2 = u^2 + 2as$ with their $u$ **or** $s = ut + \frac{1}{2}at^2$ with their $u$
$a = \frac{4}{3}\ (\text{ms}^{-2})$ | **A1** | AO1.1 | Accept 1.33 or better
## Question (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0 = -4 + \frac{4}{3}t$ | M1 | Use of $v = u + at$ with $v = 0$ and their $a$ and $u$ |
| $t = 3$ | A1 | 1.1 |
| $-s_{\max} = -4t + \frac{1}{2}\left(\frac{4}{3}\right)t^2$ | M1 | Use of $s = ut + \frac{1}{2}at^2$ with their $a$, $u$ & $t$ |
| $s_{\max} = 6 < 10$ so $P$ is never at $B$ | A1 [4] | Compare with 10 or suitable comment |
**OR**
| Answer | Mark | Guidance |
|--------|------|----------|
| $-10 = -4t + \frac{1}{2}\left(\frac{4}{3}\right)t^2$ | M1, A1 | Use of $s = ut + \frac{1}{2}at^2$ with their $u$ and $a$ and suitable $s$ |
| e.g. $\det = -24$ therefore not possible | M1, A1 | Consider $b^2 - 4ac$ or attempt to solve three term quadratic in $t$; Or $36 - 60 < 0$ therefore not possible |
**OR**
| Answer | Mark | Guidance |
|--------|------|----------|
| $0 = (\pm 4)^2 + 2\left(\frac{4}{3}\right)s$ or $v^2 = (\pm 4)^2 + 2\left(\frac{4}{3}\right)(-10)$ | M2 | Use of $v^2 = u^2 + 2as$ with their $a$ and $u$ and either $v = 0$ or $s = \pm 10$ |
| $s = -6$ or $v^2 = -\frac{32}{3}$ | A1 | |
| Suitable conclusion | A1 | Dependent on previous A mark |
---
## Question (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = at^3 + bt^2 + ct$ | | |
| $\dot{x} = 3at^2 + 2bt + c$ | M1 | 1.1 — Attempt to differentiate once; Two terms differentiated correctly |
| $\ddot{x} = 6at + 2b$ | M1 | 2.1 — Attempt to differentiate again and substitute $t = 0$ into both equations and substitute their acceleration in their second derivative and their $u$ in their first derivative; Two terms differentiated correctly following through from their first derivative |
| $c = -4$ and $b = \frac{2}{3}$ | A1ft | 1.1 — $b = 0.5 \times$ their accel. and $c = \pm 4$; Allow $b = 0.665$ from accel. $= 1.33$ |
| $18 = a(6)^3 + \frac{2}{3}(6)^2 - 4(6) \Rightarrow a = \frac{1}{12}$ | A1ft | 1.1 — Allow $a = -\frac{5}{36}, -\frac{7}{108}, \frac{1}{108}$ which come from $u = 4$ |
| Velocity of $Q = \left(\frac{1}{4}(6)^2 + \frac{4}{3}(6) - 4\right) = 13\ (\text{ms}^{-1})$ | A1 [5] | 1.1 — cao |11\\
\includegraphics[max width=\textwidth, alt={}, center]{efde7b10-b4f3-469f-ba91-b765a16ea835-7_127_1147_260_459}
A particle $P$ is moving along a straight line with constant acceleration. Initially the particle is at $O$. After 9 s , $P$ is at a point $A$, where $O A = 18 \mathrm {~m}$ (see diagram) and the velocity of $P$ at $A$ is $8 \mathrm {~ms} ^ { - 1 }$ in the direction $\overrightarrow { O A }$.
\begin{enumerate}[label=(\roman*)]
\item (a) Show that the initial speed of $P$ is $4 \mathrm {~ms} ^ { - 1 }$.\\
(b) Find the acceleration of $P$.\\
$B$ is a point on the line such that $O B = 10 \mathrm {~m}$, as shown in the diagram.
\item Show that $P$ is never at point $B$.
A second particle $Q$ moves along the same straight line, but has variable acceleration. Initially $Q$ is at $O$, and the displacement of $Q$ from $O$ at time $t$ seconds is given by
$$x = a t ^ { 3 } + b t ^ { 2 } + c t$$
where $a$, $b$ and $c$ are constants.\\
It is given that
\begin{itemize}
\item the velocity and acceleration of $Q$ at the point $O$ are the same as those of $P$ at $O$,
\item $\quad Q$ reaches the point $A$ when $t = 6$.
\item Find the velocity of $Q$ at $A$.
\end{itemize}
\section*{OCR}
\section*{Oxford Cambridge and RSA}
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q11 [13]}}