| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | December |
| Marks | 10 |
| Topic | Variable acceleration (1D) |
| Type | Maximum or minimum velocity |
| Difficulty | Moderate -0.3 This is a straightforward mechanics calculus question requiring differentiation to find acceleration, solving for k, finding minimum velocity using second derivative test, and integration to find position. All techniques are standard M1 procedures with no novel problem-solving required, making it slightly easier than average but not trivial due to the multi-step nature and algebraic manipulation involved. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dv}{dt} = 8t^3 + 2kt\) | B1 | Correct expression for the acceleration |
| \(8(2)^3 + 2k(2) = 28\) | M1 | Substitute \(t = 2\) into their \(a\) and equate to 28 |
| \(4k = 28 - 64 \Rightarrow k = -9\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dv}{dt} = 0 \Rightarrow 2t(4t^2 - 9) = 0\) | M1 | Substituting the correct value of \(k\) and equating to zero |
| \(t = 1.5\) (and \(t = 0\)) | A1 | AG Correctly finding the given value of \(t\) |
| E.g. \(\frac{d^2v}{dt^2}\bigg | _{t=1.5} = 24(1.5)^2 - 18 > 0\) so a minimum | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(s = \frac{2}{5}t^5 - 3t^3 - 4t (+c)\) | M1* | Attempt to integrate \(v\) (all powers increased by 1) |
| \(-6.4125 = 0.4 - 3 - 4 + c \Rightarrow c = K\) | M1dep* | Attempt to find \(c\) |
| \(s = 0.4(1.5)^5 - 3(1.5)^3 - 4(1.5) + 0.1875\) | M1 | Substitute 1.5 into their expression for \(s\) – dependent on both previous M marks |
| \(s = -12.9 \text{ so distance of } P \text{ from } O \text{ is } 12.9 \text{ m}\) | A1 |
## Part (a)
$\frac{dv}{dt} = 8t^3 + 2kt$ | B1 | Correct expression for the acceleration
$8(2)^3 + 2k(2) = 28$ | M1 | Substitute $t = 2$ into their $a$ and equate to 28
$4k = 28 - 64 \Rightarrow k = -9$ | A1 | AG
## Part (b)
$\frac{dv}{dt} = 0 \Rightarrow 2t(4t^2 - 9) = 0$ | M1 | Substituting the correct value of $k$ and equating to zero
$t = 1.5$ (and $t = 0$) | A1 | AG Correctly finding the given value of $t$
E.g. $\frac{d^2v}{dt^2}\bigg|_{t=1.5} = 24(1.5)^2 - 18 > 0$ so a minimum | B1 | Showing that this value of $t$ gives a minimum | Or complete argument from the shape of the curve, or from first derivatives
## Part (c)
$s = \frac{2}{5}t^5 - 3t^3 - 4t (+c)$ | M1* | Attempt to integrate $v$ (all powers increased by 1) | Constant not required for this first M mark
$-6.4125 = 0.4 - 3 - 4 + c \Rightarrow c = K$ | M1dep* | Attempt to find $c$ | $c = 0.1875$
$s = 0.4(1.5)^5 - 3(1.5)^3 - 4(1.5) + 0.1875$ | M1 | Substitute 1.5 into their expression for $s$ – dependent on both previous M marks
$s = -12.9 \text{ so distance of } P \text{ from } O \text{ is } 12.9 \text{ m}$ | A1 |A particle $P$ moves along the $x$-axis. At time $t$ seconds the velocity of $P$ is $v\text{m s}^{-1}$, where $v = 2t^4 + kt^2 - 4$.
The acceleration of $P$ when $t = 2$ is $28\text{m s}^{-2}$.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = -9$. [3]
\item Show that the velocity of $P$ has its minimum value when $t = 1.5$. [3]
\end{enumerate}
When $t = 1$, $P$ is at the point $(-6.4125, 0)$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the distance of $P$ from the origin $O$ when $P$ is moving with minimum velocity. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 2018 Q9 [10]}}