| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Applied differentiation |
| Type | Velocity-time graph modelling |
| Difficulty | Standard +0.3 This is a straightforward kinematics problem requiring differentiation to find acceleration, using given conditions to form simultaneous equations for p, q, r, and integration to find distance. All steps are standard A-level techniques with clear guidance from the graph, making it slightly easier than average. |
| Spec | 1.08d Evaluate definite integrals: between limits3.02d Constant acceleration: SUVAT formulae3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = 0, v = 18 \Rightarrow r = 18\) | B1 | — |
| \(t = 5, v = 9 \Rightarrow 25p + 5q + 18 = 9\) | M1 | Substitutes \(t = 5, v = 9\) into quadratic; allow with \(r\) |
| \(\frac{dv}{dt} = 2pt + q\) | B1 | — |
| \(t = 5, \frac{dv}{dt} = 0 \Rightarrow 10p + q = 0\) | M1 | Substitutes \(t = 5\) and sets \(\frac{dv}{dt} = 0\); dependent on one term differentiated correctly |
| \(p = \frac{9}{25},\ q = -\frac{18}{5}\) | A1 | BC (oe e.g. exact decimals) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_2^5 \left(\frac{9}{25}t^2 - \frac{18}{5}t + 18\right) dt\) | M1 | Using their values of \(p\), \(q\) and \(r\) in an attempt to find distance travelled from 2 to 5 by integration |
| \(+ 9 \times 5\) | B1 | For distance travelled from 5 to 10 |
| \(= 75.24 \text{ m}\) | A1 | BC cao (oe) |
## Question 10(a):
$v = pt^2 + qt + r$
$t = 0, v = 18 \Rightarrow r = 18$ | **B1** | —
$t = 5, v = 9 \Rightarrow 25p + 5q + 18 = 9$ | **M1** | Substitutes $t = 5, v = 9$ into quadratic; allow with $r$
$\frac{dv}{dt} = 2pt + q$ | **B1** | —
$t = 5, \frac{dv}{dt} = 0 \Rightarrow 10p + q = 0$ | **M1** | Substitutes $t = 5$ and sets $\frac{dv}{dt} = 0$; dependent on one term differentiated correctly
$p = \frac{9}{25},\ q = -\frac{18}{5}$ | **A1** | BC (oe e.g. exact decimals)
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## Question 10(b):
$\int_2^5 \left(\frac{9}{25}t^2 - \frac{18}{5}t + 18\right) dt$ | **M1** | Using their values of $p$, $q$ and $r$ in an attempt to find distance travelled from 2 to 5 by integration
$+ 9 \times 5$ | **B1** | For distance travelled from 5 to 10
$= 75.24 \text{ m}$ | **A1** | BC cao (oe)
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\includegraphics[max width=\textwidth, alt={}, center]{d6430776-0b87-4e5e-8f78-c6228ee163d5-6_670_1106_797_258}
The diagram shows the velocity-time graph modelling the velocity of a car as it approaches, and drives through, a residential area.
The velocity of the car, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, at time $t$ seconds for the time interval $0 \leqslant t \leqslant 5$ is modelled by the equation $v = p t ^ { 2 } + q t + r$, where $p , q$ and $r$ are constants.
It is given that the acceleration of the car is zero at $t = 5$ and the speed of the car then remains constant.
\begin{enumerate}[label=(\alph*)]
\item Determine the values of $p , q$ and $r$.
\item Calculate the distance travelled by the car from $t = 2$ to $t = 10$.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q10 [8]}}