OCR PURE — Question 12 7 marks

Exam BoardOCR
ModulePURE
Marks7
PaperDownload PDF ↗
TopicVariable acceleration (1D)
TypeFinding constants from motion conditions
DifficultyStandard +0.8 This question requires finding the maximum velocity time by differentiation, using it to set up a definite integral for distance, solving for constant k, then finding when v=0. It combines calculus techniques (differentiation, integration) with algebraic manipulation across multiple connected steps, going beyond routine exercises but using standard A-level methods.
Spec1.07n Stationary points: find maxima, minima using derivatives1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration
12 \includegraphics[max width=\textwidth, alt={}, center]{31b0d5b6-1593-489b-bbcd-486e7c96ff18-09_647_935_260_242} A particle \(P\) moves in a straight line. At time \(t\) seconds, where \(t \geqslant 0\), the velocity of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). It is given that \(v = - 3 t ^ { 2 } + 24 t + k\), where \(k\) is a positive constant. The diagram shows the velocity-time graph for the motion of \(P\). \(P\) attains its maximum velocity at time \(T\) seconds. Given that the distance travelled by \(P\) between times \(t = 1\) and \(t = T\) is 297 m , determine the time when \(P\) is instantaneously at rest. \section*{END OF QUESTION PAPER}
Question 12:
AnswerMarks Guidance
AnswerMark Guidance
\(-6t + 24 = 0\)M1 Either attempt to differentiate and set \(a=0\) or attempt to complete the square \(-3(t-4)^2 + \ldots\) to find \(t\). Need \(bt + 24 = 0\)
\(T = 4\)A1
\(\int(-3t^2 + 24t + k)\,dt\)M1* Attempt to integrate \(v\) (at least two terms correct)
\(= -t^3 + 12t^2 + kt\ (+c)\)A1
\(\left(-4^3 + 12(4)^2 + 4k\right) - \left(-1^3 + 12(1)^2 + k\right) = 297\)M1dep* Setting up an equation in \(k\) using 297, and \(t=1\) and \(t=4\) and attempt to solve for \(k\). If correct \(k=60\)
\(t^2 - 8t - 20 = 0\)M1 Setting \(v=0\) and attempt to solve three-term quadratic in \(t\) (perhaps BC). Dependent on both previous M marks
\((t-10)(t+2) = 0\)
As \(t \geq 0\), \(t \neq -2\ \therefore\ t = 10\)A1 Must see explicit rejection of negative value of \(t\)
[7] 
## Question 12:

| Answer | Mark | Guidance |
|--------|------|----------|
| $-6t + 24 = 0$ | M1 | Either attempt to differentiate and set $a=0$ or attempt to complete the square $-3(t-4)^2 + \ldots$ to find $t$. Need $bt + 24 = 0$ |
| $T = 4$ | A1 | |
| $\int(-3t^2 + 24t + k)\,dt$ | M1* | Attempt to integrate $v$ (at least two terms correct) |
| $= -t^3 + 12t^2 + kt\ (+c)$ | A1 | |
| $\left(-4^3 + 12(4)^2 + 4k\right) - \left(-1^3 + 12(1)^2 + k\right) = 297$ | M1dep* | Setting up an equation in $k$ using 297, and $t=1$ and $t=4$ and attempt to solve for $k$. If correct $k=60$ |
| $t^2 - 8t - 20 = 0$ | M1 | Setting $v=0$ and attempt to solve three-term quadratic in $t$ (perhaps BC). Dependent on both previous M marks |
| $(t-10)(t+2) = 0$ | | |
| As $t \geq 0$, $t \neq -2\ \therefore\ t = 10$ | A1 | Must see explicit rejection of negative value of $t$ |
| **[7]** | | |
12\\
\includegraphics[max width=\textwidth, alt={}, center]{31b0d5b6-1593-489b-bbcd-486e7c96ff18-09_647_935_260_242}

A particle $P$ moves in a straight line. At time $t$ seconds, where $t \geqslant 0$, the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. It is given that $v = - 3 t ^ { 2 } + 24 t + k$, where $k$ is a positive constant.

The diagram shows the velocity-time graph for the motion of $P$.\\
$P$ attains its maximum velocity at time $T$ seconds. Given that the distance travelled by $P$ between times $t = 1$ and $t = T$ is 297 m , determine the time when $P$ is instantaneously at rest.

\section*{END OF QUESTION PAPER}

\hfill \mbox{\textit{OCR PURE  Q12 [7]}}
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