OCR M1 2014 June — Question 5 12 marks

Exam BoardOCR
ModuleM1 (Mechanics 1)
Year2014
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeDisplacement-time graph interpretation or sketching
DifficultyModerate -0.5 This is a straightforward displacement-time graph interpretation question requiring basic gradient calculations and reading values from a piecewise linear graph. Part (i) uses simple distance = speed × time, parts (ii) involve reading gradients, part (iii) sums distances, and part (iv) requires differentiating a quadratic and solving. All techniques are standard M1 material with no novel problem-solving required, making it slightly easier than average.
Spec3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02f Non-uniform acceleration: using differentiation and integration
5 \includegraphics[max width=\textwidth, alt={}, center]{66eb8290-3a80-40bf-be40-a936ed7d5a1b-3_652_1675_959_187} A particle \(P\) can move in a straight line on a horizontal surface. At time \(t\) seconds the displacement of \(P\) from a fixed point \(A\) on the line is \(x \mathrm {~m}\). The diagram shows the \(( t , x )\) graph for \(P\). In the interval \(0 \leqslant t \leqslant 10\), either the speed of \(P\) is \(4 \mathrm {~ms} ^ { - 1 }\), or \(P\) is at rest.
  1. Show by calculation that \(T = 1.75\).
  2. State the velocity of \(P\) when
    1. \(t = 2\),
    2. \(t = 8\),
    3. \(t = 9\).
    4. Calculate the distance travelled by \(P\) in the interval \(0 \leqslant t \leqslant 10\). For \(t > 10\), the displacement of \(P\) from \(A\) is given by \(x = 20 t - t ^ { 2 } - 96\).
    5. Calculate the value of \(t\), where \(t > 10\), for which the speed of \(P\) is \(4 \mathrm {~ms} ^ { - 1 }\).
Question 5:
Part (i)
AnswerMarks Guidance
AnswerMarks Guidance
\(5/(T-3) = -4\) OR \(5/(3-T) = 4\)M1 Accept verification, \(4 \times (3 - 1.75)\) M1 \(= 5\) A1 OR \(5/(3-1.75)\) M1 \(= 4\) A1
\(T = 1.75\)A1
Part (ii)(a)
AnswerMarks Guidance
AnswerMarks Guidance
\(-4 \text{ ms}^{-1}\)B1
Part (ii)(b)
AnswerMarks Guidance
AnswerMarks Guidance
\(4 \text{ ms}^{-1}\)B1
Part (ii)(c)
AnswerMarks Guidance
AnswerMarks Guidance
\(4 \text{ ms}^{-1}\)B1
Part (iii)
AnswerMarks Guidance
AnswerMarks Guidance
\(2 \times (-)4,\ 2 \times 4,\ (1 \times)4\)M1* Calculates any one unknown distance; Allow if only one calc. correct
\(d = (-)5 + (-)8 + 8 + 4\)D*M1 Adds 5 and "3 other" distances or \(-5\) and "3 other" displacements; Note \(t=5\) to \(t=9\), \(t=5\) to \(t=10\) etc, may be one term
\(d = 25 \text{ m}\)A1 Correctly comes from \(4x(1.25+4+1)\) 3/3
Part (iv)
AnswerMarks Guidance
AnswerMarks Guidance
\(v = d(20t - t^2 - 96)/dt\)M1* Differentiates \(x\), accept \(20 - t\) as "differentiation"
\(v = 20 - 2t\)A1
\(20 - 2t = -4\)D*M1 \(20 - 2t + c = -4\) is DM0
\(t = 12\) (ignore any solutions less than 10)A1 Only from \(20 - 2t = -4\). This answer can arise fortuitously from solving \(20t - t^2 - 96 = 0\); SC Verifying that \(t=12\) gives \(v=-4\) can gain final M1A1 
# Question 5:

## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $5/(T-3) = -4$ OR $5/(3-T) = 4$ | M1 | Accept verification, $4 \times (3 - 1.75)$ M1 $= 5$ A1 OR $5/(3-1.75)$ M1 $= 4$ A1 |
| $T = 1.75$ | A1 | |

## Part (ii)(a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-4 \text{ ms}^{-1}$ | B1 | |

## Part (ii)(b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4 \text{ ms}^{-1}$ | B1 | |

## Part (ii)(c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4 \text{ ms}^{-1}$ | B1 | |

## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2 \times (-)4,\ 2 \times 4,\ (1 \times)4$ | M1* | Calculates any one unknown distance; Allow if only one calc. correct |
| $d = (-)5 + (-)8 + 8 + 4$ | D*M1 | Adds 5 and "3 other" distances or $-5$ and "3 other" displacements; Note $t=5$ to $t=9$, $t=5$ to $t=10$ etc, may be one term |
| $d = 25 \text{ m}$ | A1 | Correctly comes from $4x(1.25+4+1)$ 3/3 |

## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = d(20t - t^2 - 96)/dt$ | M1* | Differentiates $x$, accept $20 - t$ as "differentiation" |
| $v = 20 - 2t$ | A1 | |
| $20 - 2t = -4$ | D*M1 | $20 - 2t + c = -4$ is DM0 |
| $t = 12$ (ignore any solutions less than 10) | A1 | Only from $20 - 2t = -4$. This answer can arise fortuitously from solving $20t - t^2 - 96 = 0$; SC Verifying that $t=12$ gives $v=-4$ can gain final M1A1 |

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5\\
\includegraphics[max width=\textwidth, alt={}, center]{66eb8290-3a80-40bf-be40-a936ed7d5a1b-3_652_1675_959_187}

A particle $P$ can move in a straight line on a horizontal surface. At time $t$ seconds the displacement of $P$ from a fixed point $A$ on the line is $x \mathrm {~m}$. The diagram shows the $( t , x )$ graph for $P$. In the interval $0 \leqslant t \leqslant 10$, either the speed of $P$ is $4 \mathrm {~ms} ^ { - 1 }$, or $P$ is at rest.\\
(i) Show by calculation that $T = 1.75$.\\
(ii) State the velocity of $P$ when
\begin{enumerate}[label=(\alph*)]
\item $t = 2$,
\item $t = 8$,
\item $t = 9$.\\
(iii) Calculate the distance travelled by $P$ in the interval $0 \leqslant t \leqslant 10$.

For $t > 10$, the displacement of $P$ from $A$ is given by $x = 20 t - t ^ { 2 } - 96$.\\
(iv) Calculate the value of $t$, where $t > 10$, for which the speed of $P$ is $4 \mathrm {~ms} ^ { - 1 }$.
\end{enumerate}

\hfill \mbox{\textit{OCR M1 2014 Q5 [12]}}
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