OCR PURE — Question 11 10 marks

Exam BoardOCR
ModulePURE
Marks10
PaperDownload PDF ↗
TopicConstant acceleration (SUVAT)
TypeTwo vehicles: overtaking or meeting (algebraic)
DifficultyStandard +0.3 This is a standard two-particle SUVAT problem requiring setting up position equations, solving a quadratic, and finding maximum separation. While it has multiple parts, each step follows routine procedures (SUVAT equations, quadratic formula, differentiation) with no novel insight required. Slightly easier than average due to straightforward setup and clean numbers.
Spec3.02d Constant acceleration: SUVAT formulae

11 A car starts from rest at a set of traffic lights and moves along a straight road with constant acceleration \(4 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). A motorcycle, travelling parallel to the car with constant speed \(16 \mathrm {~ms} ^ { - 1 }\), passes the same traffic lights exactly 1.5 seconds after the car starts to move. The time after the car starts to move is denoted by \(t\) seconds.
  1. Determine the two values of \(t\) at which the car and motorcycle are the same distance from the traffic lights. These two values of \(t\) are denoted by \(t _ { 1 }\) and \(t _ { 2 }\), where \(t _ { 1 } < t _ { 2 }\).
  2. Describe the relative positions of the car and the motorcycle when \(t _ { 1 } < t < t _ { 2 }\).
  3. Determine the maximum distance between the car and the motorcycle when \(t _ { 1 } < t < t _ { 2 }\). \section*{END OF QUESTION PAPER}

Question 11(a):
AnswerMarks Guidance
Car: \(s = 2t^2\)A1 1.1
Motorcycle: \(s = 16(t-1.5)\)A1 1.1
\(2t^2 = 16(t-1.5)\)M1dep\* 3.1a
\(t^2 - 8t + 12 = 0 \Rightarrow t = \ldots\)M1 1.1
\(t_1 = 2\) and \(t_2 = 6\)A1 1.1
[6 marks]
Question 11(b):
AnswerMarks Guidance
The motorcycle is ahead of the carB1 3.2a
[1 mark]
Question 11(c):
AnswerMarks Guidance
Maximum distance is at \(t = 4\)B1ft 3.1b
\(f(4) = 16(4) - 24 - 2(4)^2\)M1 3.4
\(= 8\) mA1 1.1
[3 marks]
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## Question 11(a):

Car: $s = 2t^2$ | **A1** | 1.1 |

Motorcycle: $s = 16(t-1.5)$ | **A1** | 1.1 |

$2t^2 = 16(t-1.5)$ | **M1dep\*** | 3.1a | Equates their expressions for $s$ to obtain a three-term quadratic in $t$

$t^2 - 8t + 12 = 0 \Rightarrow t = \ldots$ | **M1** | 1.1 | Attempts to solve their three-term quadratic in $t$. Dependent on all previous **M** marks. Must obtain two positive values for $t$

$t_1 = 2$ and $t_2 = 6$ | **A1** | 1.1 |

**[6 marks]**

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## Question 11(b):

The motorcycle is ahead of the car | **B1** | 3.2a |

**[1 mark]**

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## Question 11(c):

Maximum distance is at $t = 4$ | **B1ft** | 3.1b | Follow through their positive values of $t$. Or from differentiating $f(t) = 16t - 24 - 2t^2$

$f(4) = 16(4) - 24 - 2(4)^2$ | **M1** | 3.4 | Method for finding maximum distance between the car and motorcycle

$= 8$ m | **A1** | 1.1 |

**[3 marks]**

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11 A car starts from rest at a set of traffic lights and moves along a straight road with constant acceleration $4 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. A motorcycle, travelling parallel to the car with constant speed $16 \mathrm {~ms} ^ { - 1 }$, passes the same traffic lights exactly 1.5 seconds after the car starts to move. The time after the car starts to move is denoted by $t$ seconds.
\begin{enumerate}[label=(\alph*)]
\item Determine the two values of $t$ at which the car and motorcycle are the same distance from the traffic lights.

These two values of $t$ are denoted by $t _ { 1 }$ and $t _ { 2 }$, where $t _ { 1 } < t _ { 2 }$.
\item Describe the relative positions of the car and the motorcycle when $t _ { 1 } < t < t _ { 2 }$.
\item Determine the maximum distance between the car and the motorcycle when $t _ { 1 } < t < t _ { 2 }$.

\section*{END OF QUESTION PAPER}
\end{enumerate}

\hfill \mbox{\textit{OCR PURE  Q11 [10]}}