OCR MEI AS Paper 2 2020 November — Question 11 10 marks

Exam BoardOCR MEI
ModuleAS Paper 2 (AS Paper 2)
Year2020
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Functions
TypeExponential model with shifted asymptote
DifficultyStandard +0.3 This is a straightforward exponential modelling question requiring substitution to find parameters, evaluation at given points, differentiation for acceleration, and limit analysis. All techniques are standard AS-level procedures with no novel problem-solving required, making it slightly easier than average.
Spec1.06a Exponential function: a^x and e^x graphs and properties1.06i Exponential growth/decay: in modelling context1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)

11 A car is travelling along a stretch of road at a steady speed of \(11 \mathrm {~ms} ^ { - 1 }\).
The driver accelerates, and \(t\) seconds after starting to accelerate the speed of the car, \(V\), is modelled by the formula \(\mathrm { V } = \mathrm { A } + \mathrm { B } \left( 1 - \mathrm { e } ^ { - 0.17 \mathrm { t } } \right)\).
When \(t = 3 , V = 13.8\).
  1. Find the values of \(A\) and \(B\), giving your answers correct to 2 significant figures. When \(t = 4 , V = 14.5\) and when \(t = 5 , V = 14.9\).
  2. Determine whether the model is a good fit for these data.
  3. Determine the acceleration of the car according to the model when \(t = 5\), giving your answer correct to 3 decimal places. The car continues to accelerate until it reaches its maximum speed.
    The speed limit on this road is \(60 \mathrm { kmh } ^ { - 1 }\). All drivers who exceed this speed limit are recorded by a speed camera and automatically fined \(\pounds 100\).
  4. Determine whether, according to the model, the driver of this car is fined \(\pounds 100\).

Question 11(a):
AnswerMarks Guidance
\(A = 11\)B1 (1.1) from \(t=0\) and \(V=11\)
Substitution of \(t=3\) and \(V=13.8\)M1 (3.3) \(13.8 = A + B(1 - e^{-0.17 \times 3})\); implied by 7.0086
\(B = 7.0\)A1 (1.1) [3] Allow 7, 7.01, 7.009, etc
Question 11(b):
AnswerMarks Guidance
\(t=4\) and \(V=14.453681053\)... to 1 or more dp good fitB1 (3.4) allow SC1 for two correct values with no comment or incorrect comment(s); allow full marks if more accurate value of \(B\) used: 14.458 and 15.013
\(t=5\) and \(V=15.008095476\).. to 1 or more dp good fitB1 (1.1) [2] If B0B0, allow sc M1 for subst with their values
Question 11(c):
AnswerMarks Guidance
\(\frac{dV}{dt} =\) their \(7.0 \times 0.17\, e^{-0.85}\)M1 (3.1a) Allow for their \(7.0 \times 0.17\, e^{-0.17t}\); condone use of 7.0086 for \(B\)
\(0.509\ \text{ms}^{-2}\)A1 (1.1) [2]
Question 11(d):
AnswerMarks Guidance
\(t \to \infty,\quad V \to 11 + 7\) oeM1 (3.3) Allow ft from (a) for M1M1
their \(18 \times 3.6\)M1 (1.1) Allow for any attempt to change to km/hr, accept use of 'their 18'
\(64.8 > 60\) so the motorist is finedA1 (3.5a) [3] Need to show \(64.8 > 60\) for full marks
## Question 11(a):

$A = 11$ | B1 (1.1) | from $t=0$ and $V=11$

Substitution of $t=3$ and $V=13.8$ | M1 (3.3) | $13.8 = A + B(1 - e^{-0.17 \times 3})$; implied by 7.0086

$B = 7.0$ | A1 (1.1) **[3]** | Allow 7, 7.01, 7.009, etc

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## Question 11(b):

$t=4$ and $V=14.453681053$... to 1 or more dp good fit | B1 (3.4) | allow SC1 for two correct values with no comment or incorrect comment(s); allow full marks if more accurate value of $B$ used: 14.458 and 15.013

$t=5$ and $V=15.008095476$.. to 1 or more dp good fit | B1 (1.1) **[2]** | If B0B0, allow sc M1 for subst with their values

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## Question 11(c):

$\frac{dV}{dt} =$ their $7.0 \times 0.17\, e^{-0.85}$ | M1 (3.1a) | Allow for their $7.0 \times 0.17\, e^{-0.17t}$; condone use of 7.0086 for $B$

$0.509\ \text{ms}^{-2}$ | A1 (1.1) **[2]** |

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## Question 11(d):

$t \to \infty,\quad V \to 11 + 7$ oe | M1 (3.3) | Allow ft from (a) for M1M1

their $18 \times 3.6$ | M1 (1.1) | Allow for any attempt to change to km/hr, accept use of 'their 18'

$64.8 > 60$ so the motorist is fined | A1 (3.5a) **[3]** | Need to show $64.8 > 60$ for full marks
11 A car is travelling along a stretch of road at a steady speed of $11 \mathrm {~ms} ^ { - 1 }$.\\
The driver accelerates, and $t$ seconds after starting to accelerate the speed of the car, $V$, is modelled by the formula\\
$\mathrm { V } = \mathrm { A } + \mathrm { B } \left( 1 - \mathrm { e } ^ { - 0.17 \mathrm { t } } \right)$.\\
When $t = 3 , V = 13.8$.
\begin{enumerate}[label=(\alph*)]
\item Find the values of $A$ and $B$, giving your answers correct to 2 significant figures.

When $t = 4 , V = 14.5$ and when $t = 5 , V = 14.9$.
\item Determine whether the model is a good fit for these data.
\item Determine the acceleration of the car according to the model when $t = 5$, giving your answer correct to 3 decimal places.

The car continues to accelerate until it reaches its maximum speed.\\
The speed limit on this road is $60 \mathrm { kmh } ^ { - 1 }$. All drivers who exceed this speed limit are recorded by a speed camera and automatically fined $\pounds 100$.
\item Determine whether, according to the model, the driver of this car is fined $\pounds 100$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI AS Paper 2 2020 Q11 [10]}}