Question 16 - A-Level Maths

AQA Paper 2 2019 June — Question 16 16 marks

Exam Board	AQA
Module	Paper 2 (Paper 2)
Year	2019
Session	June
Marks	16
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Variable acceleration (1D)
Type	Displacement from velocity by integration
Difficulty	Standard +0.8 This question requires finding a maximum by differentiating an exponential function, integrating exponentials with different constants, and then using the integral to evaluate distance at t=9.8. While the techniques are standard A-level (differentiation, integration, second derivative test), the exponential expressions are more complex than typical textbook exercises, requiring careful algebraic manipulation and numerical evaluation. The multi-step nature and need to handle multiple exponential terms elevates this above average difficulty.
Spec	1.07n Stationary points: find maxima, minima using derivatives 1.08a Fundamental theorem of calculus: integration as reverse of differentiation 3.02f Non-uniform acceleration: using differentiation and integration 3.02g Two-dimensional variable acceleration

An elite athlete runs in a straight line to complete a 100-metre race. During the race, the athlete's velocity, $v \text{ m s}^{-1}$, may be modelled by $$v = 11.71 - 11.68e^{-0.9t} - 0.03e^{0.3t}$$ where $t$ is the time, in seconds, after the starting pistol is fired.

Find the maximum value of $v$, giving your answer to one decimal place. Fully justify your answer. [8 marks]
Find an expression for the distance run in terms of $t$. [6 marks]
The athlete's actual time for this race is 9.8 seconds. Comment on the accuracy of the model. [2 marks]

Show mark scheme Show mark scheme source

Question 16:

Answer	Marks
16(a)	Differentiates to obtain with at least

𝑑𝑑𝑑𝑑

one exponent term correct

Answer	Marks	Guidance
𝑑𝑑𝑑𝑑	3.4	M1

= 10.512𝑆𝑆 −0.009𝑆𝑆

M𝑆𝑆a𝑡𝑡ximum occurs when

𝑑𝑑𝑑𝑑

𝑣𝑣 𝑑𝑑𝑑𝑑 = 0

−0.9𝑑𝑑 0.3𝑑𝑑

10.512𝑆𝑆 −0.009𝑆𝑆 = 0

𝑡𝑡 =5 .886

−0. 9 × 5.886

𝑣𝑣 =11.71−11 . 6 8 𝑆𝑆

0. 3 × 5 .886

−0.03𝑆𝑆

𝑣𝑣 = 11.5

This is the maximum value as it is

the only value which relates to

𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 = 0

Obtains fully correct expression

for

Answer	Marks	Guidance
𝑑𝑑𝑑𝑑	1.1b	A1

Exp𝑑𝑑la𝑑𝑑ins that maximum occurs when

𝑣𝑣

𝑑𝑑𝑑𝑑

Accept reference to stationary point

Answer	Marks	Guidance
𝑑𝑑𝑑𝑑 = 0	2.4	E1

Forms equation and solves to

find a value for t 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 = 0

Answer	Marks	Guidance
PI by correct t	1.1a	M1

Obtains correct value of

AWRT 5.9

Answer	Marks	Guidance
𝑡𝑡	1.1b	A1

Substitutes their into the given model

PI by correct v

Answer	Marks	Guidance
𝑡𝑡	1.1b	M1

Finds value for maximum

Answer	Marks	Guidance
AWRT 11.5	1.1b	A1

𝑣𝑣

Justifies final answer as being a

maximum value eg:

• This is the maximum value as

it is the only value which

relates to

• Evaluates 𝑑𝑑 s 𝑑𝑑 econd derivative at

𝑑𝑑𝑑𝑑 = 0

t = 5.9 where

d2v

=−9.4608e−0.9t −0.0027e0.3t

dt2

obtaining correct value of

-0.063 or explains both terms

are negative so it is less than 0

• Tests first derivative

considering gradient either side

of =5.9

• Sketches curve with maximum

ide𝑡𝑡ntified at

(5.9 , 11.5)

CSO

NB R1 can be awarded even if E1 was

Answer	Marks	Guidance
not awarded	2.4	R1

Answer	Marks	Guidance
16(b)	Integrates at least one term correct	3.4

−0.9𝑑𝑑 0.3𝑑𝑑

𝑠𝑠 = 11.71𝑡𝑡+12.978𝑆𝑆 −0.1𝑆𝑆

+𝑐𝑐

when

𝑠𝑠 = 0 𝑡𝑡 = 0

Distance =𝑐𝑐 = −12.878

−0.9𝑑𝑑

11.71𝑡𝑡+12.978𝑆𝑆

0.3𝑑𝑑

−0.1𝑆𝑆 −12.878

Answer	Marks	Guidance
Integrates at least two terms correct	1.1a	M1

Obtains a fully correct integrated

Answer	Marks	Guidance
expression including a constant	1.1b	A1

Interprets initial conditions - states

when

Answer	Marks	Guidance
PI by substitution of correct values	3.4	B1

S𝑠𝑠u=bs0titutes 𝑡𝑡 =0 and to find

their constant – must be clear

evidence of 𝑠𝑠su=b0stitutio𝑡𝑡n= se0en if

Answer	Marks	Guidance
incorrect c obtained	1.1a	M1

Obtains fully correct expression for

distance – coefficients can be in any

form and do not have to be

evaluated as a single decimal

Answer	Marks	Guidance
ACF	3.2a	A1

Answer	Marks
16(c)	Substitutes into their

expression for distance to find s

PI by sight o𝑡𝑡f =999..989 m for s

or

Substitutes s = 100 into their

expression for distance to find t

Answer	Marks	Guidance
PI by sight of 9.801 for t	1.1a	M1

Model predic𝑠𝑠ts= d9is9t.a9n9ce to be 99.99

which is very near to 100

Accurate

Compares s value with 100 metres

or

t value with 9.8 and concludes that it

is a good model

Answer	Marks	Guidance
CAO	3.5a	A1
Total	16
Q	Marking Instructions	AO

Question 16:
--- 16(a) ---
16(a) | Differentiates to obtain with at least
𝑑𝑑𝑑𝑑
one exponent term correct
𝑑𝑑𝑑𝑑 | 3.4 | M1 | 𝑆𝑆𝑣𝑣 −0.9𝑑𝑑 0.3𝑑𝑑
= 10.512𝑆𝑆 −0.009𝑆𝑆
M𝑆𝑆a𝑡𝑡ximum occurs when
𝑑𝑑𝑑𝑑
𝑣𝑣 𝑑𝑑𝑑𝑑 = 0
−0.9𝑑𝑑 0.3𝑑𝑑
10.512𝑆𝑆 −0.009𝑆𝑆 = 0
𝑡𝑡 =5 .886
−0. 9 × 5.886
𝑣𝑣 =11.71−11 . 6 8 𝑆𝑆
0. 3 × 5 .886
−0.03𝑆𝑆
𝑣𝑣 = 11.5
This is the maximum value as it is
the only value which relates to
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 = 0
Obtains fully correct expression
for
𝑑𝑑𝑑𝑑 | 1.1b | A1
Exp𝑑𝑑la𝑑𝑑ins that maximum occurs when
𝑣𝑣
𝑑𝑑𝑑𝑑
Accept reference to stationary point
𝑑𝑑𝑑𝑑 = 0 | 2.4 | E1
Forms equation and solves to
find a value for t 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 = 0
PI by correct t | 1.1a | M1
Obtains correct value of
AWRT 5.9
𝑡𝑡 | 1.1b | A1
Substitutes their into the given model
PI by correct v
𝑡𝑡 | 1.1b | M1
Finds value for maximum
AWRT 11.5 | 1.1b | A1
𝑣𝑣
Justifies final answer as being a
maximum value eg:
• This is the maximum value as
it is the only value which
relates to
• Evaluates 𝑑𝑑 s 𝑑𝑑 econd derivative at
𝑑𝑑𝑑𝑑 = 0
t = 5.9 where
d2v
=−9.4608e−0.9t −0.0027e0.3t
dt2
obtaining correct value of
-0.063 or explains both terms
are negative so it is less than 0
• Tests first derivative
considering gradient either side
of =5.9
• Sketches curve with maximum
ide𝑡𝑡ntified at
(5.9 , 11.5)
CSO
NB R1 can be awarded even if E1 was
not awarded | 2.4 | R1
--- 16(b) ---
16(b) | Integrates at least one term correct | 3.4 | M1 | 𝑠𝑠 = �𝑣𝑣 𝑆𝑆𝑡𝑡
−0.9𝑑𝑑 0.3𝑑𝑑
𝑠𝑠 = 11.71𝑡𝑡+12.978𝑆𝑆 −0.1𝑆𝑆
+𝑐𝑐
when
𝑠𝑠 = 0 𝑡𝑡 = 0
Distance =𝑐𝑐 = −12.878
−0.9𝑑𝑑
11.71𝑡𝑡+12.978𝑆𝑆
0.3𝑑𝑑
−0.1𝑆𝑆 −12.878
Integrates at least two terms correct | 1.1a | M1
Obtains a fully correct integrated
expression including a constant | 1.1b | A1
Interprets initial conditions - states
when
PI by substitution of correct values | 3.4 | B1
S𝑠𝑠u=bs0titutes 𝑡𝑡 =0 and to find
their constant – must be clear
evidence of 𝑠𝑠su=b0stitutio𝑡𝑡n= se0en if
incorrect c obtained | 1.1a | M1
Obtains fully correct expression for
distance – coefficients can be in any
form and do not have to be
evaluated as a single decimal
ACF | 3.2a | A1
--- 16(c) ---
16(c) | Substitutes into their
expression for distance to find s
PI by sight o𝑡𝑡f =999..989 m for s
or
Substitutes s = 100 into their
expression for distance to find t
PI by sight of 9.801 for t | 1.1a | M1 | m
Model predic𝑠𝑠ts= d9is9t.a9n9ce to be 99.99
which is very near to 100
Accurate
Compares s value with 100 metres
or
t value with 9.8 and concludes that it
is a good model
CAO | 3.5a | A1
Total | 16
Q | Marking Instructions | AO | Marks | Typical Solution

Show LaTeX source

An elite athlete runs in a straight line to complete a 100-metre race.

During the race, the athlete's velocity, $v \text{ m s}^{-1}$, may be modelled by
$$v = 11.71 - 11.68e^{-0.9t} - 0.03e^{0.3t}$$

where $t$ is the time, in seconds, after the starting pistol is fired.

\begin{enumerate}[label=(\alph*)]
\item Find the maximum value of $v$, giving your answer to one decimal place.

Fully justify your answer. [8 marks]

\item Find an expression for the distance run in terms of $t$. [6 marks]

\item The athlete's actual time for this race is 9.8 seconds.

Comment on the accuracy of the model. [2 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 2 2019 Q16 [16]}}

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Q1 1 Q2 1 Q3 1 Q4 4 Q5 7 Q6 6 Q7 10 Q8 11 Q9 9 Q10 1 Q11 1 Q12 1 Q13 5 Q14 8 Q15 9 Q16 16 Q17 9