Question 17 - A-Level Maths

OCR MEI Further Pure Core 2019 June — Question 17 22 marks

Exam Board	OCR MEI
Module	Further Pure Core (Further Pure Core)
Year	2019
Session	June
Marks	22
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	First order differential equations (integrating factor)
Type	Applied/modelling contexts
Difficulty	Challenging +1.2 This is a structured multi-part differential equations question with clear guidance at each step. While it requires integrating factor technique and involves a realistic modelling context, the question provides the differential equations explicitly (parts b and d say 'show that'), gives the solution form in part (e)(i), and breaks the problem into manageable steps. The integrating factor method is standard Further Maths content, and students are heavily scaffolded through what could otherwise be a challenging applied problem.
Spec	4.10b Model with differential equations: kinematics and other contexts 4.10c Integrating factor: first order equations

17 A cyclist accelerates from rest for 5 seconds then brakes for 5 seconds, coming to rest at the end of the 10 seconds. The total mass of the cycle and rider is \(m \mathrm {~kg}\), and at time \(t\) seconds, for \(0 \leqslant t \leqslant 10\), the cyclist's velocity is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). A resistance to motion, modelled by a force of magnitude 0.1 mvN , acts on the cyclist during the whole 10 seconds.

Explain why modelling the resistance to motion in this way is likely to be more realistic than assuming this force is constant. During the braking phase of the motion, for \(5 \leqslant t \leqslant 10\), the brakes apply an additional constant resistance force of magnitude \(2 m \mathrm {~N}\) and the cyclist does not provide any driving force.
Show that, for \(5 \leqslant t \leqslant 10 , \frac { \mathrm {~d} v } { \mathrm {~d} t } + 0.1 v = - 2\).
1. Solve the differential equation in part (b).
2. Hence find the velocity of the cyclist when \(t = 5\). During the acceleration phase ( \(0 \leqslant t \leqslant 5\) ), the cyclist applies a driving force of magnitude directly proportional to \(t\).
Show that, for \(0 \leqslant t \leqslant 5 , \frac { \mathrm {~d} v } { \mathrm {~d} t } + 0.1 v = \lambda t\), where \(\lambda\) is a positive constant.
1. Show by integration that, for \(0 \leqslant t \leqslant 5 , v = 10 \lambda \left( t - 10 + 10 \mathrm { e } ^ { - 0.1 t } \right)\).
2. Hence find \(\lambda\).
Find the total distance, to the nearest metre, travelled by the cyclist during the motion.

Show mark scheme Show mark scheme source

Question 17:

Answer	Marks	Guidance
17	(a)	The resistance force is likely to increase with
velocity.	B1
[1]	3.5b	allow ‘proportional to’,

‘varies with’

Answer	Marks	Guidance
17	(b)	dv dv

m 2m0.1mv  0.1v2

Answer	Marks	Guidance
dt dt	B1
[1]	3.3	[by Newton’s 2nd Law]
17	(c)	(i)

 d  ve0.1t  2e0.1t

Answer	Marks	Guidance
dt	M1	M1
 ve0.1t 2e0.1tdt 20e0.1tc	A1	1.1b
when t = 10, v = 0  c = 20e	M1	3.1b
 v = 20(e1–0.1t – 1)	A1cao	3.4

Alternative solution

dv

 dt

20.1v

M1

Answer	Marks	Guidance
 10ln(20.1v)tc	A1	A1
When t = 10, v = 0  c = 10ln 2 + 10	M1	substituting t=10, v = 0

 ln(2+0.1v) = ln 2 + 1 – 0.1t

Answer	Marks	Guidance
2 + 0.1v = e ln 2 + 1 – 0.1t = 2e1 – 0.1t	M1	anti-logging
 v = 20(e1 – 0.1t – 1)	A1cao

Alternative solution

Answer	Marks	Guidance
AE  + 0.1 = 0  cf v = Ae0.1t	M1
PI v = k  k = 20	B1
GS v = Ae0.1t  20	A1
When t = 0. v = 10: 0 =Ae120  A = 20e	M1	substituting t=10, v = 0
 v = 20(e1–0.1t – 1)	A1cao

[5]

Answer	Marks	Guidance
17	(c)	(ii)
[1]	3.5a	13 or better
17	(d)	dv dv c

m ct0.1mv  0.1vtwhere

Answer	Marks	Guidance
dt dt m	B1
[1]	3.3	[by Newton’s 2nd Law]
17	(e)	(i)

ve0.1t te0.1t

Answer	Marks	Guidance
dt	M1	2.1
 ve0.1t te0.1tdt10te0.1t10e0.1tdt	M1	2.1
ve0.1t 10te0.1t100e0.1tc	A1	2.1
When t = 0, v = 0  c = 100λ	M1	3.1b
 v10(t1010e 0.1t)*	A1	A1

Alternative solution

Answer	Marks
CF v = Ae0.1t	M1
PI v = Ct + D	M1
C + 0.1(Ct + D)=t  C = 10, D = 100	A1

GS v =Ae0.1t + 10t  100

Answer	Marks	Guidance
0=A100  A = 100	M1	substituting t = 0, v = 0
v10(t1010e 0.1t)*	v10(t1010e 0.1t)*	A1cao

[5]

Answer	Marks	Guidance
17	(e)	(ii)

their v when t = 5

Answer	Marks	Guidance
 λ = 1.218	A1	1.1b

[2]

Answer	Marks	Guidance
17	(f)	5

s  10(t1010e 0.1t)dt

1

Answer	Marks	Guidance
0	M1	3.1b

5 1 

10 t2 10t100e0.1t

 

2 

Answer	Marks	Guidance
0	B1	1.1b

t210t100e 0.1t

 

2 

Answer	Marks	Guidance
12.18(12.550100e0.5)22.49(m)	A1	1.1b

10 s  20(e10.1t1)dt

2

Answer	Marks	Guidance
5	M1	3.1b

and 10

10 20 10e10.1tt

 

5

Answer	Marks	Guidance
20(1510e0.5)29.74(m)	A1	1.1b
Total distance = 52 m	A1cao	3.2b

[6]

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CB2 8EA

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Telephone: 01223 552552

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© OCR 2019

Question 17:
17 | (a) | The resistance force is likely to increase with
velocity. | B1
[1] | 3.5b | allow ‘proportional to’,
‘varies with’
17 | (b) | dv dv
m 2m0.1mv  0.1v2
dt dt | B1
[1] | 3.3 | [by Newton’s 2nd Law]
17 | (c) | (i) | IF e0.1t | M1 | 1.1a | must be correct
 d  ve0.1t  2e0.1t
dt | M1 | M1 | 1.1b | 1.1b
 ve0.1t 2e0.1tdt 20e0.1tc | A1 | 1.1b
when t = 10, v = 0  c = 20e | M1 | 3.1b | substituting t=10, v = 0
 v = 20(e1–0.1t – 1) | A1cao | 3.4 | or 54.4e–0.1t – 20
Alternative solution
dv
 dt
20.1v
M1
 10ln(20.1v)tc | A1 | A1
When t = 10, v = 0  c = 10ln 2 + 10 | M1 | substituting t=10, v = 0
 ln(2+0.1v) = ln 2 + 1 – 0.1t
2 + 0.1v = e ln 2 + 1 – 0.1t = 2e1 – 0.1t | M1 | anti-logging
 v = 20(e1 – 0.1t – 1) | A1cao
Alternative solution
AE  + 0.1 = 0  cf v = Ae0.1t | M1
PI v = k  k = 20 | B1
GS v = Ae0.1t  20 | A1
When t = 0. v = 10: 0 =Ae120  A = 20e | M1 | substituting t=10, v = 0
 v = 20(e1–0.1t – 1) | A1cao
[5]
17 | (c) | (ii) | When t = 5, v = 20(e0.5 – 1) = 12.97 m s–1. | B1
[1] | 3.5a | 13 or better
17 | (d) | dv dv c
m ct0.1mv  0.1vtwhere
dt dt m | B1
[1] | 3.3 | [by Newton’s 2nd Law]
17 | (e) | (i) | d  
ve0.1t te0.1t
dt | M1 | 2.1
 ve0.1t te0.1tdt10te0.1t10e0.1tdt | M1 | 2.1 | integrating by parts
ve0.1t 10te0.1t100e0.1tc | A1 | 2.1
When t = 0, v = 0  c = 100λ | M1 | 3.1b | substituting t = 0, v = 0
 v10(t1010e 0.1t)* | A1 | A1 | 2.1 | 2.1 | NB AG | NB AG
Alternative solution
CF v = Ae0.1t | M1
PI v = Ct + D | M1
C + 0.1(Ct + D)=t  C = 10, D = 100 | A1
GS v =Ae0.1t + 10t  100
0=A100  A = 100 | M1 | substituting t = 0, v = 0
v10(t1010e 0.1t)* | v10(t1010e 0.1t)* | A1cao | NB AG | NB AG
[5]
17 | (e) | (ii) | When t = 5, 20(e0.5 – 1) = 10λ(10e–0.5 – 5) | M1 | 3.1b | subst t = 5 and equating to
their v when t = 5
 λ = 1.218 | A1 | 1.1b | 1.2 or better
[2]
17 | (f) | 5
s  10(t1010e 0.1t)dt
1
0 | M1 | 3.1b | integrating v between 0, 5
5
1 
10 t2 10t100e0.1t
 
2 
0 | B1 | 1.1b | 1 
t210t100e 0.1t
 
2 
12.18(12.550100e0.5)22.49(m) | A1 | 1.1b | art 22.5 (soi) | art 22.5 (soi)
10
s  20(e10.1t1)dt
2
5 | M1 | 3.1b | integrating their v between 5
and 10
10
20 10e10.1tt
 
5
20(1510e0.5)29.74(m) | A1 | 1.1b | art 29.7 (soi)
Total distance = 52 m | A1cao | 3.2b
[6]
PPMMTT
OCR (Oxford Cambridge and RSA Examinations)
The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
Education and Learning
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
www.ocr.org.uk
For staff training purposes and as part of our quality assurance
programme your call may be recorded or monitored
Oxford Cambridge and RSA Examinations
is a Company Limited by Guarantee
Registered in England
Registered Office; The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA
Registered Company Number: 3484466
OCR is an exempt Charity
OCR (Oxford Cambridge and RSA Examinations)
Head office
Telephone: 01223 552552
Facsimile: 01223 552553
© OCR 2019

Show LaTeX source

17 A cyclist accelerates from rest for 5 seconds then brakes for 5 seconds, coming to rest at the end of the 10 seconds. The total mass of the cycle and rider is $m \mathrm {~kg}$, and at time $t$ seconds, for $0 \leqslant t \leqslant 10$, the cyclist's velocity is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$.

A resistance to motion, modelled by a force of magnitude 0.1 mvN , acts on the cyclist during the whole 10 seconds.
\begin{enumerate}[label=(\alph*)]
\item Explain why modelling the resistance to motion in this way is likely to be more realistic than assuming this force is constant.

During the braking phase of the motion, for $5 \leqslant t \leqslant 10$, the brakes apply an additional constant resistance force of magnitude $2 m \mathrm {~N}$ and the cyclist does not provide any driving force.
\item Show that, for $5 \leqslant t \leqslant 10 , \frac { \mathrm {~d} v } { \mathrm {~d} t } + 0.1 v = - 2$.
\item \begin{enumerate}[label=(\roman*)]
\item Solve the differential equation in part (b).
\item Hence find the velocity of the cyclist when $t = 5$.

During the acceleration phase ( $0 \leqslant t \leqslant 5$ ), the cyclist applies a driving force of magnitude directly proportional to $t$.
\end{enumerate}\item Show that, for $0 \leqslant t \leqslant 5 , \frac { \mathrm {~d} v } { \mathrm {~d} t } + 0.1 v = \lambda t$, where $\lambda$ is a positive constant.
\item \begin{enumerate}[label=(\roman*)]
\item Show by integration that, for $0 \leqslant t \leqslant 5 , v = 10 \lambda \left( t - 10 + 10 \mathrm { e } ^ { - 0.1 t } \right)$.
\item Hence find $\lambda$.
\end{enumerate}\item Find the total distance, to the nearest metre, travelled by the cyclist during the motion.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2019 Q17 [22]}}

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