1. Extension II Mathematics: Mechanics
2012 HSC – Resisted Motion
Brett M. Bujeya
John Paul College
June 23, 2013
This work is licensed under the Creative Commons 3.0 License.
Brett M. Bujeya 2012 HSC - Resisted Motion June 23, 2013 1 / 7

2. 2012 HSC, Question 13Question 13 (15 marks) Use a SEPARATE writing booklet.
(a) An object on the surface of a liquid is released at time t = 0 and immediately
sinks. Let x be its displacement in metres in a downward direction from the
surface at time t seconds.
The equation of motion is given by
2
dv v
= 10 − ,
dt 40
where v is the velocity of the object.
(i)
(ii)
(iii)
t
20 e − 1( )Show that v = .t
e + 1
dv dv ⎛ 400
Use = v to show that x = 20logedt dx ⎝⎜
400 − v
How far does the object sink in the first 4 seconds?
⎞
⎠⎟ .
4
2
2
2
Brett M. Bujeya 2012 HSC - Resisted Motion June 23, 2013 2 / 7

3. 2012 HSC, Question 13(i) – Solution
dv
dt
= 10 −
v2
40
dv
dt
=
400 − v2
40
40
v
0
dv
400 − v2
=
t
0
dt
Using partial fractions,
40
dv
400 − v2
= 40
dv
(20 + v)(20 − v)
=
1
20 + v
+
1
20 − v
dv
∴
t
0
dt =
v
0
1
20 + v
+
1
20 − v
dv
t = ln
20 + v
20 − v
v
0
∴ t = ln
20 + v
20 − v
et
=
20 + v
20 − v
20(et
− 1) = v(1 + et
)
v =
20(et
− 1)
1 + et
Brett M. Bujeya 2012 HSC - Resisted Motion June 23, 2013 3 / 7

4. 2012 HSC, Question 13(ii) – Solution
v
dv
dx
= 10 −
v2
40
v
dv
dx
=
400 − v2
40
x
0
dx = 40
v
0
v
400 − v2
dv
x = −20 ln(400 − v2
)
v
0
x = −20 ln
400 − v2
400
∴ x = 20 ln
400
400 − v2
Brett M. Bujeya 2012 HSC - Resisted Motion June 23, 2013 4 / 7

5. 2012 HSC, Question 13(iii) – Solution
When t = 0, v = 0 and x = 0
When t = 4, v =
20(e4
− 1)
e4 + 1
and
x = 20 ln

 400
400 − 400(e4−1)2
(e4+1)2

 = 20 ln
e4
+ 1
2
4e4
Hence, the particle moves 20 ln
e4
+ 1
2
4e4
metres in the first 4 seconds.
Brett M. Bujeya 2012 HSC - Resisted Motion June 23, 2013 5 / 7

6. Notes From the Marking Centre I
i. In better responses, after finding the expression
dt
dv
=
40
400 − v2
,
candidates used partial fractions and correct integration to obtain
t = ln
20 + v
20 − v
. In better responses, candidates demonstrated their
competency by changing the subject to obtain the required expression for
v. In weaker responses, some candidates attempted to use the result for
dx
a2 − x2
without resorting to partial fractions. In many of these
responses, candidates did not obtain the correct result of this integration.
ii. Having obtained
dx
dv
=
40v
400 − v2
, the majority of candidates handled this
part well and obtained the required result.
Brett M. Bujeya 2012 HSC - Resisted Motion June 23, 2013 6 / 7

7. Notes From the Marking Centre II
iii. This part was again well done. The most successful approach was to
substitute t = 4 into v =
20(et
− 1)
et + 1
and then substitute this expression
into x = 20 ln
400
400 − v2
.
Another approach was to integrate using
dx
dt
=
20(et
− 1)
et + 1
. Only a small
number of candidates used this approach, and of them very few
successfully obtained the required result.
Brett M. Bujeya 2012 HSC - Resisted Motion June 23, 2013 7 / 7