M_BANK\YR12-2U\APPLICATIONSOFCALCULUS.CAT
Applications of Calculus to the Physical World
1)!
2)!
3)!
4)!
5)!
2U84-4iii
The rate at which water runs out of a tank is proportional to the volume of water in the tank,
dV
ie.
= kV. The tank is full to start with and has a capacity of 36 000 litres.
dt
a.
Show that V = V0ekt satisfies this equation where V0 is the volume of water in the
tank initially.
1
b.
If of the water in the tank runs out in 30 minutes, find the volume of water
4
remaining in the tank after 60 minutes.†
« a) Proof b) 20 250 litres »
2U84-8ii
A particle moves in a straight line such that its displacement, x cm, after t seconds is given by:
x = t3  6t2 + 9t + 4.
a.
When and where does the particle first change direction?
b.
What is the average speed of the particle in the first second?†
« a) 1 second, 8 cm b) 4 cm/s »
2U84-9ii
NOT TO

SCALE
R
R
If the area of this sector is 625 m2, find the values of R and  (to the nearest degree) so that
the sector has minimum perimeter.†
« 25 m, 2 radians »
2U85-8i
A train runs between 2 stations stopping at each. Its velocity, v km/min, t minutes after
1
leaving the first station is given by: v  (4  t ) . Find:
5
a.
the time taken to travel between the two stations;
b.
the maximum velocity attained;
c.
the distance between the two stations.†
4
2
km »
« a) 4 minutes b) km/min c) 2
5
15
2U85-9i
2
2·5
PRINTED AREA
1
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6)!
7)!
The area of a rectangular sheet of paper is 300 cm2. A margin of 2 cm is allowed at the top,
1cm at the bottom and 2·5 cm on each edge. What are the exact dimensions of the printed
area if it is to be a maximum?†
« (10 5  5) cm  (6 5  3) cm »
2U85-10i
dV
1
The rate at which liquid is flowing into a vessel after t minutes is given by
. If

dt t  1
(loge2) m3 of liquid flows into the vessel after 3 minutes, how much liquid flows in after
8 minutes? Give your answer to 3 significant figures.†
« 1·50 m3 »
2U86-8iv
A farmer needs to construct 2 holding paddocks, one rectangular and the other triangular, for
goats and chickens. The figure shows that he uses an existing long fence as part of the
boundary.
Existing Fence
A
C
E
B
8)!
9)!
10)!
D
If he has only 440 metres of fencing, find the maximum total area of holding paddocks that
he can construct if AB = CD = DE and  CDE = 30.†
« 17 600 m2 »
2U86-9ii
A stone is projected vertically upwards from the top of a tower. If the height of the stone
above ground level is given by: h = 25 + 20t  5t2 where h is in metres and t is the time in
seconds.
a.
Find the initial velocity of projection.
b.
Calculate the time taken for the stone to hit the ground.
c.
Show the acceleration of the stone is constant. Explain.†
d 2h
« a) 20 m/s b) 5·74 seconds c) a  2   10 , which is independent of x. »
dt
2U87-9i
The size of a colony of insects is given by the equation: P = 5000 ekt where P is the
population after t days.
a.
If there are 8500 insects after 1 day, find the value of k correct to 1 decimal place.
b.
When will the colony triple in size? (Answer to the nearest day)
c.
What is the growth rate of the population after 2 days?†
« a) 0·5 b) 2 days c) 6796 insects/day »
2U87-9ii
A particle moves in such a way that its distance, x metres, from the origin after t seconds is
given by: x = 2  2 sin 2t for 0  t  2.
a.
Find an equation for its velocity after t seconds.
b.
Where is the particle initially and what is its velocity then?
c.
Describe the motion of the particle.†
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« a) x = 4 cos 2t b) 2 m, 4 m/s c) SHM, Amplitude = 2, centre = 2, Angular velocity = 2,
Phase = 0 »
11)! 2U87-10
The diagram shows a sector OAB of a circle, centre O and radius x metres. Arc AB subtends
an angle  radians at O. An equilateral triangle BCO adjoins the sector.†
O
x

C
A
i.
ii.
iii.
B
Write down expressions for the:
a.
area of the sector OAB;
b.
area of the triangle BCO;
c.
length of the arc AB.
Hence write down an expression for the:
a.
area
b.
perimeter
of the figure bounded by ABCO.
The perimeter of this figure is (12  2 3 ) metres.
a.
Express its area in terms of .
b.
For what value of  is the area a maximum?
c.
Show that the maximum area is (6  3 ) m2.†
1 2
3 2
(6  3 ) 2 (2θ  3 )
1 2
x c) x ii) a) x (2  3 ) b) x( + 3) iii) a)
« i) a) x  b)
2
4
4
(θ  3) 2
b) 3  3 c) Proof »
12)!
13)!
2U88-9i
A particle moving with a constant acceleration of 4 ms–2 starts from rest. Find the:
a.
time taken for the particle to attain a velocity of 22 ms–1;
b.
distance travelled by the particle in this time.†
« a) 5·5 seconds b) 60·5 m »
2U88-9ii
The diagram below is an artists entry in the ‘Designing a Bicentennial Flag’ competition. It is
to consist of two green rectangular regions with a yellow vertical stripe between these
regions.
y
a
b
Y
ALL LENGTHS
E
IN CM
L
x GREEN L GREEN
O
W
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14)!
15)!
16)!
17)!
The perimeter of the entire flag is 376 cm and the green regions cover an area of 6561 cm2.
Let the width of the vertical stripe be y cm and the dimensions of the green regions be as
shown in the diagram.
a.
Show that a + b = 188  x  y and hence show that the width of the yellow vertical
stripe is given by y = (188  x  6561x1) cm.
b.
Find the dimensions of this flag if the width of the yellow vertical stripe is to be a
maximum.†
« a) Proof b) 81 cm  26 cm »
2U88-10i
The diameter of a tree (D cm), t years after the start of a particular growth period is given by
D = 70ekt.
dD
a.
Show that
 kD where k is a constant.
dt
b.
If k = 0·15, how long will it take for the diameter of the tree to measure 74 cm.?
(Answer to the nearest integer.)†
« a) Proof b) 0·37 years »
2U89-8a
A stone dropped into a still pond causes circular ripples on the water surface. The area of the
disturbed water, t seconds after the stone hits the surface of the water is given by: A(t) = 4t2
where the area is measured in m2.
i.
Find an expression to represent the rate of change of the area of the disturbed water.
ii.
Find, in terms of , the rate of change of the area of the disturbed water after one
second.†
« i) 8t ii) 8 »
2U89-8b
Two particles, A and B, move along a straight line so that A’s displacement, in metres, from
the origin after t seconds, is given by xA = 2  2 sin 2t and B’s acceleration, in metres per
d 2 xB
second per second, is given by 2  12 cos 2t .
dt
i.
What is the acceleration of particle A after 3 seconds?
ii.
Which particle is moving faster after one second if particle B is initially at rest at
xB = 6.
iii.
Find the maximum displacement of particle A.†
« i) 2·24 m/s ii) B iii) 2 m »
2U89-9b
A railway enthusiast designs a miniature railway of length 1000 metres. The route consists of
two semicircles at opposite ends of a rectangle.
ym
xm
i.
If the rectangle has a length of y metres and its width is x metres, show that:
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y = 500 
ii.
iii.
18)!
πx
.
2
2000 x  πx 2
Show that the area, A, enclosed by the railway track is given by A =
.
4
Find the maximum area, to the nearest hectare, enclosed by the railway track.†
« i) ii) Proof iii) 8 hectares »
2U90-7b
A particle moves along a straight line so that it’s position, s(t) cm, at time, t mins, is given by:
s(t) = t3  12t2 + 36t.
i.
Find an expression for the velocity, v(t), and acceleration, a(t), of the particle at any
time t.
ii.
When is the particle at rest?
iii.
Copy this table into your answer sheet and complete it.
t
s(t)
V(t)
A(t)
0
0
2
32
4
16
6
0
8
32
Describe the motion of the particle.†
« i) v(t) = 3t2  24t + 36, a(t) = 6t  24 ii) t = 2, t = 6 iii)
t
0
2
4
6
8
s(t)
0
32
16
0
32
v(t) 36
0
36
12 0
a(t) 24 12
0
12 24
iv) The particle moves away from the origin for 2 minutes, stops and returns to the origin. It then
moves away from the origin with an increasing acceleration. »
19)! 2U90-9b
A wire of length 30 cm is cut into two pieces. One piece is bent to form an equilateral
triangle and the other to form a circle.
iv.
NOT TO
SCALE
20)!
If the piece of wire used to form the triangle is of length 3x cm:
i.
show that the sum of the areas of the two figures, A cm2, is given by:
3 2
9
A
x 
(10  x) 2 ;
4
4
ii.
find the length of wire that must be used to form the triangle so that the sum of the
areas of the two figures is a minimum.†
« i) Proof ii) 18·7 cm (to 1 d.p.) »
2U91-9a
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21)!
22)!
A particle moves in a straight line so that its acceleration, f m/s2 at time t seconds is given by
f = 3(4 + t)2. Initially the particle moves with a velocity of 64 m/s from a position 3 metres to
the right of O.
i.
Show that the velocity of the particle at any time t seconds is given by: v = (4 + t)3.
ii.
Hence find the displacement, x metres, of the particle as a function of time.
iii.
Find the distance travelled by the particle during the first 2 seconds.†
1
« i) Proof ii) x  (4  t ) 4  61 iii) 260 »
4
2U91-9b
Max brews his own beer in a barrel. From past experience, he knows that the amount of
sugar, M kg, that is present in a liquid kept at a constant temperature after t minutes, satisfies
the following equation: M = M0ekt where k is a constant.
i.
He places 10 kg of sugar in the barrel for his brew. After 5 minutes he finds that
only half of this sugar is present in the barrel. Find the value of k correct to
2 decimal places.
ii.
How long will it take before 1 kg of this sugar is present in this barrel? Give your
answer correct to the nearest minute.†
« i) k = 0·14 ii) 16 minutes »
2U91-10b
A farmer, who wishes to keep his animals separate, sets up his field so that fences exist at FC,
CD and BE, as shown in the diagram below. The side FD is beside a river and no fence is
needed there.
F
x
River
z
B
E
x
C
23)!
2z
D
B is in the middle of FC and CD is twice the length of BE.
i.
If FB = x metres and BE = z metres, write down the expressions in terms of x and z
for:
.
the area, A, of the field FCD.
.
the amount of fencing, L, that the farmer would need.
ii.
If the area of the field is 1200 m2, show that the length of fencing required is given
1800
metres .
by: L  2 x 
x
iii.
Hence find the values of x and z so that the farmer uses the minimum amount of
fencing.†
« i) ) A = 2xz ) L = 2x + 3z ii) Proof iii) x = 30, z = 20 »
2U92-8b
A particle moves so that its velocity, v metres per second, at any time t is given by: v = e2t.
Initially the particle is at x = 2.
i.
Find the acceleration, a, of the particle as a function of time.
ii.
What is the acceleration of the particle after 1 second?
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iii.
iv.
v.
Find an expression for the displacement, x metres, of the particle in terms of t.
Find the distance the particle travelled during the first two seconds.
Describe what happens to the velocity of the particle for large values of t.†
1
1
« i) a = 2e–2t ii) 2e–2 iii) x   e  2 r  2
iv) 2·5 m (to 1 d.p.) v) It approaches zero. »
2
2
24)! 2U92-10b
A circular stained glass window of radius 5 metres requires metal strips for support along
AB, DC and FG. O is the centre of the window.
A
F
y
x
B
5
O
D
G
C
i.
ii.
iii.
Copy the diagram and information onto your answer sheet.
If OF = OG = y metres and FB = x metres, find an expression for y in terms of x.
The total length of the strips of metal used for support (i.e. AB + DC + FG) is L
iv.
metres. Show that: L = 4x + 2 5  x 2 .
The window will have a maximum strength when the length of the supports is a
maximum. Show that when FB = 2 metres, the window will have a maximum
strength.†
« ii) y  5  x 2 iii) iv) Proof »
25)!
2U93-7d
The graph represents the velocity (v m/s) of a particle after t seconds. The particle is moving
in a straight line starting from rest.
V
m/s
(1, 20)
20
10
0
26)!
1
2
3
4
t secs
–10
(3, –10)
–20
i.
What is the velocity of the particle after 1 second?
ii.
What is the acceleration of the particle after 3 seconds?
iii.
When does the particle change directions?
iv.
Explain what is represented by the area of the shaded region in the diagram.
« i) 20 m/s ii) 0 m/s iii) 2 seconds iv) The distance travelled in the first 2 seconds. »
2U93-8c
In a chemical experiment, the amount of crystals, x grams, that dissolved in a solution after
t minutes was given by: x = 20(1  ekt).
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i.
27)!
After 3 minutes it was found that 10 grams of the crystals had dissolved. Show the
value of k correct to 3 significant figures was 0·231.
ii.
At what rate were the crystals dissolving after 5 minutes? Give your answer to the
nearest gram/minute.
« i) Proof ii) 1 gram/min »
2U93-10c
Boxes in the shape of rectangular prisms are to be constructed from special materials. The
width (x metres) of the base is to be half the length of the base and each box is to hold a
volume of 4 cubic metres.
y
28)!
29)!
2x
x
Material that is used to build the base and top costs $15 per m2. A cheaper material at
$10 per m2 is used for the four sides.
120
i.
Show that the total cost ($C) of building each box is given by: C  60 x 
.
x
ii.
What is the width of the base of the cheapest box that can be constructed?
« i) Proof ii) x = 1 m »
2U94-3d
When heat was applied to a metallic disc for t seconds, its area, A cm2, increased at a rate
dA
 t 2  2t  1 .
given by:
dt
i.
At what rate was the area of the metallic disc increasing at the end of the third
second?
ii.
Before heat was applied, the area of a metallic disc was 10 cm2. Heat was applied to
this metallic disc for 3 seconds. What was the area of this metallic disc at the end of
three seconds?†
« i) 4 cm2/s ii) 13 cm2 »
2U94-8c
The diagram shows the displacement, x metres, of a particle from the origin O after t seconds
for 0  t  2.
x
4
3
2
1

3
0

2 t
–1
2
2
–2
The equation of this motion is of the form x = 1 + A cos nt.
i.
Use the graph to find the values of A and n.
ii.
What is the initial displacement of the particle?
iii.
When is the particle at rest for the first time after the motion begins?
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iv.
30)!
31)!
At what times is the particle 1 metre to the right of the origin?†

 3 5 7
« i) A = 3, n = 2 ii) 4 m iii) t 
iv) t  ,
»
,
,
2
4 4
4
4
2U94-9a
An isotope of carbon, C14 decays at a rate proportional to the mass present. The rate of
dM
change is given by:
  kM where k is a positive constant and M is the mass present.
dt
i.
Show M = M0ekt is a solution to this equation.
ii.
The half-life of this isotope C14 is 5600 years. This means it takes 5600 years for
100 grams of C14 to decay to 50 grams. Find the value of k correct to 3 significant
figures.
iii.
Archaeoligists use radiocarbon dating to establish the age of discoveries. Calculate
the age of an item in which only one-eighth of the original carbon remains.†
« i) Proof ii) k = 1·24  104 iii) 16 800 years »
2U94-10b
The diagram shows the curve y  9  x 2 for x  0. P is the point (t , 9  t 2 ) on the curve
and M is the foot of the perpendicular drawn from P to the x axis.
y
3
P (t , 9  t 2 )
O
i.
ii.
iii.
iv.
32)!
x
3
M
Calculate the area bounded by the curve y  9  x 2 and the co-ordinate axes.
Write down an expression in terms of t for the area, A, of the triangle OPM.
Find the co-ordinates of the point P which gives triangle OPM a maximum area.
Show the ratio of the area of the triangle OPM found in part (iii) to the area bounded
by the curve and the co-ordinate axes is 1 : .†
3
3
9
1
« i)
units2 ii) t 9  t 2 iii) P (
,
) iv) Proof »
2
4
2
2
2U95-4c
After heavy rain the flood gate of a dam was opened. Water was released from the dam at a
rate of 31·8t litres per second where t is measured in seconds after the gate was opened.
i.
Explain why the total volume of water, V, released from the dam in k seconds can be
k
found by evaluating the expression V   31  8t dt .
0
ii.
33)!
9
Calculate the time it took for 10 litres of water to be released from the dam.
Express your answer to the nearest second.†
«i) Proof ii) 7 931 secs »
2U95-6b
The rabbit population P, in a park was growing exponentially and could be calculated using
the formula P = 500ekt where t was the time in weeks after the rabbits were first counted.
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i.
ii.
iii.
34)!
At the end of the first week there were 550 rabbits in the park. Calculate the value
of k correct to 3 decimal places.
How many weeks did it take for the rabbit population to reach 1500?
At what rate was the colony increasing after 2 weeks?†
« i) k = 0·095 ii) 11·53 weeks iii) 57·7 rabbits/week »
2U95-7c
A particle moves in a straight line. The diagrams below are the displacement and velocity
graphs of this particle during the first 9 seconds of its motion.
Displacement
x metres
0·95
1·00 2·00 3·00 4·00 5·00 6·00 7·00 8·00 9·00 t seconds
–0·95
Velocity
m/s
1·5
1·5
35)!
1·00 2·00 3·00 4·00 5·00 6·00 7·00 8·00 9·00 t seconds
Use these graphs to answer the following questions.
i.
Write down the initial displacement and velocity of the particle.
ii.
When does the particle change direction for the first time?
iii.
Where does the particle reach its maximum velocity?†
« i) x = 0·95 m, v = 0 m/s ii) t = 2 seconds iii) x = 0 metres »
2U95-9c
2
The diagram shows a rectangle PQRS where P and Q are on the curve y = e  x and R and S
are on the x axis. The point O is the origin and lengths OS and OR are equal.
y
1 y  e  x2
P
S
i.
ii.
Q
O
R
x
Let the length OR = x and show the area of the rectangle PQRS is represented by the
2
expression A = 2x e  x .
Find the value of x for which PQRS has a maximum area.†
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« i) Proof ii)
36)!
37)!
38)!
39)!
40)!
1
»
2
2U96-5b
A pool is being drained and the number of litres of water, L, in the pool at time t minutes is
given by the equation: L = 120(40  t)2.
i.
At what rate is the water draining out of the pool when t = 6 minutes?
ii.
How long will it take for the pool to be completely empty?†
« i) 8160 L/m ii) 40 minutes »
2U96-6c
An island was found to be contaminated by fallout from a nuclear test explosion. One of the
most dangerous components of nuclear fallout is strontium-90. The rate at which the amount,
dA
A, of strontium-90 decays t years after and explosion is given by:
= kA where k is a
dt
constant.
i.
Show that A = A0ekt, where A0 is the amount of strontium-90 present immediately
after and explosion, satisfies the given equation.
ii.
It was found that 28 years after the explosion there was half the original amount of
strontium-90 present on the island. Find the value of k correct to 4 decimal places.
iii.
Another explosion has left 99% of the original amount of strontium-90 present on an
island. How many months ago did this explosion occur?†
« i) Proof ii) 0·0248 iii) 5 months »
2U96-7a
The price of one tonne of copper, $P, was studied over a period of t years.
dP
 0 . What does this say about the price of
i.
Throughout the period of study
dt
copper?
ii.
It was also observed that the rate of change of the price of copper decreased over the
d 2P
period of study. What does this statement imply about
?†
dt 2
d2p
« i) It was increasing ii)
0 »
dt 2
2U96-7b
The position x cm, of a particle moving along an x-axis is given by x = 3t + e2t where t is the
time in seconds.
1
i.
What is the position of the particle when t  seconds ?
2
ii.
What is the initial velocity of the particle?
iii.
Show the initial acceleration of the particle is 4 cm/s2.
iv.
Explain why the particle will never come to rest.†
3 1
« i) (  ) cm ii) 1 cm/s iii) Proof iv) When v = 0, t < 0 »
2 e
2U96-10a
ABCDE is a pentagon of fixed perimeter P cm. In the figure triangle ABE is equilateral and
BCDE is a rectangle. The length of AB is x cm.
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A
x cm
E
B
NOT TO
SCALE
D
C
P  3x
Show that the length of BC is
cm.
2
i.
41)!
ii.
Show that the area of the pentagon is given by A =
iii.
Find the value of


1
2 Px  (6  3 ) x 2 cm2..
4
P
for which the area of the pentagon is a maximum.†
x
« i) ii) Proof iii) 6  3 »
2U96-10b
Two particles A and B are moving along a straight line so that their velocities in m/s at any
time t seconds are given by VA = t2 + 1 and VB = 7  2t respectively.
i.
Sketch, on the same co-ordinate axes, the velocity-time graph for both particles.
ii.
What is the significance of the point of intersection of the two graphs?
iii.
When have the two particles travelled exactly the same distance?†
V ms1
30
25
20
15
10
5
« i)
42)!
43)!
–5
–10
A
3·5
1
2
3
4
B
t secs
ii) The magnitude of the velocities (i.e. speed) is the same.
iii) t = 3 seconds »
2U97-6a
The population of a small country town is decreasing at an increasing rate. Given that N is
the population of the town at a given time t, what does this statement imply about
dN
d 2N
and
?†
dt
dt 2
d 2N
dN
 0 and
0 »
«
dt
dt 2
2U97-6b
During a week, the expected amount of time A minutes that a high school student spends on
Maths homework is determined by the equation: A = 20ekt where t is the year the student is
in. In Year 7, students are expected to spend a total of 120 minutes on Maths homework each
week.
i.
Find the value of k correct to 3 decimal places.
ii.
At what rate is the expected amount of time spent on Maths homework increasing
when a student is in Year 10? Answer correct to the nearest minute/year.
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iii.
44)!
How many more minutes is a Year 12 student expected to spend on Maths
homework than a Year 7 student. (Answer correct to the nearest integer)†
« i) 0·256 ii) 66 m/year iii) 312 mins »
2U97-7c
V m/s
1
2
3
0
t (in seconds)
The velocity-time graph of a particle starting from rest at the origin is shown in the diagram.
i.
When is the particle at rest again?
ii.
Does the particle change its direction during the motion? Give reasons.
iii.
Give the first time that the acceleration of the particle is zero.
iv.
Give a physical interpretation of the shaded area bounded by the t axis and the curve.
v.
Copy this diagram onto your answer sheet.
vi.
On this diagram, change the vertical axis to x and sketch the displacement-time
graph of the particle.†
« i) t = 2 secs ii) No. Since V is always positive or zero. iii) t = 1 sec iv) The displacement v)
V m/s
Displacement
x
vi)
45)!
0
1
2
3
t
»
2U97-9c
45
h
NOT TO
SCALE
45
1m
a
A trough of depth h metres and length 1 metre was constructed out of stainless steel sheeting.
The cross-section of the trough was an isosceles trapezium with the acute angles being
45 o each. The width of the bottom of the trough was a metres. The area of the cross-section
measured 60 m2.
60
h.
i.
Show that a 
h
ii.
Show that the amount of stainless steel, A in m2, required to construct the trough
60
 h  2 h 2  120 .
was given by A 
h
iii.
Find the depth of the trough, to the nearest mm, if the amount of stainless steel used
is kept to a minimum.†
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46)!
« i) ii) Proof iii) 5·728 m »
2U97-10b
Two particles P and Q are moving along a horizontal line. At any time t seconds, the position
of particle P is given by x = et and the position of particle Q is given by x = 1 + 6e–t. The
diagram below shows the position of particle P, x metres, at any time t seconds.
x
x = et
NOT TO
(in metres)
SCALE
1
0
i.
ii.
iii.
iv.
t (in seconds)
As time increases indefinitely, what position does the particle Q approach?
On a pair of coordinate axes, sketch the path of particle Q.
Calculate the position where the two particles meet.
Explain why P and Q will never travel at the same velocity.†
x
7
1
« i) Q approaches 1 ii)
47)!
t
0
iii) (3, ln 3) iv) There is no simultaneous solution
for the two particle paths »
2U98-6b
x
metres
NOT TO
SCALE
(4, 3)
0 1
6
10
(8, –1)
t
seconds
5
The graph shows the displacement, x metres from the origin, at any time t seconds, of a
particle moving in a straight line.
i.
Where was the particle initially?
ii.
When was the particle at the origin?
iii.
When was the particle at rest?
iv.
Estimate the time when the acceleration was zero?
v.
How far did the particle travel during the first 10 seconds?†
« i) 5 m to the left of the origin ii) t = 1, 6 and 10 secs iii) t = 4 and 8 secs iv) t = 5 secs v) 15 m »
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48)!
2U98-7b
The volume, V, of water in a dam at time t was monitored over a period of time. When
monitoring began, the dam was 75% full. During the monitoring period, the volume
decreased at an increasing rate due to a long period of drought.
dV
d 2V
and
i.
What does this tell us about
?
dt
dt 2
ii.
Sketch the graph of V against t.†
V
0·75
dV
d 2V
 0 , 2  0 ii)
« i)
dt
dt
t »
49)!
2U98-7c
A particle moves in a straight line such that its distance, x metres, from a fixed point O at any
time t seconds is given by: x = 5 + ln(1 + 2t).
i.
Find expressions for the velocity and acceleration of the particle at any time
t seconds.
ii.
When will the particle be 10 metres from O?
iii.
Find the velocity and acceleration of the particle when it is 10 metres from O.
iv.
Show that the particle does not change direction during its motion.†
e5  1
2
4
2
4
m/s ,
m/s 2
« i) Velocity =
, Acceleration =
ii)
secs iii)
t

2
21
441
1  2t
2
(1  2t )
iv) Proof »
50)! 2U98-9c
y metres
x metres
x metres
51)!
The diagram shows a rectangular paddock which has an area of 1 hectare (10 000 m2). The
paddock requires fencing around the perimeter and also along a circular arc in one corner.
The radius of the arc is half the width of the paddock.
i.
Show that the total amount of fencing required is given by
πx 10 000
L  4x 

metres, where x is the radius of the circular arc.
2
x
ii.
Hence show that when x is approximately 42·4 metres, the paddock will require the
least amount of fencing.†
« Proof »
2U98-10c
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52)!
53)!
54)!
As soon as advertising and other promotions of a particular product ended, the sales, S, in
thousands of units of the product fell at a rate proportional to the number of sales made, that
dS
is
  kS where k is a constant and the time, t, is measured in days after the advertising
dt
and promotions ended.
dS
i.
Show that S = Soekt satisfies the equation
  kS where So is the number of
dt
sales, in thousands of units, made at the end of the promotions period.
ii.
30 days after the promotions ended the number of sales had fallen by 50%. Find the
value of k.
iii.
The decision was made to stop making this product as soon as the sales fell to 15%
of the level at the end of the promotions. How many days after the promotions
ended did that occur?†
« i) Proof ii) 0·0231049 iii) 82 days »
2U99-8c
A particle moves along a straight line about a fixed point O so that its acceleration, a ms–2, at
π
time t seconds is given by a  8 cos (2t  ) . Initially the particle is moving to the right with
6
–1
velocity of 2 ms from a position 3 metres to the left of O.
i.
Find expressions for the velocity and position of the particle at any time t.
5π
ii.
Show that the particle changes directions when t 
seconds.
12
iii.
At what time does that particle return to its initial position for the first time?†
π
π
5π
secs »
« i) v  4 sin (2t  ), x   2 cos (2t  ) ii) Proof iii) t 
6
6
12
2U99-9b
The median house price, $P, in a certain suburb is falling at an increasing rate after a recent
dp
d2p
and 2 ?†
peak. What does this tell you about
dt
dt
dp
d2p
 0,
0 »
«
dt
dt 2
2U99-9c
NOT TO
SCALE
O
50 cm h cm
x cm
A
B
The diagram shows a cone of base radius r cm and height h cm inscribed in a sphere of radius
50 cm. The centre of the sphere in O and  OAB = 90. Let OA = x cm.
i.
Show that r  2500  x 2 .
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Hence show that the volume, V cm3, of the cone is given by:

V  (2500  x 2 ) (50  x) .
3
iii.
Find the radius of the largest cone which can be inscribed in the sphere. (Give your
answer to the nearest mm.)†
« i) ii) Proof iii) 471 mm »
2U99-10a
In a fish hatchery, the fish population, N, satisfies the equation N = Noekt where No and k are
constants and t is measured in months.
i.
Initially there were 1000 fish in the hatchery and at the end of 5 months there were
5000. Find the value of k correct to three decimal places.
ii.
Find the number of fish in the hatchery at the end of 8 months. (Give your answer
correct to the nearest hundred.)
iii.
At the end of which month will the fish population exceed 50 000 for the first time?
iv.
At what rate is the population increasing at the end of six months? (Give your
answer correct to the nearest hundred fish per month.)†
« i) 0·322 ii) 13 100 iii) 13 iv) 2200 fish/month »
2U00-7a
An experimental vaccine was injected into a cat. The amount, M millilitres, of vaccine
present in the bloodstream of the cat, t hours later was given by M = e–2t + 3.
i.
How much vaccine was initially injected into the cat?
ii.
At what rate was the amount of vaccine decreasing at the end of 3 hours?
iii.
Show that there will always be more than 3 millilitres of vaccine present in the cat’s
bloodstream.
iv.
Sketch the curve of M = e–2t + 3 to show how the amount of vaccine present in the
cat’s bloodstream changes over time.†
M
4
3
ii.
55)!
56)!
« i) 4 mL ii) 2e–6 mL/hr iii) Proof iv) 0
57)!
t
»
2U00-7b
A particle moves in a straight line. At times t seconds, its displacement, x metres from a fixed
point O on the line is given by x = 1 – cos  t,
i.
What is the initial displacement of the particle?
ii.
Sketch the graph of x as a function of t.
iii.
Find an expression for the velocity of the particle at any time t.
1
iv.
What is the velocity of the particle at time t  ?
6
v.
At what time does the particle first reach its maximum speed?†
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y
2
1
« i) x = 0 ii)
58)!
0
1
2
3
4
t
iii)  sin  t iv)

2
m/s v) t 
1
sec »
2
2U00-9b
The water’s edge is a straight line ABC which runs East-West. A lighthouse is 6 km due north
of A. 10 km due east of A is the general store. To get to the general store as quickly as
possible, the lighthouse keeper rows to a point B, x km from A, and then jogs to the general
store. The lighthouse keeper’s rowing speed is 6 km/h and his jogging speed is 10 km/h.
Show that it takes the lighthouse keeper
ii.
lighthouse to B.
Show that the total time taken for the lighthouse keeper to reach the general store is
iii.
iv.
59)!
60)!
36  x 2
hours to row from the
6
i.
x 2  36 10  x

given by: T 
hours.
6
10
1
Hence show that when x  4 km , the time it takes from the lighthouse keeper to
2
travel from the lighthouse to the general store is a minimum.
Hence find the quickest time it takes the lighthouse keeper to go to the general store
from the lighthouse. Give your answer correct to the nearest minute.†
« i) ii) iii) Proof iv) 1 hour 48 minutes »
2U01-6b
The number N of bacteria in a colony is growing at a rate that is proportional to the current
number. The number at time t hours is given by N = N0ekt where N0 and k are positive
constants.
i.
If the size of the colony doubles every half hour, find the value of k.
ii.
If the colony now contains 600 million bacteria, how long ago did the colony
contain 3 million bacteria?
iii.
Show that the numbers of bacteria present at consecutive integer hours form a
geometric sequence.†
« i) k = 138629 (to 5 d.p) ii) 38219 hours (to 4 d.p) iii) Proof »
2U01-6c
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y
40
30
20
10
0
–10
–20
1
2
3
4
5
6
t
A particle moves along a straight line for 6 seconds. The particle’s velocity v at time
t seconds is shown on the graph above.
i.
When is the particle at rest?
ii.
What does the shaded region represent?
iii.
Is this particle further from its initial position at t = 3 or at t = 5? Explain briefly.†
« i) t = 3 or t = 5 ii) The distance travelled during the third second iii) The particle changes
direction after it comes to rest at t = 3 and begins to move back towards its initial position. Hence the
particle is further from its initial position at t = 3. »
61)! 2U01-9b
A particle is moving along the x axis. Its position x at time t is given by
t
 100e 5
x  60t
(t  0) .
i.
Find the initial position of the particle.
ii.
Show that the particle is always moving to the right.
iii.
What happens to the acceleration eventually?†
« i) 100 units to the right of the origin ii) Proof iii) It approaches zero »
62)! 2U02-5c
The population P of a town is growing at a rate proportional to the town’s current population.
The population at time t years is given by P = Aekt, where A and k are constants. The
population 20 years ago was 100 000 people and today the population of the town is 150 000
people.
i.
Find the value of A.
ii.
Find the value of k.
iii.
Find the population that will be present 20 years from now.†
« i) A = 100 000 ii) 00203 (to 3 s.f) iii) 225 000 »
63)! 2U02-7b
A particle is projected vertically upwards from a point 2 metres above horizontal ground. The
displacement at time t seconds is given by x = 24·5t – 4·9t2, t  0.
i.
Find an expression for the velocity of the particle.
ii.
Find when the particle comes to rest.
iii.
Hence, find the greatest height of the particle above the ground.
iv.
Find the length of time for which the particle is at least 21·6 metres above the
ground.†
« i) v = 245 – 98t ii) t = 25 seconds iii) 32625 metres iv) 3 seconds »
64)! 2U03-7b
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A particle moves in a straight line so that its velocity, v metres per second, at time t is given by
2
. The particle is initially 1 metre to the right of the origin.
v3
1t
i.
Find an expression for the position x, of the particle at time t.
ii.
Explain why the velocity of the particle is never 3 metres per second.
iii.
Find the acceleration of the particle when t = 2 seconds.†
2
2
« i) x = 3t – 2ln(1 + t) + 1 ii)
can never be zero  v will never be 3 iii) ms  2 »
9
1t
65)! 2U03-9b
A train is travelling at a constant velocity of 80 kilometres per hour as it passes through the
railway station at a town. At the same time, a second train commences its journey from rest at
the railway station. The second train accelerates uniformly for 15 minutes until it reaches
100 kilometres per hour and maintains this velocity for a further 5 minutes.
At this time each of the trains then begin to slow down at a constant rate, arriving at the next
station at the same time.
i.
By illustrating graphically the relationship between velocity and time, calculate the
time taken for the trains to travel between the two stations.
ii.
How far apart are the stations?†
km/h
100
80
60
40
20
« i)
66)!
67)!
1st train
2nd train
5 10 15 20 25 30
time
t = 55 minutes ii) 50 kilometres »
2U04-5c
A radioactive substance decays at a rate proportional to the mass present. The rate of change
dM
 kM , where k is a positive constant and M grams is the mass present at
is given by
dt
any time, t hours.
i.
Show that M = M0e–kt is a solution to this equation.
ii.
If 100 grams of this substance decays to 80 grams in 20 hours, find:
.
The value of k, correct to 2 decimal places.
.
The mass present after a further 10 hours, to the nearest gram.
iii.
The half-life time is the time taken for 100 grams of this substance to decay to
50 grams. What is the half-life time of this substance? Give your answer to the
nearest hour.†
« i) Proof ii) ) k = 0∙011157 ) 72 grams iii) 62 hours »
2U04-10a
dv
 k , where k is a constant.
The acceleration of a particle at any time t seconds is given by
dt
i.
Show that v = kt + c1, for some constant c1.
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ii.
The displacement x metres, at any time t seconds, is shown in the table below:
t (sec)
x (metres)
iii.
0
1
1
2
2
9
Show that x = 3t2 – 2t + 1.
Find when the particle comes to rest.
« i) ii) Proof iii) t =
68)!
69)!
70)!
71)!
1
second »
3
2U05-7b
The number N of bacteria in a culture at a time t seconds is given by the equation
N = 20 000e0003t
i.
What is the number of bacteria initially?
ii.
Determine the number of bacteria after 20 seconds.
iii.
After what period of time will the number of bacteria have doubled?
iv.
At what rate is the number of bacteria increasing when t = 20 seconds?†
« i) 20 000 ii) 21 237 iii) 231 seconds iv) 63∙7 bacteria/second »
2U05-8b
At time t seconds, the position x cm of a point moving in the straight line XOX is given by
x = at2 + bt cm, where a and b are constants. The particle passes through the origin O with
velocity 16 cm/s in the positive direction at time t = 0 seconds, and after 8 seconds, it is again
at O.
i.
Find the velocity of the particle at any time, in terms of a and b.
ii.
Find the values of the constants a and b.
iii.
Find the time when the object is at rest.
iv.
Find the position of the particle when it is at rest.†
« i) v = 2at + b ii) a = –2, b = 16 iii) t = 4 seconds iv) 48 cm from the origin »
2U06-8b
A 100 mg tablet is dissolved in a glass of water. After t minutes the amount of undissolved
tablet, U in mg, is given by the formula:
U = 100 e–kt , where k is a constant.
i.
Calculate the value of k, correct to 4 decimal places, given that 2 mg of the tablet
remains after 10 minutes.
ii.
Find the rate at which the tablet is dissolving in the glass of water after 12 minutes.
Give your answer correct to two decimal places.†
« i) 0∙3912 ii) –0∙36 mg/minute »
2U06-9a
Two particles, A and B, move along a straight line so that their displacements, xA and xB , in
metres, from the origin at time t seconds are given by the following equations respectively:
xA = 12t + 5
xB = 6t2 – t3
i.
Find two expressions for the velocities of particles A and B.
ii.
Which of the two particles is travelling faster at t = 1 second?
iii.
At what time does particle B come to rest?
iv.
Find the maximum positive displacement of particle B.†
« i) vA = 12, vB = 12t – 3t2 ii) A iii) t = 4 seconds iv) 32 metres »
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