Week 10.
Marks
Question 1 (Mathematics Extension 1, 3 Units)
12m/s
FIGURE NOT
TO SCALE
40°
2.4m
O
A
A tennis ball is thrown on the audience’s bench from a height of 2.4 metres above
ground level at an angle of 40° to the horizontal, and with a speed of 12 metres per
second.
Take the origin O as a point on the ground directly under the point of the throw.
The equations of motion of the tennis ball are
&&
x=0
&&
y = −g
i) Using calculus, show that the position of the tennis ball at time t is given by
x = 12t cos 40°
y = 2.4 + 12t sin 40° −
4
1 2
gt .
2
ii) The tennis ball lands at a point A on the ground. Find the length of OA to
the nearest centimeters. (Take g = 10 )
2
Preliminary & HSC Course – Mathematics & Mathematics Extension 1
Mentor’s Advice
Velocity: x& = ∫ &&
xdt
&
Displacement: x = ∫ xdt
&
Question 1 (Solutions and Answers)
a)
Let the initial y& = y& 0 and the initial x& = x&0 , then
i)
x&0
12
y&
sin 40° = 0
12
cos 40° =
12
y& 0
40°
∴ x&0 = 12 cos 40°
∴ y&0 = 12sin 40°
x&0
x = 0 (given)
Horizontally, &&
x& = ∫ 0dt = C1
x&0 = 12 cos 40° when t = 0
x& = 12cos 40°
∴
C1 = 12 cos 40°
i.e.
⋅⋅⋅⋅⋅⋅⋅
c
x = ∫ (12 cos 40° ) dt = 12t cos 40° + C2
x = 0 when t = 0
C2 = 0
i.e.
∴ x = 12t cos 40°
Vertically, &&
y = − g (given)
⋅⋅⋅⋅⋅⋅⋅
d
y& = ∫ ( − g ) dt = − gt + C3
y& 0 = 12sin 40° when t = 0
∴
y& = − gt + 12 sin 40°
C3 = 12sin 40°
i.e.
⋅⋅⋅⋅⋅⋅⋅
e
1
y = ∫ ( − gt + 12sin 40° ) dt = − gt 2 + 12t sin 40° + C4
2
y = 2.4 when t = 0 i.e. C4 = 2.4
∴
1
y = − gt 2 + 12t sin 40° + 2.4
2
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⋅⋅⋅⋅⋅⋅⋅
f
All rights reserved.
2
Preliminary & HSC Course – Mathematics & Mathematics Extension 1
The total time taken by the tennis ball, t , occurs when y = 0 .
ii)
i.e.
1
0 = − ⋅10 ⋅ t 2 + 12 sin 40° ⋅ t + 2.4 (from f)
2
0 = −5t 2 + 12 sin 40° ⋅ t + 2.4
Using the quadratic formula, t =
t=
−b ± b 2 − 4ac
,
2a
−(12sin 40°) ± (12sin 40°) 2 − 4 ⋅ (−5) ⋅ (2.4) −7.71 ± 10.37
= 1.81 ( t ≥ 0 )
=
2 ⋅ (−5)
−10
∴
Substitute t = 1.81 into d
x = 12 cos 40° ⋅ (1.81) = 16.6 (m)
Marks
Question 2 (Mathematics, 2 Units)
A pine grove was found to be infected by the vermin due to extreme climate change
in the summer season.
The rate at which the number, P , of the pine infected t days after an infection
control is given by
dP
= − kP , where k is a constant.
dt
2
i) Show that the amount of infected pine is given by
P = P0 e − kt ,
where P0 is the initial number of infected pines just before the beginning of an
infection control.
ii) It was found that 12 days after the infection control there was one-third of the
initial amount of infected pine present on the grove. Find the value of k
2
correct to two decimal places.
iii) The infection control usually decreases the number of infected pines below 1%
of the initial number of infected pines. How long would it take for this to
happen? Given your answer in nearest days.
2
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All rights reserved.
3
Preliminary & HSC Course – Mathematics & Mathematics Extension 1
Mentor’s Advice
i) Use
ii)
d f ( x)
⎡⎣e ⎤⎦ = f '( x) ⋅ e f ( x )
dt
1
of the initial amount after 12 days i.e.
3
iii) Below 1% of the initial number
P=
P<
i.e.
P0
when t = 12
3
P0
100
Question 2 (Solutions and Answers)
b)
dP d
= ⎡⎣ P0 e − kt ⎤⎦ = P0 ⋅ ( − k ) ⋅ e − kt = (− k ) ⋅ P0 e − kt
dt dt
i)
But, P0 e − kt = P
ii)
From given, P =
∴
P0
when t = 12
3
−12k = log e
1
3
∴
dP
= − kP
dt
i.e.
P0
= P0 e − k ⋅12
3
k =−
1
1
log e = 0.09
12
3
Below 1% of the initial number of infected pine
iii)
P0
,
100
P0
100
1
⎛1⎞
log e ⎜ ⎟
12
⎝3⎠
1
P0
,
100
⎛1⎞
− log e ⎜ ⎟⋅t
P
then 0 = P0 e 12 ⎝ 3 ⎠
100
( P = P0 e −0.09t since the result in (ii))
→
P<
Suppose that P =
P
then 0 = P0 e −0.09t
100
1
= e −0.09t
100
1
= e −12 k
3
Answer 2) If you use k = −
Answer 1) If you use k = 0.09
Suppose that P =
i.e.
→
−0.09t = log e
t = 51.17
∴ It will take 52 days.
∴
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1
100
( P = P0 e
−
−
1
⎛1⎞
loge ⎜ ⎟⋅t
12
⎝ 3⎠
since the result in (ii))
1
⎛1⎞
⎛ 1 ⎞
log e ⎜ ⎟ ⋅ t = log e ⎜
⎟
12
⎝3⎠
⎝ 100 ⎠
t = 50.30
∴ It will take 51 days.
∴
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4