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p(1) = 9 Kinematics
4(1) – 6(1) + c(1) = 9
2
3
c = 11 1. (a) When t = 3, v = 0.
p(x) = 4x – 6x + 11x 3a(3) – 2b(3) = 0
2
3
2
[4] 27a – 6b = 0 …… (1)
26. y = –x + 16x – 48 s = 3at – 2bt dt
2
2
dy = –2x + 16
dx = 3at 3 – 2bt 2 + c
2
3
–2x + 16 = 0 = at – bt + c
3
2
x = 8
When t = 0, s = 0, so c = 0.
When x = 8, When t = 3, s = –1
2
y = –(8) + 16(8) – 48 a(3) – b(3) = –1
3
2
= 1 = 16 27a – 9b = –1 …… (2)
2 S(8, 16)
Equation (1) – Equation (2)
= √3 When y = 0, 3b = 1
2
–x + 16x – 48 = 0
= √(7) – x = 4 or x = 12 b = 1 27Rights Reserved.
2
3
1
P(4, 0) and Q(12, 0). 27a – 6 1 2 = 0
2
2
= 1 8 + 3 Equation of line PQ, 3 a = 2
2
1 m = 16 – 0 = – 4 27 [5]
= 2 2 0 – 12 3
2
1
Penerbitan Pelangi Sdn Bhd. All
= 4 (b) V = 3 1 2 t – 2 1 2 t
2
2 y = – x + 16 3
(b) 3 2 2 2
2
3
9
1 – 4 x + 16 = –x + 16x – 48 = t – t
3
4
== –4x + 48 = –3x + 48x – 144 dv = t – 2
2
3x – 52x + 192 = 0 dt 9 3
2
x = 16 or x = 12 When dv = 0,
3 4 2 dt
When x = 16 , 9 t = 3
3 3
4 16 t = s
y = – 1 2 + 16 2
3 3 [3]
= 80
9 2. (a) When t = 0,
T 1 16 , 80 2 a = 4 – 2(0)
= 4 m s
–2
9
3
16 [1]
3
–x + 16x – 48 dx
2
8 8 (b) v = 4 – 2t dt
3
= – x 3 3 + 16x 2 – 48x] 4 16 3 = 4t – 2t 2 + c
2
2
2
1
= – 128 2 1 6400 2 = 4t – t + c
– –
3
81
When t = 0, v = 16.
2
= 2944 4(0) – 0 + c = 16
81 c = 16
2
2
Area = (8 × 16) – 3 1 16 + 80 16 1 2944 2 v = 4t – t + 16

9
3
dv
2 4 – 81 dt = 0
= 2048 4 – 2t = 0
81 2t = 4
= 25.28 unit t = 2
2
[10]
Maximum velocity = 4(2) – 2 + 16
2
= 20 m s
–2
[5]
Cambridge IGCSE
TM
210 Ace Your Additional Mathematics
Answers Add Math.indd 210 14/03/2022 12:29 PM