Page 71 - ACE YR IGCSE A TOP APPR' TO ADD MATH

(c) When v = 0, (ii) Total distance travelled
4t – t + 16 = 0 1 3
2
u
2
2
t = 2 + 2√5 = t – 4t + 3 dt + u t – 4t + 3 dt +
t = 2 – 2√5 (rejected) 0 4 1
So, P = 2 + 25 s. t – 4t + 3 dt
2
[2] 3 3 2 1 3 3
4 u3
2
(d) s = 4t – t + 16 dt = 3 t 3 – 4t + 3t + t 3 – 2t + 3t 4 u +
2
2
1
0
4t 2 t 3 3 t 3 – 2t + 3t 4 4
2
= – + 16t + d 3
2 3 4 4 4 3
t
3
= 2t – + 16t + d = 3 + + 3
2
3
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
3
When t = 0, s = 0. = 4 m
[3]
0
3
2(0) – + 16(0) + d = 0 4. (a) a = 6t – 10
2
3 d = 0 When a = 0,
3
t
s = 2t – + 16t 6t = 10
2
5
3 t = s
When t = 5, 3 [2]
5
3
s = 2(5) – + 16(5) (b) dv = 0
2
3
dt
1
= 88 m 6t – 10 = 0
3 [4]
5
t = s
3
3. (a) (i) When t = 0, v = 3.
0 – 4(0) + s = 3 s = 3t – 10t – 32 dt
2
2
s = 3 3t 3 10t 2
[1] = 3 – 2 – 32t + c
(ii) v = t – 4t + 3 = t – 5t – 32t + c
2
2
3
t – 4t + 3 , 0 When t = 0, s = 0.
2
(t – 3)(t – 1) , 0 0 – 5(0) – 32(0) + c = 0
2
3
1 < t < 3 c = 0
[2] 3 2
5
5
5
(iii) a = 2t – 4 s = 1 2 – 5 1 2 – 32 1 2
3
3
3
2t – 4 . 0 1690
2t . 4 = – 27
t . 2
[1] Distance from O = 1690 m
27
(b) (i) [5]
y
4
3
2
1
0 x
1 2 3 4 5
–1
[2]
Answers 211

Answers Add Math.indd 211 14/03/2022 12:29 PM