20[Anal Add Math CD]

CHAPTER20 Additional Mathematics  SPM  Chapter 20   Motion along a Straight Line 1. (a) Distance OA = 5 m Distance OB = 2 m (c) s , 0 t2 – 4t , 0 t(t – 4) , 0 (b) Displacement of A from O = 5 m Displacement of B from O = –2 m s 2. (a) Displacement after 5 seconds = 52 + 3(5) = 40 m t 0 (b) Distance travelled in the first 4 seconds = s4 – s0 = [42 + 3(4)] – (0) = 28 m 4 0 , t , 4 is the time when the particle is on the left of O. (d) s . 0 Based on diagram in (c), t . 4 is the time when the particle is on the right of O. (c) s5 – s3 = [52 + 3(5)] – [32 + 3(3)] = 22 m (d) Distance travelled during the 3rd second = s3 – s2 = [32 + 3(3)] – [22 + 3(2)] =8m 4. (a) s = t(t – 4) = t2 – 4t t s 3. (a) s = t2 – 4t When s = 0, t2 – 4t = 0 t(t – 4) = 0 t = 0, 4 After 4 seconds, the particle returns to O. 0 2 0 4 – 4 5 0 5 s 5 0 (b) When s = – 4, – 4 = t2 – 4t t2 – 4t + 4 = 0 (t – 2)2 = 0 t = 2 After 2 seconds, the particle is at 4 m on the left of O. t 2 4 5 –4 (b) (i) The furthest distance on the left is 4 m from O. (ii) Total distance travelled in the first 5 seconds =2×4+5 = 13 m 1 (iii) The displacement after 5 seconds is 5 m on the right of O. © Penerbitan Pelangi Sdn. Bhd. Bhd. 3 . 7. (a) s = t2 – 9 t=0 A O When t = 0. t 3 4 –9 ∫ 4 0 0 (b) When v = 10. 0 6t – 12 . 2 The time interval is 0 < t . 3 0 < t . s = –9 OA = 9 m (b) Initial velocity = 6(0) – 12 = –12 m s–1 Its direction of motion is towards left. (ii) The time interval when the particle moves towards left is 0 . (b) s = 0 t2 – 9 = 0 (t + 3)(t – 3) = 0 t = 3.  Additional Mathematics  SPM  Chapter 20 (iv) The time when the particle passes through O again is 4 s. 2 ∫ 4 (iii) v dt = 6. 0 t2 – 9 . t . (t2 – 4t) dt 4 t3 = ––– – 2t2 0 3 43 = ––– – 2(4)2 – (0) 3 2 = –10— m 3 3 4 Therefore. 3 is the time interval when the particle is on the left-hand side of O. (a) v = 2t + 6 Initial velocity = 2(0) + 6 = 6 m s–1 © Penerbitan Pelangi Sdn. the particle is moving towards right. t = –3 is ignored (c) s . (c) When t = 1. 6t – 12 = 0 t = 2 When the particle is instantaneous at rest. s = 3(2)2 – 12(2) = –12 m s t 0 –3 (c) When v = 0. 4 is the time interval when the particle is on the right-hand side of O. 2 when the particle is moving towards left. (v) t . 0 (t + 3)(t – 3) . 10 = 2t + 6 2t = 4 t = 2 After 2 seconds passing through O. (a) v = t(t – 4) t (e) s4 – s3 = (42 – 9) – (32 – 9) =7m 0 v 2 0 4 – 4 0 v (f) s = t2 – 9 s t 0 2 4 7 –4 0 (b) (i) Minimum velocity is – 4 m s–1. 4. its velocity is 10 m s–1. v = 2(1) + 6 = 8 Therefore. (d) s = 42 – 9 = 7 m 8. the time is 2 s and the displacement is 12 m on the left of O. (a) s = 3t2 – 12t ds v = ––– dt = 6t – 12 5. the distance travelled in the first 2 4 seconds is 10— m. 0 t . 0 Substitute t = 2 into s = 3t2 – 12t. (d) v . — 3 4 — 3 3 3 4 2 3  – 2(3)2 – ––– 2 4 – 2(2)2 = ––– 4 4 1 = 6— m 4 3 4 4 3 4 The distance travelled in the first 3 seconds 1 = 4 + 6— 4 1 = 10— m 4 (c) v . 0 1 5— 3 8 — 3 1 5– 3 (b) When v = 0. 0 –3t2 + 8t . 3 –3 8 3 – 3 4 – 3 –3 v 0 t 0 The time when the particle stops instantaneously 8 for the second time is — s. 2 is the time interval when the particle is moving towards left. v = 53 – 4(5) = 105 m s–1 3 © Penerbitan Pelangi Sdn.Additional Mathematics  SPM  Chapter 20   9. s = 0 \c=0 s = –t3 + 4t2 10. (a) v = t3 – 4t When v = 0. 0 . the particle stops instantaneously. 3 3 v (c) v . 0 t(–3t + 8) . (a) v = –3t2 + 8t (f) v = –3t2 + 8t ∫ = ∫ (–3t + 8t) dt s = v dt 2 = –t3 + 4t2 + c t 0 v 0 When t = 0. Velocity. t = —. 3 Maximum distance = –t3 + 4t2 8 3 8 2 =–— +4— 3 3 13 = 9––– m 27 1 2 2 0 1 24 ∫ 3 v dt + v dt = ∫ 2 0 ∫ 3 2 v dt (t3 – 4t) dt 2 t 4 – 2t2 = ––– 4 0 4 2  = ––– – 2(2)2 – (0) 4 = – 4 3 3 4 4 t 4 – 2t2 v dt = ––– 4 8 — 2 3 2 0 (–3t2 + 8t) dt 8 3 8 = –— +4— 3 3 13 = 9––– m 27 2 2 v dt = 3–t3 + 4t 4 ∫  0∫ Time interval when the particle travelled in the 8 right direction is 0 . –3t2 + 8t = 0 t(–3t + 8) = 0 8 t = 0. t . 0 t3 – 4t . Bhd. t = –2 is ignored After 2 seconds. 1 2 (d) When t = 5. (b) The distance travelled in the first 3 seconds = t 0 8 – 3 ∫ = ∫ (d) Distance travelled = 0 0 8 — 3 8 — 3 31 2 – (0) (e) The maximum distance of the particle is when 8 v = 0. 0 t(t2 – 4) . . that is. 0 From the graph. t . t3 – 4t = 0 t(t2 – 4) = 0 t = 0. t2 = 4 t = 2. —. Bhd. ∫ v dt = = 3t3 – 2t2 + t4 3 1 — 3 1 t ∫ t 1 – 3 1 1 — 3 31—13 2 3 0 1 2 1 – 2 — + — – (0) 3 3 1 2 4 1 v dt = 3t3 – 2t2 + t4 —1 3 = [1 – 2(1) + 1] – 4 = 0 – ––– 27 4 = – ––– m 27 3 2 1 2 31—13 2 3 1 2 1 –2 — +— 3 3 1 2 4 . (a) s = t3 – 2t2 + t (g) v = 3t2 – 4t + 1 = (3t – 1)(t – 1) ds Velocity. a = ––– dt = 6t – 4 t 0 v 1 1 — 3 0 1 2 0 5 v 5 (c) When v = 0. 6t – 4 = 8 t = 2 The displacement when a = 8 = 23 – 2(2)2 + 2 = 2 m on the right of O – 4 2 0 8 a 8 t 0 (e) Minimum velocity is when a = 0.  Additional Mathematics  SPM  Chapter 20 11. 6t – 4 = 0 4 t = — 6 2 = — 3 Minimum velocity = 3t2 – 4t + 1 2 2 2 =3— –4— +1 3 3 1 –1 =–—ms 3 1 2 t 0 –4 2 2 – 3 (i) Distance travelled in the first second 1 — 3 1 = v dt + 1 v dt ∫ ∫ 0  — 3 v 1 1 2 0 (f) s = t3 – 2t2 + t t 0 s 1 0 2 0 2 s 2 0 1 2 0 (3t2 – 4t + 1) dt 1 — = 4 = ––– m 27 ∫ 4 0 1 — 3 © Penerbitan Pelangi Sdn. 1 3 The acceleration of the particle when it stops at second time = 6(1) – 4 = 2 m s–2 1 1 – 3 1 2 (h) a = 6t – 4 t 2 — 3 0 a (d) When a = 8. v = ––– dt = 3t2 – 4t + 1 dv (b) Acceleration. 3t2 – 4t + 1 = 0 (3t – 1)(t – 1) = 0 1 t = —. 3t2 + 12 = 24 3t2 = 12 t2 = 4 t = 2. — 3 Acceleration when v = 24 =6×2 = 12 m s–2 (e) (i) t 10 – 3 1 2 1 2 50 25 = ––– – ––– 3 3 Distance travelled during the second second = s 2 – s1 = 32 – 13 = 19 m 5 25 = – ––– m 3 © Penerbitan Pelangi Sdn. 3 (e) v = 0 when s is the maximum 6t – 10 = 0 5 t = — 3 Maximum displacement on the left 5 2 5 = 3 — – 10 — 3 3 s = t3 + 12t Distance travelled after 2 seconds = 23 + 12(2) = 32 m (ii) s2 = 32 m s1 = 13 + 12(1) = 13 m Therefore. t . 0 10 t . 0 6t – 10 .Additional Mathematics  SPM  Chapter 20   Distance travelled in the first second 4 4 = ––– + ––– 27 27 8 = ––– m 27 (iii) s = t3 + 12t = 43 + 12(4) = 112 m 13. —. the time interval when the particle is 5 moving towards direction of left is 0 < t . 0 3t2 – 10t . 0 t(3t – 10) . \ c = 0 \ s = 3t2 – 10t ∫ = ∫ (3t + 12) dt (b) s = v dt (c) s . \c=0 s = t3 + 12t (c) (i) Displacement = 13 + 12(1) = 13 m on the right of O 0 (ii) Velocity = 3(1)2 + 12 = 15 m s–1 (iii) Acceleration = 6t = 6(1) = 6 m s–2 (d) When v = 24. ––– 6 5 t . Bhd. v = 12. s = 0. t = –2 is ignored Therefore. ––– . . 3 (d) v . the time interval for the particle on the 10 left of O is 0 . (a) a = 6t ∫ = ∫ 6t dt v = a dt ∫ s = ∫ (6t – 10) dt (b) s = v dt = 3t + c 2 Given t = 0. (a) v = 6t – 10 dv a = ––– dt a = 6 12. s = 0. 0 2 = t3 + 12t + c s Given t = 0. \ c = 12 v = 3t2 + 12 s = 3t2 – 10t + c Given t = 0. 0 t2 – 8t + 12 . the particle travels from point O to point R. 4 Total distance travelled in the first 4 seconds = Area of shaded region = From v 4 74 ––– 3 Average speed = –––– 4 1 = 6— m s–1 6 \ 2 . Bhd. 0 (6t – 10) dt 3 = 3t2 – 10t4 v 5 3 = [3(5)2 – 10(5)] – [3(3)2 – 10(3)] = 25 – (–3) = 28 m (g) (i) v = 6t – 10 0 v (4. v = 42 – 8(4) + 12 = –4 m s–1 6 2 4 – 4(4)2 + 12(4) = 10 2 + 5 1 3 3 = 16 m After 4 seconds. (a) v = –t2 + 3t + 4 At the point R. 0 (t – 2)(t – 6) . 6 (c) Total distance travelled in the first 4 seconds 5 7 1 1 = — × 10 × — + — × 14 × — 3 3 2 2 49 25 = ––– + ––– 3 3 74 ––– = 3 1 t 6 2 The acceleration at point R = –2(4) + 3 = –5 m s–2 . ∫ v = (2t – 8) dt Given t = 0. (a) Given a = 2t – 8 For minimum velocity.  Additional Mathematics  SPM  Chapter 20 (f) Distance travelled from t = 3 to t = 5 5 v dt = ∫ = ∫ 3 5 3 (b) v . 14) B 0 –10 A 5 – 3 12 t 2 1 0 2 ∫ 2 0 (t2 – 8t + 12) dt + u∫ 4 2 4 u3 t3 3 3 3 3 a = 2t – 8. a = 0 \ 2t – 8 = 0 t = 4 4 4u 4 u3 43   3 u (t2 – 8t + 12) dt – 4t2 – 12t = 2 – 4(2)2 + 12(2) + 3 3 – 2 – 4(2)2 – 12(2) 3 t 6 2 2 3 = t – 4t2 + 12t + 3 0 Total displacement (ii) Average velocity = –––––––––––––––– Time 8 =— 4 = 2 m s–1 © Penerbitan Pelangi Sdn. t . v = 12 \ c = 12 v = t2 – 8t + 12 Minimum velocity. dv (b) a = ––– dt = –2t + 3 v = t2 – 8t + c 3 4u 2. v = 0 –t2 + 3t + 4 = 0 t2 – 3t – 4 = 0 (t – 4)(t + 1) = 0 \t=4 1. \ c = 0 2 sQ = —t3 – 2t2 – 16t 3 7 © Penerbitan Pelangi Sdn.Additional Mathematics  SPM  Chapter 20   (c) For maximum velocity. (b) v = –t2 + 8t – 7 = (–t + 7)(t – 1) t 0 1 7 v –7 0 0 (b) Particle Q stops at C. . (a) vQ = 2t2 – 4t – 16 dvQ –––– = 4t – 4 dt Therefore. 0 7 3 4 3 v 0 ∫ ∫ 4 1 2 = 36 m 1 Total distance travelled = 3— + 36 3 1 = 39— m 3 t 1 7 –7 4. t = –2 is ignored vQ = 2t2 – 4t – 16 v 0 1 Therefore. dt 4t – 4 = 0 t = 1 0 0 –8 4 Therefore. 7. a = 0 –2t + 3 = 0 3 t = — 2 Maximum velocity 3 2 3 =–— +3— +4 2 2 1 = 6— m s–1 4 1 2 3. (a) v = (i) (c) The total distance travelled during the first 7 seconds = 1 2 1 0 1 0 v dt + ∫ 7 1 v dt (–t2 + 8t – 7) dt 1 t3 = – ––– + 4t2 – 7t 3 0 1 = – — + 4 – 7 – (0) 3 1 = –3— m 3 3 1 –t + 8t – 7 Initial velocity = – (0)2 + 8(0) – 7 = –7 m s–1  v dt = 2 ∫ 4 2 7 t3 v dt = – ––– + 4t2 – 7t 3 1 1 73 1 2 = – ––– + 4(7) – 7(7) – – — + 4 – 7 3 3 (ii) v . –2t + 8 . the minimum velocity of Q = 2(1)2 – 4(1) – 16 = –18 m s–1 ∫ sQ = (2t2 – 4t – 16) dt t 2 = —t3 – 2t2 – 16t + c 3 7 –7 Given t = 0. (iii) a . dvQ vQ is minimum when –––– = 0. 4. SQ = 0. the time interval during which the acceleration is negative is t . –2t . 0 (–t + 1)(t – 7) . t . the time interval during which the particle moves towards right is 1 . Bhd. t . 0 –t2 + 8t – 7 . \ vQ = 0 2t2 – 4t – 16 = 0 t2 – 2t – 8 = 0 (t – 4)(t + 2) = 0 t = 4. The two particles meet after 5 seconds they pass through A and B respectively and it is at 45 m on the right of A. t = –1 is ignored (c) When Q at C. vP = 2t + 4 When t = 4. 3t2 – 12t = 15 3t2 – 12t – 15 = 0 t2 – 4t – 5 = 0 (t + 1)(t – 5) = 0 t = 5. (a) Q P A C (b) s . it reverses its direction at a distance of 16 m on the left of B. t = 0 is ignored ds v = ––– dt = 6t – 12 sQ = t 2 – 8t sP = t2 + 4t. 6t – 12 = 0 t = 2 Displacement from O = 3(2)2 – 12(2) = –12 m 1 AC = 60 – 53— 3 2 — = 6 m 3 (e) When s = 15. 0 3t2 – 12t .  Additional Mathematics  SPM  Chapter 20 When t = 4. Difference in distance between P and Q 2 = 12 – 6— 3 1 = 5— m 3 The velocity = 6(5) – 12 = 18 m s–1 2. 4. t . the maximum distance from O is 12 m. Substitute t = 4 into sQ = t2 – 8t. (b) When Q reverses. (c) Velocity of object P. 0 3t(t – 4) . s = 0 3t2 – 12t = 0 3t(t – 4) = 0 t = 4. Bhd. sQ = 42 – 8(4) = –16 m After 4 seconds particle Q passes through B. 0 Substitute t = 5 into sP = t2 + 4t. sQ = t2 – 8t Assume P and Q meet at C after k seconds. 8 . the time interval for the particle to be on the left of O is 0 . (a) s = 3t2 – 12t Displacement = s1 – s0 = [3(1)2 – 12(1)] – (0) = –9 m The distance travelled in the first second is 9 m. vQ = 0 2t – 8 = 0 t = 4 (c) When the particle passes through O again. sp = 52 + 4(5) = 45 m s t 4 Therefore. t = 4 Displacement of P = 4 × 3 = 12 m Therefore. vP = 2(4) + 4 = 12 m s–1 The velocity when the particle passes through O again = 6(4) – 12 = 12 m s–1 © Penerbitan Pelangi Sdn. Distance travelled by P + Distance travelled by Q = 60 (k2 + 4k) + (8k – k2) = 60 12k = 60 k = 5 1. 0 B 60 m sP = t 2 + 4t . 2 BC = —(4)3 – 2(4)2 – 16(4) 3 1 = –53— m 3 (d) Maximum distance is when v = 0. ∫ ∫ Therefore. (c) v = t(4 – 3t) t 0 4 — 3 2 v 0 0 – 4 9 © Penerbitan Pelangi Sdn. \ c = 0 s = 2t2 – t3 When t = 1. (b) v = t2 – 6t + 9 When t = 0. 4 4 — 3 0 ∫ When the particle reverses. 3 4 a = 4 – 6 — 3 = – 4 m s–2 1 2 ∫ 4 — 3 0 ∫ v dt = 3.Additional Mathematics  SPM  Chapter 20   (d) sQ = t2 – 8t v t 0 s 0 8 10 0 20 t 0 s 4 – 3 2 –4 20 Distance travelled during the first 2 seconds 4 — 3 2 = v dt + 4 v dt t 0 8 10 (e) vQ . 4 is the time interval for object Q moves in the direction of left. the average velocity is 0 m s–1. Bhd. . 0 2t – 8 . when t = 3 0 = 32 – 6(3) – k k = –9 Distance travelled during the first second is 1 m. 0 t . v = 9 m s–1 The velocity at O is 9 m s–1. (a) v = t2 – 6t – k Given v = 0. a = 4 – 6t 4 At t = —. s = 2(2)2 – 23 =0 (b) s = (4t – 3t2) dt = 2t2 – t3 + c Given t = 0. v = 0 4t – 3t2 = 0 t(4 – 3t) = 0 4 t = —. t = 0 is ignored 3 Given v = 4t – 3t2. s = 0. s = 2(1)2 – (1)3 =1m 0 0 < t . (a) v = t(4 – 3t) v = 4t – 3t2 2 4 — 3  — 3 (4t – 3t2) dt 4 — = 32t2 – t34 3 0 4 2 4 = 2— – — 3 3 5 = 1––– m 27 3 3 1 2 1 2 4 – (0) 2 v dt = 32t2 – t34—4 3 4 2 4 = [2(2)2 – 23] – 2 — – — 3 3 5 = –1––– m 27 3 3 1 2 1 24 Total distance travelled in the first 2 seconds 5 = 1––– × 2 27 10 = 2––– m 27 (d) The average velocity in the first 2 seconds Displacement = –––––––––––– Time ∫ When t = 2. 4. a = 0 2t – 3 = 0 3 t = — 2 4 5 5 Total distance travelled = 45— + 3— 6 6 2 = 49— m 3 Minimum velocity = t2 – 3t – 10 3 2 3 = — – 3 — – 10 2 2 9 9 = — – — – 10 2 4 1 = –12— m s–1 4 (ii) a . Its velocity = 10(0) – 35 = –35 m s–1 The range of time for particle to decelerate 3 is 0 < t . \ c = –35 v = 10t – 35 (i) At point O. 0 3 t . (a) a = 10 m s–2 v = ∫ 10 dt v = 10t + c Given t = 2.  Additional Mathematics  SPM  Chapter 20 (c) Distance travelled during the third second ∫ = ∫ = 3 2 3 2 ∫ = ∫ = v dt (t2 – 6t + 9) dt t3 = ––– – 3t2 + 9t 3 3 3 (t2 – 3t – 10) dt 3 t3 = ––– – —t2 – 10t 2 3 3 2 3 4 3 ∫ = ∫ (2t – 3) dt = 5 v dt 6 4 3 3 5 4 3 53 – ––– – —(5)2 – 10(5) 2 3 5 = 3— m 6 3 For minimum velocity. 0 2t – 3 . v dt 0 0 4 5 5 33 23 = ––– – 3(3)2 + 9(3) – ––– – 3(2)2 + 9(2) 3 3 1 = — m 3 (ii) Distance from t = 0 to t = 5 3 –12– 1 –. 2 ∫ (b) (i) v = t2 – 3t – 10 = (t – 5)(t + 2) –10 6 3 6 = ––– – —(6)2 – 10(6) 2 3 Given t = 0. (a) (i) a = 2t – 3 (ii) s = (10t – 35) dt 3 — 2 1 –12— 4 s = 5t2 – 35t + c 5 6 0 8 Given t = 0. 10t – 35 = 25 t =6 v 8 0 5 6 t –10 © Penerbitan Pelangi Sdn. 2 4 10 The displacement of the object = 5(6)2 – 35(6) = –30 m = 30 m on the left of O . s = 0. v = –15. Bhd. v = –10. \ c = 0 s = 5t2 – 35t When v = 25. — 2 1 2 6. \ c = –10 v = t2 – 3t – 10 v ∫ 3 = t2 – 3t + c 0 0 4 3 t3 = ––– – —t2 – 10t 2 3 1 2 5 3 3 v = a dt t 4 3 5 = ––– – —(5)2 – 10(5) – (0) 2 3 5 = – 45— m 6 Distance from t = 5 to t = 6 4 5. t = 0. —. ... v = 0 0 = 4p – 16q p = 4q........... t = 3 (b) When the object reverses its direction................ 2 Substitute 1 into 2......... v = 0 10t – 35 = 0 35 t = ––– 10 7 t = — 2 0 = 33 + p(3)2 + q(3) 9p + 3q + 27 = 0 3p + q + 9 = 0... –1 + q + 7 = 0 q = –6 (c) At O................. 1 a = p – 2qt Given a = – 4 when t = 4 – 4 = p – 2q(4) – 4 = p – 8q...... q = 1 11 © Penerbitan Pelangi Sdn.. 2p + 2 = 0 p = –1 1 \ The distance travelled is 61— m..... s = 0 5t2 – 35t = 0 5t(t – 7) = 0 t = 0...... sA = 0 \ 0 = 4a – 16 a = 4 2 (b) sA = 4t – t2... vA = 0 4 – 2t = 0 t = 2 1 2 Maximum displacement of A = 4(2) – 22 =4m Substitute q = 1 into 1.... 7 Therefore... (a) (i) v = pt – qt2 When t = 4....... 1 sB = t....... 2 When A and B meet............. ........Additional Mathematics  SPM  Chapter 20   2......... (a) For sA = at – t When t = 4........ 1 s = t3 + pt2 + qt Given s = – 6 when t = 1 s = 5t2 – 35t 7 2 7 = 5 — – 35 — 2 2 1 = – 61— m 4 1 2 1 2 – 6 = 1 + p(1) + q(1) p + q + 7 = 0.. t = 0 is ignored as it is the starting time (c) sA = 4t – t2 vA = 4 – 2t 1 2 3. 4 Substitute p = –1 into 2. (a) s = t3 + pt2 + qt Given s = 0........ 2 1 – 2............. p = 4 Therefore.... – 4 = 4q – 8q – 4 = –4q q = 1 For sA to be maximum.... 6t – 2 = 0 1 t = — 3 1 2 1 Velocity = 3 — – 2 — – 6 3 3 1 2 =—–—–6 3 3 19 = – ––– 3 1 = –6— m s–1 3 1.... p = –1.......... sA = sB 4t – t2 = t t2 – 4t + t = 0 t2 – 3t = 0 t(t – 3) = 0 t = 3....... q = –6 (b) s = t3 – t2 – 6t v = 3t2 – 2t – 6 a = 6t – 2 Velocity when the object at O again = 10(7) – 35 = 35 m s–1 When a = 0................... p = 4......... Bhd.. s = 0 (ii) v = 4t – t2 1 s = 2t2 – —t3 + c 3 When t = 0. s = 0 1 2t2 – —t3 = 0 3 1 t2 2 – —t = 0 3 1 2 – —t = 0.  Additional Mathematics  SPM  Chapter 20 1 (b) s = 2t2 – —t3 3 When t = 0. v = 0 v = 4t – t2 0 = 4t – t2 t(4 – t) = 0 \ t = 4 v When t = 4. sA = 0 \c=0 4 sA = 4t + 3t2 – —t3 3 When t = 2. t = – — is ignored 2 (c) vA = 4 + 6t – 4t2 4 sA = 4t + 3t2 – —t3 + c 3 When t = 0. 4 sA = 4(2) + 3(2)2 – —(2)3 3 1 = 9— cm 3 12 . s = 0 \ c = 0 1 s = 2t2 – —t3 3 When at O again. 1 s = 2(4)2 – —(4)3 3 32 = ––– 3 t 0 4 6 s Distance travelled from t = 0 to t = 4 1 = 2(4)2 – —(4)3 – 0 3 2 = 10— m 3 3 32 – 3 4 Distance travelled from t = 4 to t = 6 1 1 = 2(6)2 – —(6)3 – 2(4)2 – —(4)3 3 3 2 = 0 – 10— 3 2 — = –10 3 2 = 10— m 3 3 4 3 4 t 4 6 4. t = 0 is ignored 3 t = 6 1 When s = 0. vB = 02 – 4 = – 4 cm s–1 (b) vA = 0 for A turning back 4 + 6t – 4t2 = 0 4t2 – 6t – 4 = 0 2t2 – 3t – 2 = 0 (2t + 1)(t – 2) = 0 t – 2 = 0 Total distance travelled in the first 6 seconds 2 = 2 × 10— 3 32 = 2 × ––– 3 64 = ––– 3 1 = 21— m 3 © Penerbitan Pelangi Sdn. 1 2t2 – —t3 = 0 3 1 2 t 2 – —t = 0 3 1 2 – —t = 0 3 t = 6 1 2 (iii) 2 For maximum s. 0 1 t = 2 s. (a) When t = 0. Bhd. 0 –2t . vA = 30 – 4t aB = –10 + 6t Therefore. . t = 0 \ c = 10 1 sB = —t3 – 4t + 10 3 When t = 2. 3 5. sA = 0 \c=0 sA = 30t – 2t2 aB = –10 + 6t vB = –10t + 3t2 + c (d) vA = 4 + 6t – 4t2 a = 6 – 8t For maximum vA. –6 t . 1 sA = 9— cm 3 The distance between A and B 2 1 = 9— – 4— 3 3 2 = 4— cm 3 (d) When A at rest. (a) v = 6 – 2t v . Bhd.Additional Mathematics  SPM  Chapter 20   Distance travelled from t = 2 to t = 3. 3 is the range of time during which the particle is moving towards B.5 m 2 1 2 1 2 6. sB = 0 \d=0 sB = –5t2 + t3 + 20t (e) vB = t2 – 4 When vB = 0. vB = 20 \ c = 20 vB = –10t + 3t2 + 20 sB = –5t2 + t3 + 20t + d When t = 0. 1 sB = —(2)3 – 4(2) + 10 3 2 = 4— cm 3 When t = 2. 0 6 – 2t . a = 0 6 – 8t = 0 6 t = — 8 3 t = — s 4 When t = 0. t2 – 4 = 0 t = 2 When A and B meet sA = sB 30t – 2t2 = –5t2 + t3 + 20t t3 – 3t2 – 10t = 0 t(t2 – 3t – 10) = 0 t(t – 5)(t + 2) = 0 t = 5 s. t = 0 and t = –2 are ignored ∫ sB = (t2 – 4) dt 1 = —t3 – 4t + c 3 When sB = 10. vA = 0 30 – 4t = 0 30 t = ––– 4 15 t = ––– 2 15 15 sA = 30 ––– – 2 ––– 2 2 = 112. 0 < t . ∫ 3 2 4 v dt = 4t + 3t2 – —t3 3 3 (b) aA = aB – 4 = –10 + 6t 6t = 6 t = 1 s 3 4 2 4 1 = 4(3) + 3(3)2 – —(3)3 – 9— 3 3 1 = – 6— cm 3 1 1 Total distance travelled = 9— + 6— 3 3 2 = 15— cm 3 3 4 1 2 ∫ (c) sA = (30 – 4t) dt = 30t – 2t2 + c When t = 0. (a) Initial velocity of A = 30 – 4(0) = 30 m s–1 13 © Penerbitan Pelangi Sdn. ........ 6 – 2t = 0 t = 3 ∫ v = k dt 6 14 Therefore... v = 18 18 = 3k + c..... 2 2 – 1. s = 35 \ c = 35 s = –5t2 + 30t + 35 t 0 (e) s = 6t – t2 ∫ s = (–10t + 30) dt 6 From (a).. t = 6 s is the time when the particle is at A again..... 1 When t = 3... (a) a = –10 v = –10t + c (d) When v = 0.. 6t – t2 = 0 t(6 – t) = 0 t = 6 s Therefore..... Bhd.... 10 = 2k k = 5 It shows that the particle never reaches B. t = –1 is ignored Total distance travelled in the first 5 seconds 1 1 = — × 3 × 6 + — × 2 × 4 2 2 = 13 m t 0 s 3 0 6 9 Substitute t = 7 into v = –10t + 30. v = 30 \ c = 30 v = –10t + 30 This is time when particle turning back When t = 5.  Additional Mathematics  SPM  Chapter 20 ∫ 7.... 0 When t = 1.. t 3 s = –5t2 + 30t + c When t = 0..... (a) a = k (b) s = (6 – 2t) dt = 6t – t2 + c = kt + c When t = 0. v = 8 8 = k + c.. 8 = 5 + c c = 3 \ v = 5t + 3 8. When t = 0. (b) (c) When s = 0. s = 0 \c=0 s = 6t – t2 10 = 6t – t2 t2 – 6t + 10 = 0 b2 – 4ac = (– 6)2 – 4(1)(10) = 36 – 40 = – 4 .. v = 6 – 2(5) = – 4 (b) For maximum height... –5t2 + 30t + 35 = 0 t2 – 6t – 7 = 0 (t – 7)(t + 1) = 0 t = 7....... v = –10(7) + 30 = – 40 m s–1 0 sp 9 0 © Penerbitan Pelangi Sdn..... v = 0 –10t + 30 = 0 t = 3 v 3 5 –4 Maximum height = –5(3)2 + 30(3) + 35 = – 45 + 90 + 35 = 80 m (c) When s = 0. the velocity just before it hits the ground is – 40 m s–1..... substitute k = 5 into 1.... . v = 0 –8t + 10 = 0 10 t = ––– 8 5 = — s 4 When t = 0.Additional Mathematics  SPM  Chapter 20   9. Bhd. p = 10. . (a) h = kt2 + pt v = 2kt + p (b) v = 2(–4)t + 10 v = –8t + 10 For maximum height. k = –4 –q = – 4(4)2 + 10(4) = –64 + 40 = –24 q = 24 15 © Penerbitan Pelangi Sdn. h = –q when t = 4 Therefore. v = 10 10 = p \ v = 2kt + 10 a = 2k Given a = –8 –8 = 2k k = –4 (c) h = – 4t2 + 10t At the surface of the sea.