Chapter Eercise A Q. 1. (i) u 0, v 10, t 5, a? 10 0 + 5a a m/s (ii) u 0, a, t 5, s? s ut + at s (0)(5) + ()(5) 5 m Q.. (i) u 0, v 4, a 3, t? 4 0 + 3t t 8 s (ii) u 0, a 3, t 8, s? s ut + at s (0)(8) + (3)(64) 96 m Q. 3. u 0, a 3, s 6, v? v u + as v 0 + (3)(6) v 6 m/s Q. 4. u 50, v 70, s 300, a? v u + as 4,900,500 + (a)(300) a 4 m/s u 50, v 70, a 4, t? 70 50 + 4t t 5 s Q. 5. a 0.5, s 600, t 40, u? s ut + at 600 u(40) + (0.5)(1,600) 600 40u + 400 00 40u u 5 m/s Q. 6. u 3, v 11, t 6, s? s ( u + v ) t s ( 3 + 11 ) (6) 4 m Q. 7. u 3, v 0, s 6, a? v u + as 0 9 + a(6) a 3 4 m/s u 3, v 0, a 3 4, t? 0 3 + ( 3 4 ) t t 4 s Q. 8. (i) u 70, v 50, t 8, a? 70 50 + a(8) a m/s u 70, t 8, a, s? s ut + at s 70(8) + ( ) (64) 560 80 480 s (ii) u 50, v 0, a, s? v u + as 0,500 5s s 500 m Q. 9. (i) u 4, v 0, a 8, s? v u + as 0 576 + ( 8)s s 36 m (ii) v u + as 0,304 + ( 8)s s 144 m Q. 10. (i) 7 km/hr 7,000 m 3,600 s 0 m/s (ii) 48 km/hr 48,000 m 3,600 s v u + as 1,600 400 + a(500) 9 a 9 m/s 40 3 0 + ( 9 ) t t 30 s 40 3 m/s 1
(iii) v u + as 0 1,600 + 9 ( 9 ) s s 400 m Q. 11. (i) 1 km/hr 1,000 m 3,600 s 5 18 m/s 7 km/hr 0 m/s ad 54 km/hr 15 m/s v u + as 5 400 + (a)(35) a m/s (ii) v u + as 0 5 + ( ) s s 45 m Q. 1. Let t be the time of meetig. s 1 5t + (3)t 5t + 3 t s 7t + ()t 7t + t s 1 + s 16 1t + 5 t 16 t 6 ( t 54 5 rejected ) At t 6, v 1 u + at 5 + (3)(6) 3 m/s v 7 + ()(6) 19 m/s Q. 13. (i) They will meet whe s 1 + s 400. 3t + (4)t + 7t + ()t 400 10t + 3t 400 3t + 10t 400 0 (3t + 40)(t 10) 0 t 40 OR t 10 3 t 10 s (t 40 is rejected as impossible) 3 (ii) s 1 3(10) + (4)(10) 30 m s 7(10) + ()(10) 170 m Q. 14. (i) 30 km/hr 30,000 m 3,600 s 5 3 0 + t t 4 6 s (ii) v u + as ( 50 3 ) 0 + ()s s 65 9 m (iii) v u + as Eercise B 0 ( 50 3 ) + (a)() a 65 9 m/s Q. 1. (i) u 0, a, v 30 30 0 + t t 15 s (ii) Speed (m/s) 30 A B 5 3 m/s 15 1 10 Time (s) Total distace covered Area uder graph A + B + C (15)(30)+ (1)(30) + (10)(30) 5 + 360 + 150 735 m (iii) u 30, v 0, t 10 0 30 + 10a 10a 30 a 3 m/s Magitude of deceleratio is 3 m/s Q.. (i) 0 0 + a(5) a 4 m/s C
Q. 3. (i) (ii) s ut + at s 0(5) + (4)(5) 50 m (iii) Remaiig distace 00 m. 00 m at 0 m/s takes 10 secods. The total time take 5 + 10 15 s Speed (m/s) 50 (i) Durig deceleratio Area uder graph 150 Area C 150 (t)(60) 150 30t 150 t 5 s (ii) Durig acceleratio u 0, v 60, a 3 Q. 4. 5 Durig acceleratio u 0, v 50, a 5 50 0 + 5t t 10 s Durig deceleratio u 50, v 0, a 10 0 50 10t 10t 50 t 5 s Time (s) Total distace travelled Area uder graph (10)(50) + (5)(50) + (5)(50) 50 + 1,50 + 15 1,65 m Total distace (ii) Average speed Total time 1,65 10 + 5 + 5 Speed (m/s) 60 1,65 40 40.65 m/s A B C 60 0 + 3t t 0 s Total distace covered Area uder graph (0)(60) + (75)(60) + 150 600 + 4,500 + 150 5,50 m (iii) Average speed 5,50 100 Q. 5. (i) v u + as (40) 0 + ()s s 400 m 40 0 + t t 0 s (ii) v u + as 0 (40) + ( 5)s s 160 m 0 40 + ( 5)t t 8 s 5.5 m/s (iii) Remaider 1,000 400 160 440 m Q. 6. (i) 440 m at 40 m/s takes 11 secods. Total time 0 + 11 + 8 39 s 7 0 + a(9) 75 t Time (s) a 3 m/s 3
(ii) v u + as 0 (7) + (a)(54) a 6 3 4 m/s (iii) d part: 0 7 + ( 6 3 4 ) t t 4 s 17 9 4 Distace Area uder the curve (13)(7) 175 m Average speed Total distace Total time Q. 8. Q. 9. 5 15 + (a)(4) 10 a m/s t 1 t Let t time spet acceleratig (t)(10) + (1 t)(10) 100 5t + 10 10t 100 a 10.5 m/s 4 5 0 5t t 4 (iv) First part: 15 0 + 3t t 5 s d part: 175 13 13 m/s Q. 10. t 18 t (t)(5) + (18 t)(5) 350 1.5t + 450 5t 350 40 a 100 1.5t t 8 s d 8 4 Q. 7. 5 15 15 7 + ( 6 3 4 ) t t 1 7 9 s Aswer: After 5 secods ad after ( 9 + 1 7 9 ) 10 7 9 s B A C 40 Area uder the curve 980 ()(10) + (15)() + (40 )(5) 980 4 s y z 40 0 z 40 8 5 Area 700 Q. 11. (i) 40 (0)(40) + 40y + (5)(40) 700 400 + 40y + 100 700 Time 0 + 5 + 5 30 s a 60 y y 5 z 3a
a 40 60 3 m/s Sice 60a 40 3az z 60a 3a 60 3 0 s Area 8,800 (60)(40) + 40y + (0)(40) 8,800 1,00 + 40y + 400 8,800 40y 7,00 Distaces are 1,00, 7,00, 400 metres (ii) 40 y 180 Total time 60 + 180 + 0 60 s 30 0 10 60 a b c d Shaded regio 1 km 1,000 m The deceleratio 3a 3 ( 3 ) m/s (from (i)) d 0 10 s Shaded regio 1,000 0c + (d)(0) 1,000 0c + (10)(0) 1,000 0c + 100 1,000 0c 900 c 45 Dotted regio: u 40, v 0, a, t b 0 40 b b 10 Area 10(0) + (10)(0) 300 m Total area 8,800 (60)(40) + 40a + 300 + 900 + 100 8,800 40a 6,300 a 157.5 Total time 60 + 157.5 + 10 + 45 + 10 8.5 Eercise C Q. 1. 10 Etra time 8.5 60 10 0 + t t 5 s.5 secods more tha the first time QED 5 t 5 Area uder the curve 100 (5)(10) + (t 5)(10) 100 t 1.5 s Q.. (i) 1 15 4 1 v 0 + ( 1 ) (15) m/s 0 + ( 5)t t 4 s (ii) Distace Area uder the curve Q. 3. (i) 4 0 + t t 1 s s ( u + v ) t 48 ( 4 + 0 ) t t 4 s ( 19 ) ( ) 19 3 8 m 5
(ii) First part: a ut + at 0(1) + ()(1) 144 m Total distace 144 + 48 19 m Total time 1 + 4 16 s Average speed 19 1 m/s 16 Q. 4. s 1 ut + at 0(t) + (4)t t s 0t s ut + at 48 u(4) + (a)(16) u + a 1 (1st ad d parts) Solvig these gives (i) a 3 m/s (ii) u 6 m/s (iii) First 6 secods: s ut + at s (6)(6) + (3)(36) 90 m s 1 s t 0t The distace travelled 90 48 4 m t 10 s s 00 m Q. 8. O A 6s B s C 105 m 63 m Q. 5. (i) v 1 10 + 3t; v 0 + t (ii) s 1 10t + 3 t ; s 0t + t (iii) v 1 v 10 + 3t 0 + t t 10 s (i) A to B: u u, s 105, t 6, a a s ut + at 105 6u + 18a A to C: u u, s 168, t 8, a a s ut + at Q. 6. (iv) s 1 s 10t + 3 t 0t + t t 0 OR t 0 s A 4s B 6s C 48 m 10 m A to B: u u, t 4, s 48, a a 168 8a + 3a Solvig gives a 3.5, u 7 Aswer: a 3.5 m/s (ii) O to A: u 0, v 7, a 3.5, s s v u + as 49 0 + 7s s 7 m 6 s ut + at 48 4u + 8a u + a 1 A to C: u u, t 10, s 150, a a s ut + at 150 10u + 50a u + 5a 15 Solvig gives a 1, u 10 Aswer: 1 m/s Q. 7. (i) s ut + at 18 u() + (a)(4) u + a 9 (1st part) Q. 9. 1st part: s ut + at 39 u(1) + a(1) u + a 78 1st ad d parts: 76 u() + a() u + a 76 First three parts: 111 u(3) + a(3) 6u + 9a u + 3a 74 Solvig the first two equatios gives a, u 40 Solvig the last two equatios gives a, u 40 They are cosistet.
Stoppig: v u + as 0 (40) + ( )s s 400 m It will travel a further 400 111 89 m Q. 13. Speed (m/s) Q. 10. a to b: s ut + at Q. 11. 0 u(5) + (a)(5) u + 5a 8 a to c: 40 u(8) + a(8) u + 8a 10 Solvig these gives u 7 3 m/s, a 3 m/s a to d: a ut + at 60 7 3 t + ( 3 ) t t + 7t 180 0 (use formula) t 10.4 s ( 17.4 is rejected). The time take 10.4 8.4 s 10 3 6 Area (10) + 3 + (6) 1,000 5 m/s Distaces are: (10)(5) 15; 3(5) 800; (6)(5) 75 m Q. 1. (i) : t 3 : 1 3 4 : 4 t T Let the top speed v v t v 7 The time take 90 secods T T v + v 7 Q. 14. (i) 7v + v 14 9v 14 9v 14 90 9v 1,60 v 140 m/s Distace (90)(140) 6,300 m 6.3 km 1 40 8 Time (s) 3 (0) 15 s 4 t 4 (0) 5 s (ii) (iii) v 0 + (1)(15) 15 m/s 15 15 5 s Area uder curve (0)(15) 150 m (ii) Area (1) + 40 + (8) 1,000 0 m/s 0 14 1 T 1 Area 1 Area 14T (1)(0) + 0(T 1) T 0 s s 14T 80 m 7
Q. 15. 1st part: s ut + at Q. 16. (i) 4 u() + a() u + a 1 1st ad d parts: 48 u(3) + a(3) u + 3a 3 Solvig these gives a 8, u 4 First three parts: 7 4t + (8)t t + t 18 0 t 3.77 (t 4.77 is rejected) Time take 3.77 3 0.77 s 16 Q Q. 17. (i) s 1 0 + ( ) t 4 t s 0 + (1)t t Distace apart, s s 1 + s 3 4 t Whe t 10, s 3 (100) 75 m 4 (ii) Whe s 108 3 4 t 108 t 144 t 1 s This is 1 10 secods later Q. 18. Let v top speed A B C 16 8 Distace i A (16)(16) 18 m 4 8 (ii) Distace i C (8)(16) 64 m Total distace travelled 19 m Remaider for B 300 19 108 m Time take 108 16 6.75 s Total time 16 + 6.75 + 8 30.75 s 4 ( v 4 + v 8 ) v 1,00 ( 3v 8 ) v 1,00 3v 16 1,00 v 6,400 8 v 80 m/s Time v 4 + v 8 80 4 + 80 8 30 s Q. 19. Let t time after cyclist passes P 8 t T Let T the time take : t d : a : 1 3 : 3 3 T ad t 3 T v 0 + (1) ( 3 T ) 3 T Area 300 (T) ( 3 T ) 300 T 30 s Greatest gap v 1 v 1 0 + 1(t 5) t 17 at t 17, s 1 1(17) 04 m s (1)(17 5) 7 m Gap 04 7 13 m Q. 0. Greatest gap v 1 v 8 + 4t 30 + 3(t ) 8 + 4t 30 + 3t 6 t 16 at t 16, s 1 8(16) + (4)(16) 640 at t 16 14, s 30 (14) + (3)(14) 714 Gap 714 640 74 m QED
Q. 1. Q.. A G P 1 P After t secods, Alberto has travelled s 1 1t + t After t secods, Gustav has travelled s t At both P 1 ad P, s 1 s + 1t + t t + t 4t + 44 0 t, Aswer: (i) After secods 10 (ii) After 0 secods more 50 6 Area uder the curve 696 (10) + 50 + (6) 696 1 1 0 + a 1 (10) a 1 1. m/s Similarily, a m/s Q. 3. Take speeds, acceleratios, distaces relative to the goods trai. Let p the passeger trai ad g the goods trai. The iitial relative speed, u pg u p u g 80 30 50 m/s The relative distace 1,500 m The fial relative speed is zero, sice the two trais must evetually be travellig at the same speed to avoid a crash. v u + as 0 (50) + a(1,500) a 5 6 m/s The relative deceleratio is, therefore, 5 6 m/s The actual deceleratio of the passeger trai is 5 6 m/s, sice the goods trai does ot decelerate at all. Q. 4. (i) Iitial relative speed, u 0 8 1 m/s Relative distace 10 m Fial relative speed 0 m/s v u + as 0 (1) + a(10) a 3 5 m/s t : t : 1. 5 : 3 5 8 : 3 8 5 8 T, t 3 8 T (ii) (i) u 1 a 1 s 10 66 54 t? s ut + at v 0 + (1.) ( 5 8 T ) 3 4 T Area 696 (T ) ( 3 4 T ) 696 T 1,856 T 8 9 54 1t + ( 1)t t 4t + 108 0 (t 6)(t 18) 0 t 6, 18 Aswer: After 6 secods 9
10 (ii) u 1 a 1 s s t t s ut + at s 1t t ds dt 1 t Eercise D 0 (Sice s is a miimum) t 1 At t 1, s 1(1) (1) 7 m This meas that they have travelled a distace of 7 m towards each other, ad so the distace betwee the will be 10 7 48 m. Q. 1. (i) s 35t 4.9t 0 t(35 4.9 t) 0 t 0 OR t 350 49 50 7 s (ii) v 0 35 9.8t 0 s 35 ( 5 t 350 98 50 14 5 7 7 ) 4.9 ( 5 7 ) 15 6.5 6.5 m Q.. (i) u(1) 4.9(1) 16.1 u 4.9 16.1 u 1 m/s (ii) v 0 1 9.8t 0 t 1 9.8 10 98 15 7 at t 15 7, s y 1 ( 15 (iii) s y 0 7 ) 4.9 ( 15 7 ) 45.5.5 m 1t 4.9t 0 t 0 OR t 1 4.9 10 49 30 7 s Q. 3. (i) s 1 s 0(t) + 4.9t 14.7 (t 1) + 4.9 (t 1) t 3(t 1) + (t 1) t 3t 3 + t t + 1 t (ii) 4.9() 19.6 m (iii) 14.7 0 1st Q. 4. (i) Let t time Q is i motio. 1 d t + time P is i motio. s P s Q 47(t + ) 4.9(t + ) 64.6t 4.9t 47t + 94 4.9t 19.6t 19.6 64.6t 4.9t 74.4 37.t t s (ii) 64.6() 4.9() 109.6 m Q. 5. First t secods u u, s 70, a 9.8, t t 70 ut 4.9t Equatio 1 First t secods u u, s 70 + 50 10, a 9.8, t t 10 ut 4.9 (t) 10 ut 19.6t Equatio Eq : 10 ut 19.6t Eq. 1: 140 ut + 9.8t 0 9.8t 00 98t t 100 49 t 10 7 s
70 u ( 10 7 ) 100 4.9 49 70 10u 7 10 u 56 m/s Q. 6. (i) u u, a 9.8, s 30, t 5 s ut + at 30 5u + ( 9.8)(5) 9.5 5u u 18.5 m/s (ii) 18.5 + ( 9.8)(5) 30.5 Speed 30.5 m/s Q. 7. 49t 4.9t 78.4 10t t 16 t 10t + 16 0 (t )(t 8) 0 t, 8, t 8 + t 10 QED OR + t b a + t ( 10) 1 + t 10 Q. 8. 70t 4.9t 10 700t 49t,100 7t 100t + 300 0 Product of roots a c t 300 7 7 t 300 QED Eercise E Q. 1. s d, t, u 0, a a s ut + at d a Equatio A Q.. (i) (ii) a d + k, t, u 0, a a s ut + at d + k a() d + k a Equatio B 4 Equatio A a 4d But a d + k d + k 4d 15ak k 3d QED 5k A B C 3k Distace i A (5k)(15ak) 37 ak Distace i C (3k)(15ak) ak Remaider 90ak 37 ak ak 30ak Time for B 30ak 15ak k s Total time 5k + k + 3k 10k s T t Let T the total time : t d : a 5a : 3a 5 8 : 3 8 5 8 T ad t 3 8 T v 0 + (3a) ( 5 8 T ) 15 8 at Area 90ak (T) ( 15 8 at ) 90ak T 96k T 4 6 k 11
Q. 3. h ut + ( g)t h ut gt gt ut + h 0 Product of roots a c t h g QED Q. 4. Let PQ QR P Q R (i) The jourey P R: u u, v 7u, s, a a v u + as 49u u + (a)() 48u 4a 1u a a 1u The jourey P Q: u u, v v, s, a 1u v u + as v u + ( 1u )() v u + 4u v 5u v 5u (ii) P to Q: u u, v 5u, s, t, a a 5u u + a 4u a Q to R: u 5u, v 7u, t t, a a (ii) Distace A ( v a ) v v a Distace C ( v b ) v v Remaider s v b a v b Time take i B ( s v a v b ) v v s v a v b Total time take a v + ( v s v a v T v a + v b + v s t Let T the total time : t b : a b a + b : a a + b bt a + b, t at a + b v 0 + (a)( bt a + b ) abt a + b Area s ( (T) abt a + b ) s T s ( a + b ab ) QED Q. 6. (a) t 1, u u, a a, s s 1 s ut + at s 1 u( 1) + a( 1) b) + v b QED Q. 5. (i) 7u 5u + at t u a t QED u u + a a + a t, u u, a a, s s s ut + at s u + a 1 a A B C b Distace travelled, s s s 1 u + a a QED
Q. 7. (b) Whe, s 17 17 u + a a u + 1 a 17 Whe 7, s 47 47 u + 7a a u + 6 a 47 Solvig these gives a 6, u 8 (i) 10 s u + a a 8 + (6)(10) (6) 65 m (ii) s 8 + 6 (6) (6 + 5) m (c) s + s + 1 56 6 + 5 + 6( + 1) + 5 56 0 Aswer: I the 0th ad 1st secods y z Average speed total distace total time v + yv + zv + y + z 5v 6 5v + 5yv + 5zv 3v + 6yv + 3zv 5 + 5y + 5z 3 + 6y + 3z + z y + z y Fractio of distace travelled at costat yv speed v + yv + zv y + y + z y + y + z y ( + z) + y y y + y y y 4 5 QED Q8 Q. 8. (i) 4 4 y z (ii) v 4 4z z ad v 4 Area uder curve d ( t y ) ( ((t y)) + y()(t y) + t y (t y) + yt y + (t y) d (t y) + yt y d t yt + y + yt y d )((t y)) d t y d y t y t d y t d QED + y t t y v 4 (t y) 13