Chapter 11 Exercise 11.1

Chapter 11 Exercise 11.1 Q. 1. Departure Arrival taken for journey 09:2 1:2 4 hrs 0 mins 09:0 1:7 hrs 2 mins 14:47 21: hrs 48 mins 0:7 1:1 12 hrs 9 mins 12: 2:11 hrs mins 00:29 18:14 17 hrs 4 mins 14:1 18:0 hrs 47 mins 07:2 22:18 14 hrs mins Q. 2. Distance (km) Speed (km/hr) (hrs and minutes) 0 0 2 hrs 7 10 1_ 2 hr 1. 2 hrs 1 mins 0 0 9 hrs mins 0 minutes 0 80 4 hrs 22. mins 80 7 1 hr 4 mins 1,0 7 hrs 12 mins 90.4 1 hrs 40 minutes 12 1 48 mins 0 0 hrs 0 mins 12. 18 41 2 mins Q.. (i) 12 km/h 1,000 = 12,000 m/h,00 = 1 m/s (ii) 800 m/s,00 = 2,880,000 m/hr 1,000 = 2,880 km/hr (iii) km/h 1,000 =,000 m/h,00 = 17. m/s (iv) cm/s 0 = 0.0 m/s. (v) 18 cm/s,00 = 4,800 cm/hr 0,000 = 0.48 km/hr Q. 4. D = S T = 11 2 = 22 km Q.. S = D T 00 km = = 0 km/hr hrs Active Maths 2 (Strands 1 ): Ch 11 Solutions 1

Q.. (i) :, 12:, : (ii) The 08:00 trains gets to Dublin in 2 hrs. (iii) William should leave home by :4 at the latest. He will catch the 07:01 train to Portadown. In Portadown he will catch the 07:21 which arrives in Drogheda at 08:2. The next train to arrive in Drogheda at 12: which is too late. (iv) 18:1 1:21 = 1 hr 2 mins 9 mins + 1 mins = 22 mins 22 0 = 19.4 % 112 Q. 7. T = D S = 10 = 1 hr 44 mins 7 19:12 + 1 hr 44 mins = : Q. 8. (i) 40 m/s,00 = 1,224,000 m/hr 1,000 = 1,224 km/hr (ii) 00,000 m/s,00 = 1,080,000,000 m/hr 1,000 = 1,080,000 km/hr Q. 9. T = D S = 18 = 1 hr = mins 8:4 am + mins = 9:0 a.m. Q.. (i) 17:1 bus from Athlone (ii) 04 (iii) 19:0 1:0 = 4 hrs 2 mins (iv) The 14:00 and 1:1 buses on Sunday are the fastest, taking 2 1 2 hours. Q. 11. T = D S = 00 = hrs 40 mins 7 T = D S = 00 = hrs 1 mins 80 It takes 2 mins more. Q. 12. (i) S = D T = = 0 km/hr 1.1 (ii) T = D S = 1 = 1 hr 22. mins 80 (iii) Total distance = + 1 = 1 km (iv) Total time = 1 hr mins + 1 hr 22. mins = 2 hrs 28. mins (v) S = D T = 1 2.47 = 2 km/hr (vi) S = D T = 270 = 4 km/hr 4 9 = 8 = 48 km/hr Q. 1. (i) D = S T = 0 1 = 9 km 0 (ii) T = D S = 180 = 2 hrs 1 mins 80 2 Active Maths 2 (Strands 1 ): Ch 11 Solutions

(iii) 9 + 180 = 27 km (iv) 1 hr mins + 2 hrs 1 mins = hrs 0 mins (v) 12: + hrs 0 mins = 1:2 (vi) S = D T = 27 _ = 71.791 = 71 km/hr (vii) 70 mph 2 = 140 miles mph 4 = 140 miles Total distance = 280 miles 8 = 448 km S = D T = 448 = 7 km/hr Q. 14. 1. km @ km/h = 0 min distance 1. km 40 km @ 0 km/h = 1 hr min 40.0 km km in 42 m = 42 min +.0 km total time = 2 hr 2 min 1. km 1. km Average = =.289 km/h =. km/h 2 hr 2 min Q. 1. (a) S = 4 = km/hr 1.2 (b) Cycle: m/s Swim: 2 m/s Run: 12 km/hr or 1_ m/s Trevor s fastest speed was in the cycling portion of the triathlon. (c) Swim: Distance = S T = 7.2 1 = 0.72 km Cycle: 4 km Run: km Total = 4 + + 0.72 =.72 km =,7 m (d) Total time = mins + 1 hr 1 mins + 0 mins = 11 mins 0 = 7,80 seconds Speed =,7 = 7.09 m/s 7,80 Q. 1. T = 17 = 8 hrs 4 mins 17 2 = 7 rest stops Total time = 8 hrs 4 mins + 12 mins = 8 hrs 7 mins Active Maths 2 (Strands 1 ): Ch 11 Solutions

Q. 17. (i) Daniel (ii) Ellie (iii) Daniel: 2 = 1.2 m/s Ellie: 2 2 = 1 m/s (iv) Daniel: 2 0 = m/s Ellie: 2 = 1.2 m/s (v) No, each length they swim is 2 m. (vi) At 22. secs they passed for the first time. Daniel was winning as he had already completed a full length. They passed again after secs. At this point Ellie was winning as she was passing Daniel. Q. 18. Graph A Graph B Graph C Graph D (i) 40 m 4 km 4 m 0 m (ii) s hrs mins s (iii) 40 4 4 = 8 m/s = 0.8 km/hr = 1 1 m/min or 0.022 m/s 0 = m/s Q. 19. (i) 8:00 (ii) 8:0 (iii) 2. km (iv) It occurs between C and D as the distance travelled remains the same between these two points. (v) Speed = 0. 1_ = 1. km/hr (vi) He moves fastest between points D and E because the slope of the graph between these two points is the greatest. He may be moving faster because he was late for school. 2. km (vii) = km/hr Q.. (a) (i) Graph C (iii) Graph B (ii) Graph A (b) Distance from home (km) taken Q. 21. (a) A to B Upward-sloping straight line constant speed. B to C A slower constant speed than A to B. 4 Active Maths 2 (Strands 1 ): Ch 11 Solutions C to D A faster constant speed than A to B and B to C. D to E The object is stationary slope = 0. E to F The object has started travelling at a constant speed slower than previously. F to G The object is travelling at its fastest constant speed. (b) Between A and B: Travelling at a constant speed of 18 km/hr. B and C: Travelling at a constant speed of 12 km/hr. C and D: Constant speed of km/hr. D and E: Object is not moving. E and F: Consant speed of km/hr. F and G: Constant speed of km/hr. Slope between F and G is highest indicating the part of the graph where the object is travelling the fastest. Q. 22. D = 80 1 = 80 km. S = D 80 T = = 0 km/hr. 2 0 Q. 2. T = 10 0 = hrs 0 km/h 2. = 7 km/h T = 10 7 = 2 hrs They would have saved hours.

Q. 24. (i) A B The object remains km away from its starting point for one hour. It may be stationary. B C The object moves closer to the fixed point travelling km in 1 hour at a constant speed. C D The object travels away from the fixed point again. It moves km at constant speed away from the fixed point until it is km away in total. D E The object travels 0 km in 1 hour at a constant speed. It is now 0 km away. E F The object travels a further km at a constant speed. It is now 0 km away from its starting point. (ii) Between points P and Q the graph is curved, which means the object is accelerating (average acceleration from P to Q is 2 m/s 2 ). After point Q the graph curves more steeply, suggesting that the speed of the object is increasing more rapidly as the graph approaches point R (average acceleration from Q to R is 1 m/s 2 ). (iii) At point X the object is 0 km away from the fixed point. It starts to travel towards the fixed point. The graph is curved which suggests the object is not travelling at a constant speed. The slope at any point is negative which means the object is decelerating as the graph approaches point Y. At point Y. The object begins to move away from the starting point. The graph is curved between point Y and point Z which suggests it is not travelling at constant speed. The slope at any point is positive which means the object is accelerating as the graph approaches point Z. Q. 2. (i) Graph B (iii) Graph D (ii) Graph A (iv) Graph C Q. 2. Race 1: Distance 0 m m m Race 2: Distance (m) 0 40 0 Race : Distance (m) 2 1 Race 4: Distance (m) 40 0 1 2 4 1 2 4 1 1 2 4 2 4 : T = 40 = 8 secs : T = 40 8 = secs Active Maths 2 (Strands 1 ): Ch 11 Solutions 7 7 8 8

Q. 27. Student report should include the following information: Adam won the race (Barry was 2nd and Carl was rd). Carl was given a head start (1 km). Barry stopped for a rest. At the start of the race Carl was ahead. Adam overtook Carl (at approx. 7 minutes), then Barry overtook Carl (at approx. 19 minutes). The runners overall average speeds (approx.) were: Adam 11 km/hr Barry km/hr Carl 9 km/hr Q. 28. (i) Child on a Ferris wheel. (iii) Ball thrown vertically in the air. Distance from ground Distance from the ground (ii) Swimming four lengths of a pool. (iv) Satellite orbiting the Earth. Distance from starting point (Constant speed was assumed.) Distance from the earth s surface Q. 29. Distance (km) 40 0 1 2 4 0 40 0 80 0 1 140 (i) Total distance: 40 km (ii) Total time: 2 hrs mins (iii) 40 = 19.2 km/hr 2 1 12 (mins) Active Maths 2 (Strands 1 ): Ch 11 Solutions

1 Q. 0. (i) 0,00 = 1 4 km = 20 m (ii) Distance travelled = 0 m 0 m = m/s = 118.8 km/hr (iii) Distance travelled = 0 m 72 km/hr = m/s T = 0 = s (iv) 0 km/hr = 1 2 m/s. D = 1 2 2 = 41 2 m Q. 1. (i) After 7 secs Chang has travelled 140 m. Henry can cover m more than Chang every second. 140 = 14 secs Henry will overtake Chang after 14 seconds. (ii) Chang covers 0 m more than Henry. His speed is % greater than Henry s. 0 0. = 00 m Length of race = 00 m D = 1 2 = 1 m Assuming the height of the man (while standing) is negligible then height of apartment is 8 1_ m. Active Maths 2 (Strands 1 ): Ch 11 Solutions 7