1New Additional Mathematics by Pan Pacific Publishing EXERCISE 24.2 (PAGE 549) QUESTION NO. 1 A river is flowing at 4 m/sec due south. A boat, whose speed in still water is 3 m/sec, is steered in the direction due east. Find the true speed and direction of the motion of the boat. SOLUTION VB = VB/W + VW Where θ VB/W = 3m/s VB is the true velocity of the boat VW = 4m/s VB/W is the velocity of the boat in still water VB and Vw is the velocity of the water By using Pythagoras theorem (VB)2 = (VB/W) 2+ (VW)2 (VB)2 = 32 + 42 = 25 VB = 5m/s Velocity Diagram Now tan θ = 4/3 θ = tan-1(4/3) θ = 53.1o Hence True velocity of the boat is 5m/s in the direction (90+53.1) 143.1o OR True velocity of the boat is 5m/s by making an angle of 36.9o with the bank downstream. Nadeem Ahmad, Resource Academia, Ph. 0333-4404244. E-mail:
[email protected] 2 QUESTION NO. 2 A river is flowing at 3 m/sec due east. A speedboat, whose speed in still water is 5 m/sec, is steered in the direction on a bearing of 330o. Find the resultant velocity of the speedboat. SOLUTION: VW = 3m/s 60O VB/W = 5m/s VB = VB/W + VW VB 60O x Where Velocity Diagram VB is the true velocity of the boat, VB/W is the velocity of the boat in still water and Vw is the velocity of the water. EXPLANATION: (i) The boat which is steered in the direction of 330o will move by making an angle of 60o with the bank upstream as the bearing of upstream bank in this case is 270o and hence 270o + 60o = 330o. (ii) The angle between VB/W and VW = 60o (alternate angles) CALCULATIONS: Applying cosine Rule in the velocity diagram triangle: VB2 = 52+32 -2(5)(3) cos60o VB2 = 25+9 -30 cos60o VB2 = 34 -15 = 19 VB = 4.36m/s Nadeem Ahmad, Resource Academia, Ph. 0333-4404244. E-mail:
[email protected] 3 To find angle x, VB makes with the bank downstream, we will apply sine rule sinx /5 = sin60o/4.36 x = 83.2o Hence Velocity (true) of the boat is 4.36 m/s by making an angle of 83.2o with the bank downstream. OR velocity (true) of the boat is 4.36 m/s in the direction 006.8o as (90– 83.2 = 6.8). QUESTION NO. 3 In the diagram, a river is flowing at a speed of 2.5 m/sec due east. A boat, whose speed in still water is 6 m/sec, is steered in the direction due north. Find the true velocity of the boat. SOLUTION VW = 2.5m/s VB/W = 6m/s VB = VB/W + VW Where VB θ Velocity Diagram VB is the true velocity of the boat, VB/W is the velocity of the boat in still water and Vw is the velocity of the water. CALCULATIONS: (VB)2 = (VB/W) 2+ (VW)2 (VB)2 = 62 + 2.52 VB = 6.5m/s Nadeem Ahmad, Resource Academia, Ph. 0333-4404244. E-mail:
[email protected] 4 Now tan θ = 2.5/6 θ = tan-1(2.5/6) θ = 22.6o Hence Velocity (true) of the boat is 6.5 m/s by making an angle of 67.4o with the bank downstream. QUESTION NO. 4 A soldier who swims at 1.2 m/sec in still water wishes to cross a river 20 m wide. The water is flowing between straight parallel banks at 1.8 m/sec. He swims upstream in a direction making an angle of 70o with the bank. Find a) The resultant velocity b) The time taken for the crossing, to the nearest second. SOLUTION: θ VW = 1.8m/s θ 70O 20 m VB/W = 1.2m/s VB VB = VB/W + VW 70O θ Where Velocity Diagram VB is the true velocity of the boat VB/W is the velocity of the boat in still water and Vw is the velocity of the water. a) Calculations for true velocity of the boat: Applying cosine Rule in the velocity diagram triangle: VB2 = 1.22+1.82 -2(1.2)(1.8) cos70o VB = 1.79m/s Nadeem Ahmad, Resource Academia, Ph. 0333-4404244. E-mail:
[email protected] 2 = sin 70o/1. 20 m 20 d This is shown by a fold faced line here.78m so time taken for crossing = 31.nadeem78@yahoo. VB makes with the bank downstream.79 θ = 39.0o we will find the component of distance along which VB is calculated. 0333-4404244.0o is obtained by producing the line segment representing VB in Distance Diagram Velocity Diagram.78/1. Θ =39.5 To find angle θ. we will apply sine rule sin θ /1.79 ≈ 18 sec. It is the same line which Θ =39. E-mail: ahmad.0o Hence Velocity (true) of the boat is 1. b) In this distance diagram.79 m/s by making an angle of 39. Nadeem Ahmad.0o with the bank downstream.com . Ph. (Shown as a dotted line there) Now sin 39o = 20/d d = 31. Resource Academia. what is the course taken by the boat.126 ≈ 18 sec.0 = 1. The component of VB which makes the boat cross the river is VB sinθ. Resource Academia.0o VB sin θ = 1. b) The resultant speed of the boat.79 sin 39. 0333-4404244. 20 VB sinθ VB= 1. between straight parallel banks.79 Velocity component diagram QUESTION NO.5 m/sec. In order to travel directly from A to B. A boat. 5 m upstream. 5 The diagram shows a river. crosses the river from a point A on one bank to a point B on the opposite bank. In order to make the return journey from B to A. Ph. whose speed in still water is 6 m/sec. the boat is steered in a direction making an angle α to the bank as shown.nadeem78@yahoo. Nadeem Ahmad. This is shown by a fold faced line in velocity component diagram (need not to be shown in the answer script) VB cosθ Θ =39.com . Find a) The value of α.126 m/sec so time taken for crossing = 20/1. 30 m wide. flowing at a speed of 3.6 ALTERNATIVE METHOD Time taken for crossing the river can also be calculated if we split VB into its horizontal and vertical components. c) The time taken for the crossing. E-mail: ahmad. to the nearest second. Resource Academia. VB/W is the velocity of the boat in still water and Vw is the velocity of the water. E-mail: ahmad.5m/s course y 80.5o α VB 30 m VB/w= 6m/sec VB/w = 6m/sec 80.5o Vw = 3.5o 99.5o)/6 θ = 35.5 = (sin 99.5o + 35.5o Vw = 3.5o.4o Nadeem Ahmad.com .7 SOLUTION: 5m B B x =80. To find x tan x = 30/5 x = tan-1(6) = 80.5o 80.5o so the angle between VW and VB is 99. a) (sin θ) /3.5 m/sec Θ α A Velocity Diagram (Outward Journey) A Velocity Diagram (Return Journey) VB = VB/W + VW Where VB is the true velocity of the boat. Ph.
[email protected]) = 45.1o α = 180o – (99. 1o Hence course taken by the boat in the return journey is 180o – (80.33 ≈ 7 sec RETURN JOURNEY (sin 80.4/4.4o with the bank upstream.5o + 35. 0333-4404244.8 b) (sin 45.5 y = 35.1o) = 64.nadeem78@yahoo. we will find the distance to be covered along VB which is shown as a partially dotted line in the Velocity Diagram (Outward Journey). Resource Academia.4 m Time taken = 30.33m/sec c) To find the time taken. Ph.5o) /6 = (sin y)/3. E-mail: ahmad.4o) /VB = (sin 99. Distance to be covered = √302 + 52 = 30.com . Nadeem Ahmad.5)/6 VB = 4. nadeem78@yahoo. vectors are added by the equation VA = VA / W + VW VA2 = 3002+602 -2(300)(60) cos30o VA = 250 km/h To find θ (sin θ) / 60 = (sin 30o) /250 Θ = 6. Resource Academia. Ph. 6 The speed of an aircraft in still air is 300 km/h.9o Hence True velocity of the aircraft is 250 km/hr in the direction 053. 0333-4404244.com . Nadeem Ahmad. The aircraft is steered on the course in the direction 060o.9 QUESTION NO.1o. The wind velocity is 60 km/h from the east. E-mail: ahmad. Find the true velocity of the aircraft 120o SOLUTION N Vw = 60km/h VA/w = 300km/h 60o Vw = 60km/h 30o N 60o VA VA/w = 300km/h θ In the diagram.
[email protected] = 0. VA N A B B Vw = 60km/h 300 50o VA/w = 300km/h 30o θ Vw = 60km/h 90o x VA In the diagram. for the journey from A to B = 200/ 265. The wind is blowing from the direction 030o at 60 km/h. The speed of the aircraft in still air is 300 km/h and the pilot sets the course on the bearing θo.4 km/h Hence time taken.com . Ph. 7 An aircraft flies due east from A to B where AB = 200 km. in minutes. for the journey from A to B. vectors are added by the equation VA = VA / W + VW a) (sin x) /60 = (sin 120o)/300 x = 9. in minutes. Resource Academia. E-mail: ahmad.97o ≈ 10o Hence θ = 080o b) (sin 50o) / VA = (sin 120o)/300 VA = 265. Find: a) the value of θ. 0333-4404244.754 hr ≈ 45 min Nadeem Ahmad. b) the time taken.10 QUESTION NO.
[email protected] b) (sin 50. SOLUTION: N N 170o 170o 10o VA/W VA = 350 km/h θ VW = 70 km/h VA/W VA = 350 x Θ VW = 70 In the diagram.7o + 10o) = 50. 8 a) An aircraft is flying due south at 350 km/h.3o) / VA/W = (sin 119. Ph. vectors are added by the equation VA = VA / W + VW a) (sin 10o) /70 = (sin x)/350 x = 60. The wind is blowing at 70 km/h from the direction θo. Find i) the value of θ ii) the speed of the aircraft in still air.7o If θ is supposed to be acute. 0333-4404244.3o or 119.11 QUESTION NO. E-mail: ahmad.7o θ = 180o – (119. Given that the pilot is steering the aircraft in the direction 170o. Resource Academia.com .7)/350 VA/W = 310 km/h Nadeem Ahmad. where θo is acute. then x will be obtuse so x = 119. 2 m/sec wishes to cross a river which is flowing between straight parallel banks at 2 m/sec.12 b) A man who swims at 1. Find: i) the speed at which he travels.8 o with the bank downstream.nadeem78@yahoo. Ph. VW = 2m/s 120o VM/W = 1. VM makes with the bank downstream. we will apply sine rule sin θ /1.2m/s VM 60o θ Velocity Diagram VM = VM/W + VW Where VM is the true velocity of the man. ii) the angle which his resultant velocity makes with the bank. VM / W is the velocity of the man in still water and Vw is the velocity of the water. E-mail: ahmad. 0333-4404244.8 θ = 21.com .8 m/s To find angle θ.22+22 -2(1.8 m/s by making an angle of 21. He aims downstream in a direction making an angle of 60o with the bank.2 = sin 120o/2. Resource Academia. VM2 = 1. Nadeem Ahmad.8o Hence Velocity (true) of the boat is 2.2)(2) cos120o VM = 2. 1o Velocity Diagram (Return Journey) VB = VB/W + VW Where VB is the true velocity of the boat. Resource Academia.1o VB α θ Velocity Diagram (Outward Journey) Θ= 53. 9 The diagram shows a river.6 m/sec Vw = 2. Find the resultant speed of the boat on this return journey.6 m/sec x y 33.1o Vw = 2.6 m/sec.1o B Θ= 53.com . between straight parallel banks. The speed of the boat in still water is 5.nadeem78@yahoo. C 120 m B C 120 m Θ= 53. 0333-4404244.4 m/sec. flowing at a speed of 2. the boat is steered in a direction making an angle of α to the bank as shown. 120 m downstream. VB/W is the velocity of the boat in still water and Vw is the velocity of the water. Find a) the value of α b) the resultant speed of the boat c) the time taken for the crossing The boat then makes the return journey from B to A.13 QUESTION NO.1o VB/w = 5. A boat crosses the river from a point A on one bank to a point B on the opposite bank.4 126.9o θ = 53. Nadeem Ahmad. Ph.4 m/sec 106.9o VB θ 160 m 160 m VB/w = 5. 160 m wide. E-mail: ahmad. In order to travel directly from A to B. 1o (sin x) /2.9o) /VB = (sin 53. Resource Academia.com .6 VB = 3.7 =
[email protected] m/sec Nadeem Ahmad.1o)/5.6 (θ = 53. Ph.6 y = 20o (sin 33.4 = (sin 126.1o) /VB = (sin 126.9o)/5.1o + 20o = 73.14 a) AC = 160 m and BC = 120 m tan θ = 160/120 θ = 53.1o)/5.1o) VB = 6.4 = (sin 53.1o b) (sin 106.6 x = 20o α = 53.9o)/5. 0333-4404244.8 m/sec RETURN JOURNEY (sin y) /2.7 m/sec c) AB = √1602 + 1202 = 200 m Time taken for crossing = 200/6. E-mail: ahmad. 1 Two particles P and Q.3o Hence velocity of Q relative to P is 6. Nadeem Ahmad.24 m/sec in the direction 166.3 (PAGE 549) QUESTION NO. VQ/P is the velocity of the particle Q relative to the particle P and VP is the velocity of the particle P.24 m/s (sin θ) /7 = (sin 60o)/6. i) VQ/P2 = 52+72 -2(5)(7) cos60o VQ/P2 = 25 + 49 – 70(0.5) VQ/P2 = 74 – 35 = 39 VQ/P = 6.24 θ = 76. Particle Q is moving at 5 m/sec in the direction 090o and Q is moving at 7 m/sec in a direction 030o. Ph. E-mail: ahmad. Find a) the magnitude and direction of the velocity of Q relative to P. Resource Academia. are 30 m apart with Q due north of P.15 New Additional Mathematics by Pan Pacific Publishing EXERCISE 24. to the nearest
[email protected]. 0333-4404244.com . 60o VQ = 5 m/s θ 60o VP = 7 m/s VQ/P VP = 7 m/s 30o 30o P ii) Velocity Diagram i) Initial Diagram VQ = VQ/P + VP Where VP is the velocity of the particle P. Q VQ = 5 m/s b) the time taken for Q to be due east of P. 7o QQ´ = 30.nadeem78@yahoo. 0333-4404244.7o 13.7o = 30/QQ´ QQ´ = 30 / cos 13. Resource Academia.24 ≈ 5 sec P Nadeem Ahmad.16 Q ii) the time taken for Q to be due east of P cos 13. Ph.com . Q´ E-mail: ahmad.88 m The time taken for Q to be due east of P = 30.88/6. 0o Hence distance apart.1 km/h (sin θ) /10 = (sin 30o)/17. ii) To find the distance P and Q when P is due north of Q.0o with the initial line PQ. Ph.17 QUESTION NO. when P is due north of Q.1 θ = 17o so x = 30o – 17. the angle between them will be 30o.nadeem78@yahoo. if the vectors representing VP and VQ are produced. VP = VP/Q + VQ Where VP is the velocity of the ship P.2(25)(10) cos30o VP/Q2 = 625 + 100 – 500(0. in meters.154 km = 1154 m 13. Find: a) the speed and direction of P relative to Q.com P . we will stop the ship Q at its initial position and move the ship P with the relative speed VP/Q tan 13. Q Nadeem Ahmad. 5km E-mail: ahmad.0o = 1. VP/Q is the velocity of the ship P relative to the ship Q and VQ is the velocity of the ship Q. 0333-4404244.0o = 13. when P is due north of Q = 1154 m. Resource Academia.866) VP/Q2 = 292 VP/Q = 17. 30o VP = 25 km/h VQ = 10 km/h VP/Q θ x VP/Q2 = 252+102 .0o =d/5 d d = 5tan 13. SOLUTION: In the velocity diagram. 2 VP = 25 km/h VQ = 10 km/h 120o 30o Q 5 km P At a particular instant.0o Hence velocity of P relative to Q is 17.1 km/h with an angle of 13. two ships P and Q are 5 km apart and move with constant speeds and directions as shown. b) the distance apart. we will stop boat P at its initial position and move boat Q with the relative speed VQ/P. 0333-4404244. VQ = VQ/P + VP Where VP is the velocity of the boat P.0o so direction of VQ/P = 60o – 47. 3 At a particular instant.866) VQ/P = 2. when Q is due east of P = 462 m 13o Q Nadeem
[email protected] QUESTION NO.462 km = 462 m 2 km Hence the distance apart. VQ/P2 = 32+42 .com . Resource Academia. Ph. when Q is due east of P.05 m/s in the direction 013. the angle between them will be 30o. Find a) the speed and direction of Q relative to P. SOLUTION: P N VP = 3m/s VP = 3m/s 30o VQ/P θ 2 km VQ = 4m/s VQ = 4m/s 60o Velocity diagram Q In the velocity diagram.0o. two boats P and Q are 2 km apart and P is due north of Q.0o = 013.0o Hence velocity of Q relative to P is 2. E-mail: ahmad.05 m/s (sin θ) /3 = (sin 30o)/2. P Q´ tan 13o = PQ´/2 PQ´ = 2 tan 13o = 0. in meters.05 θ = 47. in meters.2(3)(4) cos30o VQ/P2 = 25 – 24(0. The boats move with constant speeds and directions as shown. if the vectors representing VP and VQ are produced. ii) To find the distance P and Q when Q is due east of P. VQ/P is the velocity of the boat Q relative to the boat P and VQ is the velocity of the boat Q. b) the distance apart. Nadeem Ahmad.0o Hence velocity of A relative to B is 9. Ship A is sailing due south at 5 km/h and ship B is sailing due west at 8 km/h.0 with vertical so it makes an angle of 32.72 km Hence the distance between the two ships when A is on the bearing of 225o from B = 2. Ph.0o B 45o 225o x X = 180o – (32o + 45o) = 103o (sin 103o) /5 = (sin 32o)/BA´ N A´ BA´ = 2.43 km/h tan θ = 8/5 θ =
[email protected] QUESTION NO. ii) To find the distance the two ships when A is on the bearing of 225o from B. two ships A and B are 5 km apart with A due west of B.0 with the horizontal which is shown in the diagram. As As VA/B is making an angle of 58.com . Find a) the velocity of A relative to B. we will stop ship B at its initial position and move ship A with the relative speed VA/B. 0333-4404244.0o.43 km/h in the direction 122. b) the distance between the two ships when A is on the bearing of 225o from B. VA/B is the velocity of the ship A relative to the ship B and VB is the velocity of the ship B.72 km. SOLUTION: A 5 km B θ VB = 8 km/h VA = 5 km/h VA = 5 km/h VA = VA/ B + VB VA/B VB = 8 km/h Where VA is the velocity of the ship A. VA/ B = √52 + 82 VA/ B = 9. 4 At a particular moment. Resource Academia. E-mail: ahmad. 5 km A 32. 9o BB´ = 51.5 = 0. Aircraft A is flying on the bearing 120o at 300 km/h.9o i.5 km/h in the direction 281. SOLUTION: A 50 km 30o B VB = 450 km/h VB/A VA = 300 km/h VA = 300 km/h 150o 150o θ VB = 450 km/h VB = VB/A + VA Where VA is the velocity of the aircraft A. Resource Academia.20 QUESTION NO.9o) b) B´ d θ=11.098 Time taken for B to reach north of A = 51.9 VB/A = 725. (270o+11. 5 Two aircraft A and B fly at the same height with constant velocities. Ph. a) VB/A 2 = 3002+4502 -2(300)(450) cos150o VB/A 2 = 90000 + 202500 + 233826.5 km/h (sin θ) /300 = (sin 150o)/725. 0333-4404244. E-mail: ahmad.5 θ = 11. At noon.nadeem78@yahoo. Find a) the velocity of B relative to A b) the time when B is due north of A. VB/A is the velocity of the aircraft B relative to the aircraft A and VB is the velocity of the aircraft B.9o Hence the velocity of B relative to A is 725.e. aircraft B is 50 km due east of aircraft A and is flying due west at 450 km/h.098/725.com .9o = 50/BB´ BB´= 50 /cos 11.0704 hrs = 4 min 14 sec Time when B is due north of A = 12 00 00 + 00 04 14 = 12 04 14 Nadeem Ahmad.9o A B 50 km cos 11. 4o (sin
[email protected] QUESTION NO. Resource Academia. two particles P and Q. Particles P and Q are moving in the direction as shown. 0333-4404244. leave simultaneously when they are 50 m apart with P due west of Q.1 = 3106 sec Nadeem Ahmad.com . 6 VP= 8m/s P VQ = 10m/s θ 30o Q 50 km In the diagram. Given that P and Q are on the path of collision.6o b) α = 180o – (30o +23. moving with speeds 8 m/sec and 10 m/sec respectively. α VP= 8m/s VQ = 10m/s θ 30o VQ = VQ/P + VP VQ/P Where VP is the velocity of the particle P.1 Time that elapses before the collision = 50. the direction of VP/Q or VQ/P will always be considered along the initial line PQ. SOLUTION: Note: When P and Q are on the path of collision.000/16. to the nearest second.4o) / VQ/P = (sin 30o)/10 VQ/P = 16. Find a) the value of θ b) the time that elapses before the collision. a) (sin θ) /8 = (sin 30o)/10 θ = 23. E-mail: ahmad. Ph.6o) = 126. VQ/P is the velocity of the particle Q relative to the particle P and VQ is the velocity of the particle Q. 62) θ VA/H = 10. A helicopter. VA/H is the velocity of the airship relative to the helicopter and VH is the velocity of the helicopter.4 m/s VA/H Time that elapses before the interception = 500/10. flies at a speed of 12 m/sec and steers on a bearing θ in order to intercept the airship. SOLUTION: VA = 6m/s VH = 12m/s θ A H 500 m In case of interception.com .22 QUESTION NO. Resource Academia.nadeem78@yahoo. Find a) the value of θ b) the time that elapses before the interception.4 ≈ 48 sec Nadeem Ahmad. E-mail: ahmad. Ph. VA/H will be directed along the initial line AH. VA = VA/H + VH Where VA is the velocity of the airship. 500 m due east of the airship. an airship is moving due north with a speed of 6 m/sec. 7 At a given instant. 0333-4404244. a) sin θ = 6/12 θ = 30o VA = 6m/s b) VH = 12m/s VA/H = √(122 . a) Given that V = 20.23 QUESTION NO. VB/A is the velocity of the B relative to the particle A and VA is the velocity of the particle A.com . Resource Academia. E-mail: ahmad. Ph. 8 Two particles.nadeem78@yahoo. N SOLUTION (a) When V = 20 m/sec VA = 10 m/s 60o 75o N 75o VA 60o 105o VB/W VB = 20 m/sec N 50 m 15o VB = V =20 m/sec 15o SOLUTION: VB = VB/A + VA Where VB is the velocity of the particle B. Nadeem Ahmad. b) Given that B collides with A. particles A is travelling at 10 m/sec in a direction 075o and B is travelling at V m/sec in a direction 015o. Find i) The magnitude and direction of the velocity of B relative to A. are 50 m apart with A due north of B. Find i) the value of V ii) the time taken for B to collide with A. ii) The time taken for B to be due west of A. 0333-4404244. A and B. then the line segment for VB should be approximately double of VA. Resource Academia.com . the tails of VB and VB/A will meet and VB/A and VA will be added by head to tail rule. Nadeem Ahmad. CALCULATIONS: Applying cosine Rule in the velocity diagram triangle: VB/A2 = 202+102 -2(20)(10) cos60o VB/A = 17. In this case it is 180o. ii) THE TIME TAKEN FOR “B” TO BE DUE WEST OF “A”. To find the distance between B and A when B is due west of A. the vector are added in such a way that the head of VB and VA will coincide. it must be taken into account that the lengths of line segments representing different vectors should be proportionate to the original length. sin θ/10 = sin 60o/17.nadeem78@yahoo. If VB = 20 and VA = 10.(105o + 15o) = 60o. E-mail: ahmad.3 θ = 30o So the direction of VB/A will be 360o – 15o = 345o. This is real explanation of VB/A.3 m/sec in the direction 345o.3m/s The angle between VB/A and VB will be taken as θ. Ph. Hence the velocity of B relative to A is 17. While working on the velocity diagram. The angle between VB and VA will be calculated by producing the initial line segments (vectors) in the initial diagram.24 DESCRIPTION: (Not to be written in the paper) In the question like this. 0333-4404244. A part of θ is 15o which is the original direction of VB. we will consider VB/A. which explains that A is kept at its initial position and its effect of speed and direction is transferred to B. For example. E-mail: ahmad.3 15o = 2.76/VB/A = 51. the direction of the VB/A will be towards original direction BA.3 m/sec Nadeem Ahmad.76/17.com .nadeem78@yahoo. N VA = 10 m/s N 105o VA = 10 m/s VB/A 105o 50 m VB = V 15o 15o Original diagram (i) VB = V Velocity Diagram VB = VB/A + VA sin 105o/V = sin 15o/10 V = 37. Ph. 0333-4404244.76 m 50 m Hence time taken for B to be due west of A = 51.99 sec b) B IN CASE OF COLLISION In case of collision. Resource Academia.25 cos 15o = 50/BB´ B´ A BB´ = 50 / cos15o BB´ = 51. 26 ii) sin 105o/37.3 = sin 60o/VB/A VB/A = 33.4 = 1. Resource
[email protected] . Ph.5 sec Nadeem Ahmad. E-mail: ahmad. 0333-4404244.4 m/sec time taken for B to collide with A = 50/33. 10 y • P (0.nadeem78@yahoo. whether or not P and Q will meet. 50) • Q (80. 0333-4404244. as shown in the diagram. Ship P is travelling with velocity (20 i +10 j) km/ h and ship Q is travelling with velocity (–10 i+30 j) km/ h ( i ) Find an expression for the position vector of P and of Q at time t hours after 1200 hours. with full working. Resource Academia. Ph.com . ship P is at the point with position vector 50j km and ship Q is at the point with position vector (80i+20j) km. E-mail: ahmad.27 RELATIVE VELOCITY IN VECTOR FORMAT OCTOBER NOVEMBER 2002 PAPER 1 Q. (iii) Determine. 20) x O At 1200 hours. SOLUTION: i) VP = 20 i +10 j = 20 10 VQ = –10 i+30 j = 10 30 20 20 Distance travelled by P in t hours = v×t = t = 10 10 Distance travelled by Q in t hours = v×t = t 10 10 = 30 30 20 0 20 Position vectors of P at time t hours after 1200 hours= OP = = 50 10 50 10 80 10 80 10 Position vectors of Q at time t hours after 1200 hours= OQ = = 20 30 20 30 Nadeem Ahmad. (ii) Use your answers to part ( i ) to determine the distance apart of P and Q at 1400 hours. com . 0333-4404244. Resource Academia. Nadeem Ahmad. Ph.67 Similarly 50 + 10t = 20 + 30t 20t = 30 t = 1. E-mail: ahmad. at t = 2 20 2 40 = 50 10 2 70 Position vectors of P = OP = 80 10 2 60 Position vectors of Q = OQ = = 20 30 2 80 60 40 20 - = 80 70 10 PQ = OQ – OP = Distance between P and Q at 1400 hrs = |PQ| = √202 + 102 = √500 =
[email protected] If P and Q meet then the value of t by comparing x and y component should be same.28 ii) at 1400 hrs i. So in this case P and Q will not meet.e.4 m iii) If P and Q meet then for some value of t OP = OQ 20 80 10 = 50 10 20 30 20 t = 80 – 10t 30t = 80 t = 8/3 = 2. 4. The lifeboat sets off immediately and travels in a straight line at constant speed.4 /1.29 MAY JUNE 2003 PAPER 1 QUESTION NO. which is 90 km from a lifeboat and on a bearing of 315° from the lifeboat.5 = 42. sends a message for assistance. An ocean liner is travelling at 36 km/h on a bearing of 090°. intercepting the liner at 0730 hours.com . SOLUTION: N VL = 36 km/h 45o 36×1. Find the speed at which the lifeboat travels.4 Speed of the lifeboat =
[email protected] km/h Nadeem Ahmad. Resource Academia.5 hrs (from 0600 to 0730) = 36 × 1. E-mail: ahmad. d2 = 542+902 -2(54) (90) cos45o d2 = 2916 + 8100 – 6873 d2 = 4143 d = 64. Ph. At 0600 hours the liner. 0333-4404244.5 = 54 km Distance covered by the life boat will be calculated by using cosine rule.5 = 54 km 90 km d 45o 315o Speed of liner = 36 km/hr Distance covered by the liner in 1. Resource Academia. The velocity. VP/W is the velocity of the plane relative to the wind and VW is the velocity of the wind. Find (i) the bearing of Q from P.110j 40 110 70 VP = The true velocity of the plane i. E-mail: ahmad. 0333-4404244.40j) km/h and there is a constant wind blowing with velocity (50i -70j) km/h.40j = 50 70 VW = 50i . i is a unit vector due east and j is a unit vector due north.9 = 0. of the plane is (280i .9 km/h Time of flight = 273/347. VP is directed from P to Q N 330 P x VP = 330i . In this question.70j = VP = VP/W + VW Where VP is the velocity of the plane.com . 6.110j tan x = 110/330 -110 so x = tan-1(110/330) = 18. given that the distance PQ is 273 km. in still air. to the nearest minute. Ph. A plane flies from P to Q.nadeem78@yahoo. (ii) the time of flight. 280 330 50 + = = 330i .4o Q Bearing of Q from P = 90o + 18. [2] SOLUTION: 280 40 VP/W = 280i .785 hrs ≈ 47 min Nadeem Ahmad.e.30 RELATIVE VELOCITY IN VECTOR FORMAT OCTOBER NOVEMBER 2003 PAPER 1 QUESTION NO.4o = 108.4o Speed of the plane = √3302 +1102= √121000 = 347. To a cyclist travelling due south on a straight horizontal road at 7 m/s. find the direction from which the wind is blowing.nadeem78@yahoo. VW/C is the velocity of the wind relative to the cyclist and VW is the velocity of the wind.6o Hence the wind is blowing from the direction 020. N N x 45o Vc = 7m/s VW/C 45o E E 135o Vc = 7m/s sin x/7 = sin 135o/12 VW/C VW = 12m/s θ Velocity Diagram sin x = 0. Resource Academia.31 MAY JUNE 2004 PAPER 1 QUESTION NO. Nadeem Ahmad.4o θ = 180o – (135o + 24.412 x = sin-1(0. Given that the wind has a constant speed of 12 m/s. 4. Ph.412) x = 24. E-mail: ahmad.com . the wind appears to be blowing from the north-east.4o) = 20.6o. 0333-4404244. SOLUTION: VW = VW/C + VC Where VC is the velocity of the cyclist. Resource Academia. Ph.1o sin x/3 = sin 53.com .30 ≈ 34 sec Nadeem Ahmad.1o) = 103. 8. B 150m θ VW = 3m/s α 200 m VB/W = 6m/s θ VB x A tan θ = 200/150 so θ = tan-1(200/150) θ = 53.6o + 53. 0333-4404244. SOLUTION VB = VB/W + VW Where VB is the true velocity of the boat.30 m/s Distance to be covered = √2002 + 1502 = 250 m time taken to travel from A to B = 250/7. 150 m downstream of A. VB/W is the velocity of the boat in still water and Vw is the velocity of the water.1o/6 x = 23. The motor boat.6o α = 180o . find. Assuming that the motor boat is travelling at top
[email protected] OCTOBER NOVEMBER 2004 PAPER 2 QUESTION NO.1o/6 VB = 7.3o sin 103. travels directly from a point A on one bank to a point B. on the opposite bank. E-mail: ahmad. A motor boat travels in a straight line across a river which flows at 3 m/s between straight parallel banks 200 m apart. to the nearest second. which has a top speed of 6 m/s in still water.3o/VB = sin 53.(23. the time it takes to travel from A to B. 3o) = 39. SOLUTION: VP = VP/W + VW Where VP is the true velocity of the plane.3 = 3.3o x = 180o – (120o + 20. Find the time taken for the flight. Given that Y is 720 km from X on a bearing of 150o and there is a constant wind of 120 km/h blowing towards the west. 0333-4404244. 9. A plane. Resource Academia.3464 θ = 20.7o/VP = sin 120o/300 VP = 221.com .nadeem78@yahoo. E-mail: ahmad. whose speed in still air is 300km/h.33 MAY JUNE 2005 PAPER 2 QUESTION NO.3 km Time taken for the flight = 720/221.25 hrs Nadeem Ahmad.7o sin 39. Ph. VP/W is the velocity of the plane in still air and Vw is the velocity of the wind. flies directly from X to Y. N X 150o VW = 120 km/h X 150o θ VP/W = 300 km/h VP VP 30o x VW = 120 km/h Y Y sin 120o/300 = sin θo/120 sin θ = 0. OF´ = OS´ 1 3 85 5 = 12 2 5 1 + 3t = 85 – 5t => 8t = 84 12 + 2(10. which is not drawn according to scale. One corner of the surface is taken as the origin O and i and j are unit vectors along the edges of the surface. S. At the instant that the fly starts crawling. Given that the spider catches the fly. The diagram.5) = 5 + k (10. a spider. shows a horizontal rectangular surface. 5.com .5k = 28 => Nadeem Ahmad. calculate the value of
[email protected] 10.5) => => t = 10. 0333-4404244. F. at the point with position vector (85i + 5j) cm. where k is a constant. k= 2. starts at the point with position vector (i+12j) cm and crawls across the surface with a velocity of (3i +2j) cm/s.34 OCTOBER NOVEMBER 2005 PAPER 1 QUESTION NO. 3 3 Distance travelled by F in t sec = v×t = t = 2 2 5 5 = Distance travelled by S in t sec = v×t = t 1 3 3 1 Position vectors of F after t sec = OF´ = = 12 2 2 12 85 5 85 5 = 5 5 Position vectors of S after t sec = OS´ = If the spider catches the fly then for some value of t. sets off across the surface with a velocity of (-5i + kj) cm/s. Ph. Resource Academia. A fly.67 E-mail: ahmad. SOLUTION: 1 12 Initial Position vector of F = OF = i+12j = 85 5 Initial Position vector of S = OS = 85i + 5j = 3 VF = 3i +2j = 2 and VS = – 5i + kj = 5 Let the spider catches the fly after “t” seconds. (150000) cos45o VP/W = 408 km/h ii) sin θ/150 = sin 45o/408 θ = 15. Ph. a distance of 1000 km. in a time of 2 hours. flies due north from A to B. i) VP = 1000/2 = 500 km/h 45o VW = 150 km/h VP = 500 km/h VP = 500 km/h VW =150km/h VP/W θ VP/W2 = 5002+1502 -2(500) (150) cos45o VP/W2 = 250000+22500 . Nadeem Ahmad. SOLUTION: VP = VP/W + VW Where VP is the true velocity of the plane. Resource Academia. E-mail: ahmad. 3. 0333-4404244. ii) the direction in which the plane must be steered.1o Hence the plane must be steered in the direction 015. During this time a steady wind. with a speed of 150 km/h.35 MAY JUNE 2006 PAPER 1 QUESTION NO. VP/W is the velocity of the plane in still air and Vw is the velocity of the wind. is blowing from the south east. Find: i) the speed of the plane in still air. A
[email protected] .1o. 5 m/s Now tan x = 1. A ferry travels in a straight line from a point A to a point B directly opposite A.5 m/s VF = VF/W + VW Where VF is the true velocity of the ferry.com . 0333-4404244.9o with the bank upstream.9o Hence the ferry must be steered by making an angle of 36.5/2 => x = 36. Given that the ferry takes exactly one minute to cross the river. E-mail: ahmad. 4.nadeem78@yahoo. Resource Academia. VW = 2 m/s 90 m x VF/W VF = 1. VF/W is the velocity of the ferry in still water and Vw is the velocity of the water. The diagram shows a river 90 m wide.5 m/s x By using Pythagoras theorem (VF/W) 2 = (VW)2 + (VF)2 = 4 + 2. ii) the angle to the bank at which the ferry must be steered.25 VF/W = 2. Ph. find i) the speed of the ferry in still water. Nadeem Ahmad. SOLUTION: VF = 90/60 = 1.36 OCTOBER NOVEMBER 2006 PAPER 2 QUESTION NO. flowing at 2 m/sec between parallel banks. 5 = 30 30 15 Distance travelled by missile in (t – 0.5 = 0 30 15 30 15 Nadeem Ahmad. (i) show that the spacecraft moved across the screen for 1. SOLUTION: Initial Position vector of spacecraft = OA = 12j = 0 12 46 Initial Position vector of missile = OB = 46i = 0 VS = 40i +15j = 40 15 and 30 VM = ki + 30j = Let the spacecraft is hit by the missile t sec after its appearance.com .8 seconds before impact. Ph. A player fires a missile from a point B. 6. E-mail: ahmad.5 sec after the appearance of the spacecraft so the time taken by the missile to hit the spacecraft is (t – 0. Resource Academia. In a game. The diagram shows a large rectangular television screen in which one corner is taken as the origin O and i and j are unit vectors along two of the edges. 0333-4404244. The point B has position vector 46i cm and the velocity of the missile is (ki +30j) cm per second. the missile is fired 0. (ii) find the value of k.5) sec.5) sec = v×t = (t . an alien spacecraft appears at the point A with position vector 12j cm and moves across the screen with velocity (40i +15j) cm per second. As the missile is fired 0.37 MAY JUNE 2007 PAPER 1 QUESTION NO.5 40 40 0 = 12 15 15 12 Position vectors of spacecraft after t sec = OS = Position vectors of missile after t sec = OM = 46 0.nadeem78@yahoo. where k is a constant.0.5 seconds after the spacecraft appears on the screen. Given that the missile hits the spacecraft.5 46 0. 40 40 = 15 15 Distance travelled by spacecraft in t sec = v×t = t 0. 5 40(1. E-mail: ahmad.com .nadeem78@yahoo. Resource Academia. 0333-4404244. Ph. ii) 40 = 46 0.5k 72 – 46 = 1.3k 26 = 1.8 sec Hence the spacecraft has moved across the screen for 1.3k k = 20 Nadeem Ahmad.8 seconds before impact.8) – 0.38 i) As the missile hits the spacecraft so for some value of t.8) = 46 + k(1. OS= OM 40 46 0.5 = 12 15 30 15 12 + 15t = 30t – 15 15t = 27 t = 1. Ph. i is a unit vector due east. A constant wind is blowing with velocity (–60i +60j) km/h.60o The bearing. SOLUTION: 960 400 PQ = 960i +400 j = 60 60 VW = –60i +60j = As the plane takes 4 hours to travel from P to Q so 960 240 = 400 100 VP = ¼ PQ = ¼ i) VP = VP/W + VW Where VP is the true velocity of the plane. to the nearest degree. to the nearest degree.VW 240 60 300 - = = 300i + 40j 100 60 40 VP/W = ii) N VP/W = 300i + 40j θ 40 x 300 tan x = 40/300 x = tan-1(40/300) x = 7. on which the plane must be directed = θ = 082o Nadeem Ahmad. Given that the plane takes 4 hours to travel from P to Q. 5. giving your answer in the form (ai + bj) km/ h.39 OCTOBER NOVEMBER 2007 PAPER 1 QUESTION NO. on which the plane must be directed. Resource Academia. VP/W = VP . find: (i) the velocity.nadeem78@yahoo. A plane flies from P to Q where PQ = (960i +400 j) km. and j is a unit vector due north. in still air. 0333-4404244. In this question. of the plane. VP/W is the velocity of the plane in still air and Vw is velocity of the wind.com . E-mail: ahmad. (ii) the bearing. 5 hrs (from 0900 to 1030) = OS = 1. E-mail: ahmad. In this question i is a unit vector due east and j is a unit vector due north. 0333-4404244.40 MAY JUNE 2008 PAPER 1 QUESTION NO. the velocity of the ship. in terms of i and j. cos θ = Vx /V |V|= 15√ 2 km/ h Vx = V cos θ Vy Vx = 15√ 2 cos 45o = 15 Θ = 45o Vx Similarly Vy = V sin θ = 15√ 2 sin 45o = 15 Hence V = 15i + 15j ii) 2 Initial Position vector of the ship = OP = 2i + 3j = 3 15 Velocity of the ship = VS = 15i + 15j = 15 2 15 24. At 0900 hours a ship sails from the point P with position vector (2i + 3j) km relative to an origin O.5 = 3 15 25. North.5 i + 25.5 j) km at 1030 hours. (ii) Show that the ship will be at the point with position vector (24.5 = 24. SOLUTION: i) The component of velocity along the direction due east is Vx and towards north. At the same time as the ship leaves P. the velocity of the ship relative to the submarine. Ph.5 j Nadeem Ahmad. (v) Find the position vector of the point where the submarine meets the
[email protected] Position vectors of ship after 1. the position of the ship t hours after leaving P.com . Vy. The submarine proceeds with a speed of 25 km /h due north to meet the ship.5i + 25. (iii) Find.east direction means. (iv) Find. in terms of i. an angle of 45o between North and East. The ship sails north-east with a speed of 15√ 2 km/ h. in terms of i and j. 10. Resource Academia. (i) Find. j and t. a submarine leaves the point Q with position vector (47i – 27 j) km. nadeem78@yahoo. E-mail: ahmad. Ph.41 iii) 2 2 15 15 Position vectors of ship t hrs after leaving P = = 3 3 15 15 = (2 + 15t)i +(3+15t)j iv) 47 27 Initial Position vector of the submarine = OQ = 47i – 27j = 0 25 Velocity of the submarine = VSUB = 0i + 25j = 0 15 15 - = = 15i – 10j 25 15 10 VS/SUB = VS . 0333-4404244.com .VSUB = v) 2 2 15 15 Position vectors of ship t hrs after leaving P = = 3 3 15 15 47 47 0 = 27 25 27 25 Position vectors of submarine t hrs after leaving Q = If the submarine meets the ship then for some value of t 47 2 15 = 27 25 3 15 2 + 15t = 47 t=3 47 Hence position vector of the point where submarine meets the ship = 27 25 3 47 = = 47i + 48j 48 Nadeem Ahmad. Resource Academia. Resource Academia.7o with the bank upstream. The river is 48 m wide and is flowing with a speed of 1.4 m/s. Ph.4 => x = 73.8 m/s x By using Pythagoras theorem (VB/W) 2 = (VW)2 + (VB)2 = 4.82 + 1. 7. VB/W is the velocity of the boat in still water and Vw is the velocity of the water.7o Hence the ferry must be steered by making an angle of 73.com . Given that the boat takes 10 seconds to cross the river. (ii) the angle to the bank at which the boat should be steered. 0333-4404244.42 OCTOBER NOVEMBER 2008 PAPER 2 QUESTION NO. The diagram shows a river with parallel banks. VW = 1. SOLUTION: VB = 48/10 = 4.nadeem78@yahoo. A boat travels in a straight line from a point P on one bank to a point Q which is on the other bank directly opposite P. E-mail: ahmad.42 VB/W = 5 m/s Now tan x = 4.8 m/s VB = VB/W + VW Where VB is the true velocity of the boat.8/1.4 m/s 48 m x VB/W VB = 4. find: (i) the speed of the boat in still water. Nadeem Ahmad. E-mail: ahmad. find the time at which P and Q meet and the position vector of the point where this happens.east direction means. or otherwise. an angle of 45o between North and East. 9.nadeem78@yahoo. The ship sails north-east with a speed of 10√2 km/h .com . Vy. Ph. Find (i) the velocity vector of P (ii) the position vector of P at 1200 hours. (iv) Hence. (iii) Find the velocity of P relative to Q.4i + 8j = 10 10 Velocity of the ship P = VP = 10i + 10j = 4 10 16 Position vectors of ship P after 2 hrs (from 1000 to 1200) = OP´ = 2 = 8 10 28 = 16 i + 28 j Nadeem Ahmad. a ship P leaves a point A with position vector (– 4i + 8j) km relative to an origin O. Resource Academia. cos θ = Vx /V |V|= 10√ 2 km/ h Vx = V cos θ Vy Vx = 10√ 2 cos 45o = 10 Θ = 45o Vx Similarly Vy = V sin θ = 10√ 2 sin 45o = 10 Hence V = 10i + 10j ii) 4 8 Initial Position vector of the ship P = OP = . where i is a unit vector due East and j is a unit vector due North. a second ship Q leaves a point B with position vector (19i +34j) km travelling with velocity vector (8i + 6j ) km/h. At 1000 hours.43 MAY JUNE 2009 PAPER 1 QUESTION NO. North. 0333-4404244. At 1200 hours. SOLUTION: i) The component of velocity along the direction due east is Vx and towards north. 0333-4404244.44 iii) Initial Position vector of the ship Q = OQ = 19i + 34j = 19 34 8 Velocity of the ship Q = VQ = 8i + 6j = 6 10 8 2 VP/Q = VP .5 31 = = 31i + 43j 43 Nadeem Ahmad. Resource Academia.com . t hrs after leaving A = Position vectors of ship Q. 4 10
[email protected] Position vector of the point where the ship P meets the ship Q = 8 10 3. Ph. = = 2i + 4j 10 6 4 iv) Let the ship P and Q meet each other t hours after 1000 4 10 4 10 = 8 10 8 10 Position vectors of ship P. t hrs after leaving B = 19 8 16 19 8 2 = 34 6 12 34 6 If the ship P meets the ship Q then for some value of t 4 10 19 8 16 = 8 10 34 6 12 -4 + 10t = 19 + 8t -16 2t = 7 => t = 3.VQ = . E-mail: ahmad.5 hrs Hence the ship P will meet ship Q at 1330. Ph. 1 0 In this question. in hours. after 1200. 12 6t i) Show that at 1400 the boat is 25 km from the lighthouse.com . E-mail: ahmad. 0333-4404244. Resource Academia. ii) 27 48 Position vector of the lighthouse =OL = 4 8t Position vector of the boat = OB = 12 6t 4 8t 27 23 8t - = 12 6t 48 36 6t BL = OL – OB = If the boat is less than 25 km from the lighthouse then |BL|< 25 √{(23-8t)2 + (36-6t)2} < 25 {(23-8t)2 + (36-6t)2} < 625 529 – 368t + 64t2+ 1296 – 432t + 36t2 < 625 100t2 – 800t +1200 < 0 Nadeem Ahmad. where t is the time. ii) Find the length of time for which the boat is less than 25 km from the lighthouse. is a unit vector due east and is a unit vector due north 0 1 27 A lighthouse has position vector km relative to an origin O. 10. = 48 24 24 BL = OL – OB = |BL|= √72 + 242 = 25 Hence at 1400. the boat is 25 km from the lighthouse.45 MAY JUNE 2010 PAPER 1 (4037/21/M/J/10) QUESTION NO. SOLUTION: i) 4 8 2 20 Position vector of the boat at 1400 hrs =OB = = 12 6 2 24 27 48 Position vector of the lighthouse =OL = 27 7 20 .nadeem78@yahoo. A boat moves in such a way 48 4 8t that its position vector is given by km. 6 (t – 2) < 0 (t – 2) (t – 6) < 0 2<t<6 2 4 6 t Hence length of time for which the boat is less than 25 km from the lighthouse = 6 -2 = 4 hrs. Resource Academia. E-mail: ahmad. Ph. 0333-4404244. Nadeem Ahmad.46 t2 – 8t +12 < 0 t2 – 2t – 6t +12 < 0 t(t – 2) .com .nadeem78@yahoo. 1o θ = 180o – (60o + 16.1o/80 VP = 280.com . Ph. to the nearest minute.0 km/h Time of flight = 500/280. 0333-4404244.79 hrs = 107 min Nadeem Ahmad.277 x = 16. flies directly from A to B. the time taken for the flight.9o sin 103.0 = 1. E-mail: ahmad. whose speed in still air is 250 km/
[email protected]/VP = sin 16. Find.1o) = 103. There is a constant wind of 80 km/h blowing from the south. SOLUTION: B 60o VW = 80 N θ N VP 60o VP VW = 80 km/h 60o VP/W = 250 x A Velocity Diagram sin 60o/250 = sin xo/80 sin x = 0. Resource Academia.47 OCTOBER NOVEMBER 2010 PAPER 2 (4037/22/M/J/10) QUESTION 9: A plane. where B is 500 km from A on a bearing of 060o.