EXERCISE 24.2 (PAGE 549)

1 New 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 V B/W = 3m/s VB is the true velocity of the boat V W = 4m/s VB/W is the velocity of the boat in still water V B and Vw is the velocity of the water By using Pythagoras theorem (VB) 2 = (VB/W) 2 + (VW) 2 (VB) 2 = 3 2 + 4 2 = 25 VB Now = 5m/s Velocity Diagram tan θ = 4/3 θ = tan -1 (4/3) θ = 53.1 o Hence True velocity of the boat is 5m/s in the direction (90+53.1) 143.1 o OR True velocity of the boat is 5m/s by making an angle of 36.9 o with the bank downstream.

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 330 o. Find the resultant velocity of the speedboat. V W = 3m/s 60 O V B/W = 5m/s V B VB = VB/W + VW Where VB is the true velocity of the boat, 60 O x Velocity Diagram VB/W is the velocity of the boat in still water and Vw is the velocity of the water. EXPLANATION: (i) (ii) The boat which is steered in the direction of 330 o will move by making an angle of 60 o with the bank upstream as the bearing of upstream bank in this case is 270 o and hence 270 o + 60 o = 330 o. The angle between VB/W and VW = 60 o (alternate angles) CALCULATIONS: Applying cosine Rule in the velocity diagram triangle: VB 2 = 5 2 +3 2-2(5)(3) cos60 o VB 2 = 25+9-30 cos60 o VB 2 = 34-15 = 19 VB = 4.36m/s

3 To find angle x, VB makes with the bank downstream, we will apply sine rule sinx /5 = sin60 o /4.36 x = 83.2 o Hence Velocity (true) of the boat is 4.36 m/s by making an angle of 83.2 o with the bank downstream. velocity (true) of the boat is 4.36 m/s in the direction 006.8 o as (90 83.2 = 6.8). OR 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 V W = 2.5m/s V B/W = 6m/s V B VB = VB/W + VW 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. CALCULATIONS: (VB) 2 = (VB/W) 2 + (VW) 2 (VB) 2 = 6 2 + 2.5 2 VB = 6.5m/s

4 Now tan θ = 2.5/6 θ = tan -1 (2.5/6) θ = 22.6 o Hence Velocity (true) of the boat is 6.5 m/s by making an angle of 67.4 o 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 70 o with the bank. Find a) The resultant velocity b) The time taken for the crossing, to the nearest second. 20 m 70 O V W = 1.8m/s θ θ V B/W = 1.2m/s V B VB = VB/W + VW Where VB is the true velocity of the boat 70 O θ Velocity Diagram 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: VB 2 = 1.2 2 +1.8 2-2(1.2)(1.8) cos70 o VB = 1.79m/s

5 To find angle θ, VB makes with the bank downstream, we will apply sine rule sin θ /1.2 = sin 70 o /1.79 θ = 39.0 o Hence Velocity (true) of the boat is 1.79 m/s by making an angle of 39.0 o with the bank downstream. b) In this distance diagram, Θ =39.0 o we will find the component of distance along which VB is calculated. 20 m 20 d This is shown by a fold faced line here. It is the same line which is obtained by producing the line segment representing VB in Θ =39.0 o Distance Diagram Velocity Diagram. (Shown as a dotted line there) Now sin 39 o = 20/d d = 31.78m so time taken for crossing = 31.78/1.79 18 sec.

6 ALTERNATIVE METHOD Time taken for crossing the river can also be calculated if we split VB into its horizontal and vertical components. The component of VB which makes the boat cross the river is VB sinθ. This is shown by a fold faced line in velocity component diagram (need not to be shown in the answer script) VB sin θ = 1.79 sin 39.0 = 1.126 m/sec VB cosθ Θ =39.0 o so time taken for crossing = 20/1.126 18 sec. 20 VB sinθ VB= 1.79 Velocity component diagram QUESTION NO. 5 The diagram shows a river, 30 m wide, flowing at a speed of 3.5 m/sec, between straight parallel banks. A boat, whose speed in still water is 6 m/sec, crosses the river from a point A on one bank to a point B on the opposite bank, 5 m upstream. In order to travel directly from A to B, the boat is steered in a direction making an angle α to the bank as shown. Find a) The value of α, b) The resultant speed of the boat, c) The time taken for the crossing, to the nearest second. In order to make the return journey from B to A, what is the course taken by the boat.

7 B 5 m B V w = 3.5m/s α 80.5 o x =80.5 o course 80.5 o y 99.5 o V B 30 m V B/w= 6m/sec V B/w = 6m/sec α Θ A 80.5o Vw = 3.5 m/sec A Velocity Diagram (Outward Journey) Velocity Diagram (Return Journey) VB = VB/W + VW Where 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. To find x tan x = 30/5 x = tan -1 (6) = 80.5 o so the angle between VW and VB is 99.5 o. a) (sin θ) /3.5 = (sin 99.5 o )/6 θ = 35.1 o α = 180 o (99.5 o + 35.1 o ) = 45.4 o

8 b) (sin 45.4 o ) /VB = (sin 99.5)/6 VB = 4.33m/sec c) To find the time taken, we will find the distance to be covered along VB which is shown as a partially dotted line in the Velocity Diagram (Outward Journey). Distance to be covered = 30 2 + 5 2 = 30.4 m Time taken = 30.4/4.33 7 sec RETURN JOURNEY (sin 80.5 o ) /6 = (sin y)/3.5 y = 35.1 o Hence course taken by the boat in the return journey is 180 o (80.5 o + 35.1 o ) = 64.4 o with the bank upstream.

9 QUESTION NO. 6 The speed of an aircraft in still air is 300 km/h. The wind velocity is 60 km/h from the east. The aircraft is steered on the course in the direction 060 o. Find the true velocity of the aircraft SOLUTION 120 o N Vw = 60km/h VA/w = 300km/h 60 o Vw = 60km/h 30 o N VA VA/w = 300km/h 60 o θ In the diagram, vectors are added by the equation VA = VA / W + VW VA 2 = 300 2 +60 2-2(300)(60) cos30 o VA = 250 km/h To find θ (sin θ) / 60 = (sin 30 o ) /250 Θ = 6.9 o Hence True velocity of the aircraft is 250 km/hr in the direction 053.1 o.

10 QUESTION NO. 7 An aircraft flies due east from A to B where AB = 200 km. The wind is blowing from the direction 030 o 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 θ o. Find: a) the value of θ, b) the time taken, in minutes, for the journey from A to B. N A VA B B 30 0 V w = 60km/h θ x V A/w = 300km/h 90 o 30 o 50 o V w = 60km/h V A In the diagram, vectors are added by the equation VA = VA / W + VW a) (sin x) /60 = (sin 120 o )/300 x = 9.97 o 10 o Hence θ = 080 o b) (sin 50 o ) / VA = (sin 120 o )/300 VA = 265.4 km/h Hence time taken, in minutes, for the journey from A to B = 200/ 265.4 = 0.754 hr 45 min

11 QUESTION NO. 8 a) An aircraft is flying due south at 350 km/h. The wind is blowing at 70 km/h from the direction θ o, where θ o is acute. Given that the pilot is steering the aircraft in the direction 170 o. Find i) the value of θ ii) the speed of the aircraft in still air. N N 170 o 170 o 10 o V A/W V A/W V A = 350 km/h θ V W = 70 km/h V A = 350 x Θ V W = 70 In the diagram, vectors are added by the equation VA = VA / W + VW a) (sin 10 o ) /70 = (sin x)/350 x = 60.3 o or 119.7 o If θ is supposed to be acute, then x will be obtuse so x = 119.7 o θ = 180 o (119.7 o + 10 o ) = 50.3 o b) (sin 50.3 o ) / V A/W = (sin 119.7)/350 VA/W = 310 km/h

12 b) A man who swims at 1.2 m/sec wishes to cross a river which is flowing between straight parallel banks at 2 m/sec. He aims downstream in a direction making an angle of 60o with the bank. Find: i) the speed at which he travels, ii) the angle which his resultant velocity makes with the bank. V W = 2m/s 120 o V M/W = 1.2m/s V M 60 o θ Velocity Diagram VM = VM/W + VW Where VM is the true velocity of the man; VM / W is the velocity of the man in still water and Vw is the velocity of the water. VM 2 = 1.2 2 +2 2-2(1.2)(2) cos120 o VM = 2.8 m/s To find angle θ, VM makes with the bank downstream, we will apply sine rule sin θ /1.2 = sin 120 o /2.8 θ = 21.8 o Hence Velocity (true) of the boat is 2.8 m/s by making an angle of 21.8 o with the bank downstream.

13 QUESTION NO. 9 The diagram shows a river, 160 m wide, flowing at a speed of 2.4 m/sec, between straight parallel banks. A boat crosses the river from a point A on one bank to a point B on the opposite bank, 120 m downstream. The speed of the boat in still water is 5.6 m/sec. In order to travel directly from A to B, the boat is steered in a direction making an angle of α to the bank as shown. 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. Find the resultant speed of the boat on this return journey. C 120 m B C 120 m B Θ= 53.1 o Θ= 53.1 o V B/w = 5.6 m/sec y Vw = 2.4 m/sec V B 106.9 o θ 33.1 o 126.9 o 160 m 160 m Vw = 2.4 θ = 53.1 o V B/w = 5.6 m/sec V B x θ α Θ= 53.1 o Velocity Diagram (Outward Journey) Velocity Diagram (Return Journey) VB = VB/W + VW Where 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.

14 a) AC = 160 m and BC = 120 m tan θ = 160/120 θ = 53.1 o (sin x) /2.4 = (sin 53.1 o )/5.6 x = 20 o α = 53.1 o + 20 o = 73.1 o b) (sin 106.9 o ) /VB = (sin 53.1 o )/5.6 (θ = 53.1 o ) VB = 6.7 m/sec c) AB = 160 2 + 120 2 = 200 m Time taken for crossing = 200/6.7 = 29.8 m/sec RETURN JOURNEY (sin y) /2.4 = (sin 126.9 o )/5.6 y = 20 o (sin 33.1 o ) /VB = (sin 126.9 o )/5.6 VB = 3.82 m/sec

15 New Additional Mathematics by Pan Pacific Publishing EXERCISE 24.3 (PAGE 549) QUESTION NO. 1 Two particles P and Q, are 30 m apart with Q due north of P. Particle Q is moving at 5 m/sec in the direction 090 o and Q is moving at 7 m/sec in a direction 030 o. Find a) the magnitude and direction of the velocity of Q relative to P, Q VQ = 5 m/s b) the time taken for Q to be due east of P, to the nearest second. 60 o VQ = 5 m/s θ 60 o VP = 7 m/s VQ/P VP = 7 m/s 30 o 30 o P ii) Velocity Diagram i) Initial Diagram VQ = VQ/P + VP Where VP is the velocity of the particle P, VQ/P is the velocity of the particle Q relative to the particle P and VP is the velocity of the particle P. i) VQ/P 2 = 5 2 +7 2-2(5)(7) cos60 o VQ/P 2 = 25 + 49 70(0.5) VQ/P 2 = 74 35 = 39 VQ/P = 6.24 m/s (sin θ) /7 = (sin 60 o )/6.24 θ = 76.3 o Hence velocity of Q relative to P is 6.24 m/sec in the direction 166.3 o.

16 ii) Q the time taken for Q to be due east of P cos 13.7 o = 30/QQ QQ = 30 / cos 13.7 o 13.7 o QQ = 30.88 m The time taken for Q to be due east of P = 30.88/6.24 5 sec P Q

17 QUESTION NO. 2 V Q = 10 km/h V P = 25 km/h 120 o 30 o Q 5 km P At a particular instant, two ships P and Q are 5 km apart and move with constant speeds and directions as shown. Find: a) the speed and direction of P relative to Q, b) the distance apart, in meters, when P is due north of Q. In the velocity diagram, if the vectors representing VP and VQ are produced, the angle between them will be 30 o. VP = VP/Q + VQ Where VP is the velocity of the ship P, VP/Q is the velocity of the ship P relative to the ship Q and VQ is the velocity of the ship Q. V Q = 10 km/h 30 o V P = 25 km/h VP/Q 2 = 25 2 +10 2-2(25)(10) cos30 o VP/Q 2 = 625 + 100 500(0.866) VP/Q 2 = 292 VP/Q = 17.1 km/h (sin θ) /10 = (sin 30 o )/17.1 θ = 17 o so x = 30 o 17.0 o = 13.0 o V P/Q x θ Hence velocity of P relative to Q is 17.1 km/h with an angle of 13.0 o with the initial line PQ. ii) To find the distance P and Q when P is due north of Q, we will stop the ship Q at its initial position and move the ship P with the relative speed VP/Q tan 13.0 o =d/5 d d = 5tan 13.0 o = 1.154 km = 1154 m 13.0 o Hence distance apart, when P is due north of Q = 1154 m. Q 5km P

18 QUESTION NO. 3 At a particular instant, two boats P and Q are 2 km apart and P is due north of Q. The boats move with constant speeds and directions as shown. Find a) the speed and direction of Q relative to P, b) the distance apart, in meters, when Q is due east of P. N P V P = 3m/s V P = 3m/s 30 o V Q/P 2 km θ V Q = 4m/s Q V Q = 4m/s 60 o Velocity diagram In the velocity diagram, if the vectors representing VP and VQ are produced, the angle between them will be 30 o. VQ = VQ/P + VP Where VP is the velocity of the boat P, VQ/P is the velocity of the boat Q relative to the boat P and VQ is the velocity of the boat Q. VQ/P 2 = 3 2 +4 2-2(3)(4) cos30 o VQ/P 2 = 25 24(0.866) VQ/P = 2.05 m/s (sin θ) /3 = (sin 30 o )/2.05 θ = 47.0 o so direction of VQ/P = 60 o 47.0 o = 013.0 o Hence velocity of Q relative to P is 2.05 m/s in the direction 013.0 o. ii) To find the distance P and Q when Q is due east of P, we will stop boat P at its initial position and move boat Q with the relative speed VQ/P. P Q tan 13 o = PQ /2 PQ = 2 tan 13 o = 0.462 km = 462 m 2 km Hence the distance apart, in meters, when Q is due east of P = 462 m 13 o Q

19 QUESTION NO. 4 At a particular moment, two ships A and B are 5 km apart with A due west of B. Ship A is sailing due south at 5 km/h and ship B is sailing due west at 8 km/h. Find a) the velocity of A relative to B, b) the distance between the two ships when A is on the bearing of 225 o from B. A 5 km B V B = 8 km/h θ V A = 5 km/h V A = 5 km/h V A/B VA = VA/ B + VB V B = 8 km/h Where VA is the velocity of the ship A, VA/B is the velocity of the ship A relative to the ship B and VB is the velocity of the ship B. VA/ B = 5 2 + 8 2 VA/ B = 9.43 km/h tan θ = 8/5 θ = 58.0 o Hence velocity of A relative to B is 9.43 km/h in the direction 122.0 o. ii) To find the distance the two ships when A is on the bearing of 225 o from B, we will stop ship B at its initial position and move ship A with the relative speed VA/B. N As A 5 km B As VA/B is making an angle of 58.0 with vertical so it makes an angle of 32.0 with the horizontal which is shown in the diagram. X = 180 o (32 o + 45 o ) = 103 o 32.0 o 45 o 225 o x (sin 103 o ) /5 = (sin 32 o )/BA A BA = 2.72 km Hence the distance between the two ships when A is on the bearing of 225 o from B = 2.72 km.

20 QUESTION NO. 5 Two aircraft A and B fly at the same height with constant velocities. At noon, aircraft B is 50 km due east of aircraft A and is flying due west at 450 km/h. Aircraft A is flying on the bearing 120 o at 300 km/h. Find a) the velocity of B relative to A b) the time when B is due north of A. A 50 km B 30 o V B = 450 km/h V B/A V A = 300 km/h V A = 300 km/h 150 o 150 o θ V B = 450 km/h VB = VB/A + VA Where VA is the velocity of the aircraft A, VB/A is the velocity of the aircraft B relative to the aircraft A and VB is the velocity of the aircraft B. a) VB/A 2 = 300 2 +450 2-2(300)(450) cos150 o VB/A 2 = 90000 + 202500 + 233826.9 VB/A = 725.5 km/h (sin θ) /300 = (sin 150 o )/725.5 θ = 11.9 o Hence the velocity of B relative to A is 725.5 km/h in the direction 281.9 o i.e. (270 o +11.9 o ) b) B d A θ=11.9 o B 50 km cos 11.9 o = 50/BB BB = 50 /cos 11.9 o BB = 51.098 Time taken for B to reach north of A = 51.098/725.5 = 0.0704 hrs = 4 min 14 sec Time when B is due north of A = 12 00 00 + 00 04 14 = 12 04 14

21 QUESTION NO. 6 V P= 8m/s V Q = 10m/s P 30 o θ Q 50 km In the diagram, two particles P and Q, moving with speeds 8 m/sec and 10 m/sec respectively, leave simultaneously when they are 50 m apart with P due west of Q. Particles P and Q are moving in the direction as shown. Given that P and Q are on the path of collision. Find a) the value of θ b) the time that elapses before the collision, to the nearest second. Note: When P and Q are on the path of collision, the direction of VP/Q or VQ/P will always be considered along the initial line PQ. V P= 8m/s α V Q = 10m/s 30 o θ VQ = VQ/P + VP VQ/P Where VP is the velocity of the particle P, VQ/P is the velocity of the particle Q relative to the particle P and VQ is the velocity of the particle Q. a) (sin θ) /8 = (sin 30 o )/10 θ = 23.6 o b) α = 180 o (30 o +23.6 o ) = 126.4 o (sin 126.4 o ) / VQ/P = (sin 30 o )/10 VQ/P = 16.1 Time that elapses before the collision = 50,000/16.1 = 3106 sec

22 QUESTION NO. 7 At a given instant, an airship is moving due north with a speed of 6 m/sec. A helicopter, 500 m due east of the airship, flies at a speed of 12 m/sec and steers on a bearing θ in order to intercept the airship. Find a) the value of θ b) the time that elapses before the interception. V A = 6m/s V H = 12m/s A θ H 500 m In case of interception, VA/H will be directed along the initial line AH. VA = VA/H + VH Where VA is the velocity of the airship, VA/H is the velocity of the airship relative to the helicopter and VH is the velocity of the helicopter. a) sin θ = 6/12 θ = 30 o V A = 6m/s V H = 12m/s b) VA/H = (12 2-6 2 ) VA/H = 10.4 m/s θ V A/H Time that elapses before the interception = 500/10.4 48 sec

23 QUESTION NO. 8 Two particles, A and B, are 50 m apart with A due north of B. particles A is travelling at 10 m/sec in a direction 075 o and B is travelling at V m/sec in a direction 015 o. a) Given that V = 20. Find i) The magnitude and direction of the velocity of B relative to A. ii) The time taken for B to be due west of A. b) Given that B collides with A, Find i) the value of V ii) the time taken for B to collide with A. N SOLUTION (a) When V = 20 m/sec N 75 o V A = 10 m/s 60 o 75 o V A 60o 105 o V B/W V B = 20 m/sec 50 m N 15 o V B = V =20 m/sec 15 o VB = VB/A + VA Where VB is the velocity of the particle B, VB/A is the velocity of the B relative to the particle A and VA is the velocity of the particle A.

24 DESCRIPTION: (Not to be written in the paper) In the question like this, the vector are added in such a way that the head of VB and VA will coincide, the tails of VB and VB/A will meet and VB/A and VA will be added by head to tail rule. The angle between VB and VA will be calculated by producing the initial line segments (vectors) in the initial diagram. In this case it is 180 o - (105 o + 15 o ) = 60 o. While working on the velocity diagram, it must be taken into account that the lengths of line segments representing different vectors should be proportionate to the original length. For example; If VB = 20 and VA = 10, then the line segment for VB should be approximately double of VA. CALCULATIONS: Applying cosine Rule in the velocity diagram triangle: VB/A 2 = 20 2 +10 2-2(20)(10) cos60 o VB/A = 17.3m/s The angle between VB/A and VB will be taken as θ. A part of θ is 15 o which is the original direction of VB. sin θ/10 = sin 60 o /17.3 θ = 30 o So the direction of VB/A will be 360 o 15 o = 345 o. Hence the velocity of B relative to A is 17.3 m/sec in the direction 345 o. 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, 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. This is real explanation of VB/A.

25 cos 15 o = 50/BB B A BB = 50 / cos15 o BB = 51.76 m 50 m Hence time taken for B to be due west of A = 51.76/VB/A = 51.76/17.3 15 o = 2.99 sec B b) IN CASE OF COLLISION In case of collision, the direction of the VB/A will be towards original direction BA. N V A = 10 m/s N 105 o V A = 10 m/s 105 o V B/A V B = V 50 m V B = V 15 o 15 o Original diagram Velocity Diagram (i) VB = VB/A + VA sin 105 o /V = sin 15 o /10 V = 37.3 m/sec

26 ii) sin 105 o /37.3 = sin 60 o /VB/A VB/A = 33.4 m/sec time taken for B to collide with A = 50/33.4 = 1.5 sec

27 RELATIVE VELOCITY IN VECTOR FORMAT OCTOBER NOVEMBER 2002 PAPER 1 Q.10 y P (0, 50) Q (80, 20) O x At 1200 hours, ship P is at the point with position vector 50j km and ship Q is at the point with position vector (80i+20j) km, 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. (ii) Use your answers to part ( i ) to determine the distance apart of P and Q at 1400 hours. (iii) Determine, with full working, whether or not P and Q will meet. i) VP = 20 i +10 j = 20 10 VQ = 10 i+30 j = 10 30 Distance travelled by P in t hours = v t = t 20 = 20 10 10 Distance travelled by Q in t hours = v t = t 10 = 10 30 30 Position vectors of P at time t hours after 1200 hours= OP = 0 20 50 10 = 20 50 10 Position vectors of Q at time t hours after 1200 hours= OQ = 80 10 10 = 80 20 30 20 30

28 ii) at 1400 hrs i.e. at t = 2 20 2 Position vectors of P = OP = = 40 50 10 2 70 80 10 2 Position vectors of Q = OQ = = 60 20 30 2 80 PQ = OQ OP = 60-40 = 20 80 70 10 Distance between P and Q at 1400 hrs = PQ = 20 2 + 10 2 = 500 = 22.4 m iii) If P and Q meet then for some value of t OP = OQ 20 10 = 80 50 10 20 30 20 t = 80 10t 30t = 80 t = 8/3 = 2.67 Similarly 50 + 10t = 20 + 30t 20t = 30 t = 1.5 If P and Q meet then the value of t by comparing x and y component should be same. So in this case P and Q will not meet.

29 MAY JUNE 2003 PAPER 1 QUESTION NO.4. An ocean liner is travelling at 36 km/h on a bearing of 090. At 0600 hours the liner, which is 90 km from a lifeboat and on a bearing of 315 from the lifeboat, sends a message for assistance. The lifeboat sets off immediately and travels in a straight line at constant speed, intercepting the liner at 0730 hours. Find the speed at which the lifeboat travels. N V L = 36 km/h 45 o 36 1.5 = 54 km 90 km d 45 o 315 o Speed of liner = 36 km/hr Distance covered by the liner in 1.5 hrs (from 0600 to 0730) = 36 1.5 = 54 km Distance covered by the life boat will be calculated by using cosine rule. d 2 = 54 2 +90 2-2(54) (90) cos45 o d 2 = 2916 + 8100 6873 d 2 = 4143 d = 64.4 Speed of the lifeboat = 64.4 /1.5 = 42.9 km/h

30 RELATIVE VELOCITY IN VECTOR FORMAT OCTOBER NOVEMBER 2003 PAPER 1 QUESTION NO. 6. In this question, i is a unit vector due east and j is a unit vector due north. A plane flies from P to Q. The velocity, in still air, of the plane is (280i - 40j) km/h and there is a constant wind blowing with velocity (50i -70j) km/h. Find (i) the bearing of Q from P, (ii) the time of flight, to the nearest minute, given that the distance PQ is 273 km. [2] VP/W = 280i - 40j = 280 40 VW = 50i - 70j = 50 70 VP = VP/W + VW Where VP is the velocity of the plane, VP/W is the velocity of the plane relative to the wind and VW is the velocity of the wind. VP = 280 50 330 + = = 330i - 110j 40 70 110 The true velocity of the plane i.e. VP is directed from P to Q N 330 P x V P = 330i - 110j -110 tan x = 110/330 so x = tan -1 (110/330) = 18.4 o Q Bearing of Q from P = 90 o + 18.4 o = 108.4 o Speed of the plane = 330 2 +110 2 = 121000 = 347.9 km/h Time of flight = 273/347.9 = 0.785 hrs 47 min

31 MAY JUNE 2004 PAPER 1 QUESTION NO. 4. To a cyclist travelling due south on a straight horizontal road at 7 m/s, the wind appears to be blowing from the north-east. Given that the wind has a constant speed of 12 m/s, find the direction from which the wind is blowing. VW = VW/C + VC Where VC is the velocity of the cyclist, VW/C is the velocity of the wind relative to the cyclist and VW is the velocity of the wind. N N x 45 o V W/C 45 o V W/C Vc = 7m/s E 135 o E Vc = 7m/s V W = 12m/s sin x/7 = sin 135 o /12 θ Velocity Diagram sin x = 0.412 x = sin -1 (0.412) x = 24.4 o θ = 180 o (135 o + 24.4 o ) = 20.6 o Hence the wind is blowing from the direction 020.6 o.

32 OCTOBER NOVEMBER 2004 PAPER 2 QUESTION NO. 8. A motor boat travels in a straight line across a river which flows at 3 m/s between straight parallel banks 200 m apart. The motor boat, which has a top speed of 6 m/s in still water, travels directly from a point A on one bank to a point B, 150 m downstream of A, on the opposite bank. Assuming that the motor boat is travelling at top speed, find, to the nearest second, the time it takes to travel from A to B. SOLUTION VB = VB/W + VW Where 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. 150m B θ V W = 3m/s 200 m α θ V B/W = 6m/s V B x A tan θ = 200/150 so θ = tan -1 (200/150) θ = 53.1 o sin x/3 = sin 53.1 o /6 x = 23.6 o α = 180 o - (23.6 o + 53.1 o ) = 103.3 o sin 103.3 o /VB = sin 53.1 o /6 VB = 7.30 m/s Distance to be covered = 200 2 + 150 2 = 250 m time taken to travel from A to B = 250/7.30 34 sec

33 MAY JUNE 2005 PAPER 2 QUESTION NO. 9. A plane, whose speed in still air is 300km/h, flies directly from X to Y. Given that Y is 720 km from X on a bearing of 150 o and there is a constant wind of 120 km/h blowing towards the west. Find the time taken for the flight. VP = VP/W + VW Where VP is the true velocity of the plane, VP/W is the velocity of the plane in still air and Vw is the velocity of the wind. N X 150 o V W = 120 km/h X 150 o θ V P/W = 300 km/h V P V P 30 o x V W = 120 km/h Y Y sin 120 o /300 = sin θ o /120 sin θ = 0.3464 θ = 20.3 o x = 180 o (120 o + 20.3 o ) = 39.7 o sin 39.7 o /VP = sin 120 o /300 V P = 221.3 km Time taken for the flight = 720/221.3 = 3.25 hrs

34 OCTOBER NOVEMBER 2005 PAPER 1 QUESTION NO. 5. The diagram, which is not drawn according to scale, shows a horizontal rectangular surface. One corner of the surface is taken as the origin O and i and j are unit vectors along the edges of the surface. A fly, F, starts at the point with position vector (i+12j) cm and crawls across the surface with a velocity of (3i +2j) cm/s. At the instant that the fly starts crawling, a spider, S, at the point with position vector (85i + 5j) cm, sets off across the surface with a velocity of (-5i + kj) cm/s, where k is a constant. Given that the spider catches the fly, calculate the value of k. Initial Position vector of F = OF = i+12j = 1 12 Initial Position vector of S = OS = 85i + 5j = 85 5 VF = 3i +2j = 3 2 and VS = 5i + kj = 5 Let the spider catches the fly after t seconds. Distance travelled by F in t sec = v t = t 3 = 3 2 2 Distance travelled by S in t sec = v t = t 5 = 5 Position vectors of F after t sec = OF = 1 1 3 3 = 12 2 12 2 Position vectors of S after t sec = OS = 85 5 5 5 = 85 5 If the spider catches the fly then for some value of t, OF = OS 1 3 5 = 85 12 2 5 1 + 3t = 85 5t => 8t = 84 => t = 10.5 12 + 2(10.5) = 5 + k (10.5) => 10.5k = 28 => k= 2.67

35 MAY JUNE 2006 PAPER 1 QUESTION NO. 3. A plane, flies due north from A to B, a distance of 1000 km, in a time of 2 hours. During this time a steady wind, with a speed of 150 km/h, is blowing from the south east. Find: i) the speed of the plane in still air, ii) the direction in which the plane must be steered. VP = VP/W + VW Where VP is the true velocity of the plane, VP/W is the velocity of the plane in still air and Vw is the velocity of the wind. i) VP = 1000/2 = 500 km/h 45 o V W = 150 km/h V P = 500 km/h V P = 500 km/h V W =150km/h V P/W θ VP/W 2 = 500 2 +150 2-2(500) (150) cos45 o VP/W 2 = 250000+22500 - (150000) cos45 o VP/W = 408 km/h ii) sin θ/150 = sin 45 o /408 θ = 15.1 o Hence the plane must be steered in the direction 015.1 o.

36 OCTOBER NOVEMBER 2006 PAPER 2 QUESTION NO. 4. The diagram shows a river 90 m wide, flowing at 2 m/sec between parallel banks. A ferry travels in a straight line from a point A to a point B directly opposite A. Given that the ferry takes exactly one minute to cross the river, find i) the speed of the ferry in still water, ii) the angle to the bank at which the ferry must be steered. VF = 90/60 = 1.5 m/s VF = VF/W + VW Where VF is the true velocity of the ferry, VF/W is the velocity of the ferry in still water and Vw is the velocity of the water. 90 m V W = 2 m/s x V F/W V F = 1.5 m/s x By using Pythagoras theorem (VF/W) 2 = (VW) 2 + (VF) 2 = 4 + 2.25 VF/W = 2.5 m/s Now tan x = 1.5/2 => x = 36.9 o Hence the ferry must be steered by making an angle of 36.9 o with the bank upstream.

37 MAY JUNE 2007 PAPER 1 QUESTION NO. 6. 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. In a game, 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. A player fires a missile from a point B; the missile is fired 0.5 seconds after the spacecraft appears on the screen. The point B has position vector 46i cm and the velocity of the missile is (ki +30j) cm per second, where k is a constant. Given that the missile hits the spacecraft, (i) show that the spacecraft moved across the screen for 1.8 seconds before impact, (ii) find the value of k. Initial Position vector of spacecraft = OA = 12j = 0 12 Initial Position vector of missile = OB = 46i = 46 0 VS = 40i +15j = 40 15 and VM = ki + 30j = 30 Let the spacecraft is hit by the missile t sec after its appearance. As the missile is fired 0.5 sec after the appearance of the spacecraft so the time taken by the missile to hit the spacecraft is (t 0.5) sec. Distance travelled by spacecraft in t sec = v t = t 40 = 40 15 15 Distance travelled by missile in (t 0.5) sec = v t = (t - 0.5 0.5 = 30 30 15 Position vectors of spacecraft after t sec = OS = 0 40 40 = 12 15 12 15 Position vectors of missile after t sec = OM = 46 0.5 0.5 = 46 0 30 15 30 15

38 i) As the missile hits the spacecraft so for some value of t, OS= OM 40 0.5 = 46 12 15 30 15 12 + 15t = 30t 15 15t = 27 t = 1.8 sec Hence the spacecraft has moved across the screen for 1.8 seconds before impact. ii) 40 = 46 0.5 40(1.8) = 46 + k(1.8) 0.5k 72 46 = 1.3k 26 = 1.3k k = 20

39 OCTOBER NOVEMBER 2007 PAPER 1 QUESTION NO. 5. In this question, i is a unit vector due east, and j is a unit vector due north. A plane flies from P to Q where PQ = (960i +400 j) km. A constant wind is blowing with velocity ( 60i +60j) km/h. Given that the plane takes 4 hours to travel from P to Q, find: (i) the velocity, in still air, of the plane, giving your answer in the form (ai + bj) km/ h, (ii) the bearing, to the nearest degree, on which the plane must be directed. PQ = 960i +400 j = 960 400 VW = 60i +60j = 60 60 As the plane takes 4 hours to travel from P to Q so VP = ¼ PQ = ¼ 960 = 240 400 100 i) VP = VP/W + VW Where VP is the true velocity of the plane, VP/W is the velocity of the plane in still air and Vw is velocity of the wind. VP/W = VP - VW VP/W = 240-60 = 300 = 300i + 40j 100 60 40 ii) N V P/W = 300i + 40j θ 40 x 300 tan x = 40/300 x = tan -1 (40/300) x = 7.60 o The bearing, to the nearest degree, on which the plane must be directed = θ = 082 o

40 MAY JUNE 2008 PAPER 1 QUESTION NO. 10. In this question i is a unit vector due east and j is a unit vector due north. At 0900 hours a ship sails from the point P with position vector (2i + 3j) km relative to an origin O. The ship sails north-east with a speed of 15 2 km/ h. (i) Find, in terms of i and j, the velocity of the ship. (ii) Show that the ship will be at the point with position vector (24.5i + 25.5 j) km at 1030 hours. (iii) Find, in terms of i, j and t, the position of the ship t hours after leaving P. At the same time as the ship leaves P, a submarine leaves the point Q with position vector (47i 27 j) km. The submarine proceeds with a speed of 25 km /h due north to meet the ship. (iv) Find, in terms of i and j, the velocity of the ship relative to the submarine. (v) Find the position vector of the point where the submarine meets the ship. i) The component of velocity along the direction due east is Vx and towards north, Vy. North- east direction means, an angle of 45 o between North and East. cos θ = Vx /V V = 15 2 km/ h Vx = V cos θ Vy Vx = 15 2 cos 45 o = 15 Θ = 45 o Vx Similarly Vy = V sin θ = 15 2 sin 45 o = 15 Hence V = 15i + 15j ii) Initial Position vector of the ship = OP = 2i + 3j = 2 3 Velocity of the ship = VS = 15i + 15j = 15 15 Position vectors of ship after 1.5 hrs (from 0900 to 1030) = OS = 2 1.5 15 = 24.5 3 15 25.5 = 24.5 i + 25.5 j

41 iii) Position vectors of ship t hrs after leaving P = 2 15 15 = 2 3 15 3 15 = (2 + 15t)i +(3+15t)j iv) Initial Position vector of the submarine = OQ = 47i 27j = 47 27 Velocity of the submarine = VSUB = 0i + 25j = 0 25 VS/SUB = VS - VSUB = 15 15-0 15 = = 15i 10j 25 10 v) Position vectors of ship t hrs after leaving P = 2 15 15 = 2 3 15 3 15 Position vectors of submarine t hrs after leaving Q = 47 0 27 25 = 47 27 25 If the submarine meets the ship then for some value of t 2 15 3 15 = 47 27 25 2 + 15t = 47 t = 3 47 Hence position vector of the point where submarine meets the ship = 27 25 3 = 47 = 47i + 48j 48

42 OCTOBER NOVEMBER 2008 PAPER 2 QUESTION NO. 7. The diagram shows a river with parallel banks. The river is 48 m wide and is flowing with a speed of 1.4 m/s. 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. Given that the boat takes 10 seconds to cross the river, find: (i) the speed of the boat in still water, (ii) the angle to the bank at which the boat should be steered. VB = 48/10 = 4.8 m/s VB = VB/W + VW Where 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. 48 m V W = 1.4 m/s x V B/W V B = 4.8 m/s x By using Pythagoras theorem (VB/W) 2 = (VW) 2 + (VB) 2 = 4.8 2 + 1.4 2 VB/W = 5 m/s Now tan x = 4.8/1.4 => x = 73.7 o Hence the ferry must be steered by making an angle of 73.7 o with the bank upstream.

43 MAY JUNE 2009 PAPER 1 QUESTION NO. 9. At 1000 hours, a ship P leaves a point A with position vector ( 4i + 8j) km relative to an origin O, where i is a unit vector due East and j is a unit vector due North. The ship sails north-east with a speed of 10 2 km/h. Find (i) the velocity vector of P (ii) the position vector of P at 1200 hours. At 1200 hours, a second ship Q leaves a point B with position vector (19i +34j) km travelling with velocity vector (8i + 6j ) km/h. (iii) Find the velocity of P relative to Q. (iv) Hence, or otherwise, find the time at which P and Q meet and the position vector of the point where this happens. i) The component of velocity along the direction due east is Vx and towards north, Vy. North- east direction means, an angle of 45 o between North and East. cos θ = Vx /V V = 10 2 km/ h Vx = V cos θ Vy Vx = 10 2 cos 45 o = 10 Θ = 45 o Vx Similarly Vy = V sin θ = 10 2 sin 45 o = 10 Hence V = 10i + 10j ii) Initial Position vector of the ship P = OP = - 4i + 8j = 4 8 Velocity of the ship P = VP = 10i + 10j = 10 10 Position vectors of ship P after 2 hrs (from 1000 to 1200) = OP = 4 8 = 16 i + 28 j 2 10 = 16 10 28

44 iii) Initial Position vector of the ship Q = OQ = 19i + 34j = 19 34 Velocity of the ship Q = VQ = 8i + 6j = 8 6 VP/Q = VP - VQ = 10 10-8 6 = 2 = 2i + 4j 4 iv) Let the ship P and Q meet each other t hours after 1000 Position vectors of ship P, t hrs after leaving A = 4 8 10 10 = 4 10 8 10 Position vectors of ship Q, t hrs after leaving B = 19 34 2 8 8 16 = 19 6 34 6 12 If the ship P meets the ship Q then for some value of t 4 10 8 16 = 19 8 10 34 6 12-4 + 10t = 19 + 8t -16 2t = 7 => t = 3.5 hrs Hence the ship P will meet ship Q at 1330. 4 10 3.5 Position vector of the point where the ship P meets the ship Q = 8 10 3.5 = 31 = 31i + 43j 43

45 MAY JUNE 2010 PAPER 1 (4037/21/M/J/10) QUESTION NO. 10. In this question, 1 0 is a unit vector due east and 0 is a unit vector due north 1 A lighthouse has position vector 27 km relative to an origin O. A boat moves in such a way 48 4 8t that its position vector is given by km, where t is the time, in hours, after 1200. 12 6t i) Show that at 1400 the boat is 25 km from the lighthouse. ii) Find the length of time for which the boat is less than 25 km from the lighthouse. 4 8 2 i) Position vector of the boat at 1400 hrs =OB = = 20 12 6 2 24 Position vector of the lighthouse =OL = 27 48 BL = OL OB = 27-20 48 24 = 7 24 BL = 7 2 + 24 2 = 25 Hence at 1400, the boat is 25 km from the lighthouse. ii) Position vector of the lighthouse =OL = 27 48 4 8t Position vector of the boat = OB = 12 6t BL = OL OB = 27 4 8t 8t - = 23 48 12 6t 36 6t 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 + 64t 2 + 1296 432t + 36t 2 < 625 100t 2 800t +1200 < 0

46 t 2 8t +12 < 0 t 2 2t 6t +12 < 0 t(t 2) - 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.

47 OCTOBER NOVEMBER 2010 PAPER 2 (4037/22/M/J/10) QUESTION 9: A plane, whose speed in still air is 250 km/h, flies directly from A to B, where B is 500 km from A on a bearing of 060 o. There is a constant wind of 80 km/h blowing from the south. Find, to the nearest minute, the time taken for the flight. B 60 o V W = 80 N N θ V P V P 60 o V W = 80 km/h 60 o V P/W = 250 A x Velocity Diagram sin 60 o /250 = sin x o /80 sin x = 0.277 x = 16.1 o θ = 180 o (60 o + 16.1 o ) = 103.9 o sin 103.9 o /VP = sin 16.1 o /80 VP = 280.0 km/h Time of flight = 500/280.0 = 1.79 hrs = 107 min