RELATIVE VELOCITY - WELCOME IGCSE · PDF fileHence 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, we will

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Nadeem Ahmad, Resource Academia, Ph. 0333-4404244. E-mail: [email protected]

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 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

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.

Velocity Diagram

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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:

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.

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

VW = 3m/s

VB/W = 5m/s VB

60O

60O x

Velocity Diagram

3


VW = 2.5m/s

VB/W = 6m/s VB

Velocity Diagram

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

VB = VB/W + VW θ

Where

VB is the true velocity of the boat,


CALCULATIONS:

(VB)2 = (VB/W) 2+ (VW)2

(VB)2 = 62 + 2.52

VB = 6.5m/s

4


VW = 1.8m/s

20 m

VB/W = 1.2m/s VB

θ

θ

Now

tan θ = 2.5/6

θ = tan-1(2.5/6)

θ = 22.6o


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:

VB = VB/W + VW

Where

VB is the true velocity of the boat


a) Calculations for true velocity of the boat:


VB2 = 1.22+1.82 -2(1.2)(1.8) cos70o

VB = 1.79m/s

θ

Velocity Diagram

70O

70O

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Θ =39.0o

To find angle θ, VB makes with the bank downstream, we will apply sine rule

sin θ /1.2 = sin 70o/1.79

θ = 39.0o


b) In this distance diagram,

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

Velocity Diagram. (Shown as a dotted line there)

Now

sin 39o = 20/d

d = 31.78m

so time taken for crossing = 31.78/1.79 ≈ 18 sec.

Θ =39.0o

Distance Diagram

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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 cosθ

VB sin θ = 1.79 sin 39.0 = 1.126 m/sec

so time taken for crossing = 20/1.126 ≈ 18 sec. 20 VB sinθ VB= 1.79

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.

Θ =39.0o

Velocity component diagram

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Θ

80.5o

y

A

SOLUTION:

Vw = 3.5m/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.

To find x

tan x = 30/5

x = tan-1(6) = 80.5o

so the angle between VW and VB is 99.5o.

a) (sin θ) /3.5 = (sin 99.5o)/6

θ = 35.1o

α = 180o – (99.5o + 35.1o) = 45.4o

α

Velocity Diagram (Outward Journey)

5 m

x =80.5o course 80.5o

80.5o

α 99.5o

VB 30 m VB/w= 6m/sec

VB/w = 6m/sec

Vw = 3.5 m/sec

Velocity Diagram (Return Journey)

A

B B

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b) (sin 45.4o) /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 = √302 + 52 = 30.4 m

Time taken = 30.4/4.33 ≈ 7 sec

RETURN JOURNEY

(sin 80.5o) /6 = (sin y)/3.5

y = 35.1o

Hence course taken by the boat in the return journey is 180o – (80.5o + 35.1o) = 64.4o with the bank

upstream.

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120o

θ

30o

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 060o. Find the true velocity of the aircraft

SOLUTION

N Vw = 60km/h

VA/w = 300km/h

60o Vw = 60km/h

N VA VA/w = 300km/h

60o

In the diagram, 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 = 6.9o

Hence True velocity of the aircraft is 250 km/hr in the direction 053.1o.

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A

VA/w = 300km/h

VA

Vw = 60km/h

Vw = 60km/h

x

θ

50o

90o

QUESTION NO. 7

An aircraft flies due east from A to B where AB = 200 km. 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. Find:

a) the value of θ,

b) the time taken, in minutes, for the journey from A to B.

VA

300

30o


a) (sin x) /60 = (sin 120o)/300

x = 9.97o ≈ 10o

Hence θ = 080o

b) (sin 50o) / VA = (sin 120o)/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

N B

B

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N N

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

θo, where θo is acute. Given that the pilot is steering the aircraft in the direction 170o. Find

i) the value of θ

ii) the speed of the aircraft in still air.

SOLUTION: 170o 170o

10o

VA/W VA/W

VA = 350 km/h θ VW = 70 km/h VA = 350

x

Θ VW = 70


a) (sin 10o) /70 = (sin x)/350

x = 60.3o or 119.7o

If θ is supposed to be acute, then x will be obtuse so x = 119.7o

θ = 180o – (119.7o + 10o) = 50.3o

b) (sin 50.3o) / VA/W = (sin 119.7)/350

VA/W = 310 km/h

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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.

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.

VM2 = 1.22+22 -2(1.2)(2) cos120o

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 120o/2.8

θ = 21.8o

Hence Velocity (true) of the boat is 2.8 m/s by making an angle of 21.8 o with the bank downstream.

Velocity Diagram

VW = 2m/s

VM/W = 1.2m/s VM

θ 60o

120o

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C 120 m B C 120 m B

Θ= 53.1o Θ= 53.1o

VB/w = 5.6 m/sec y

Vw = 2.4 m/sec VB

160 m 160 m Vw = 2.4 θ = 53.1o

VB/w = 5.6 m/sec VB

α

θ

θ

x

Θ= 53.1o

126.9o 106.9o 33.1o

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.

VB = VB/W + VW



Velocity Diagram (Outward Journey) Velocity Diagram (Return Journey)

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a) AC = 160 m and BC = 120 m

tan θ = 160/120

θ = 53.1o

(sin x) /2.4 = (sin 53.1o)/5.6

x = 20o

α = 53.1o + 20o = 73.1o

b) (sin 106.9o) /VB = (sin 53.1o)/5.6 (θ = 53.1o)

VB = 6.7 m/sec

c) AB = √1602 + 1202 = 200 m

Time taken for crossing = 200/6.7 = 29.8 m/sec

RETURN JOURNEY

(sin y) /2.4 = (sin 126.9o)/5.6

y = 20o

(sin 33.1o) /VB = (sin 126.9o)/5.6

VB = 3.82 m/sec

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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 090o and Q is moving at 7 m/sec in a direction 030o. 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. 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, 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/P2 = 52+72 -2(5)(7) cos60o

VQ/P2 = 25 + 49 – 70(0.5)

VQ/P2 = 74 – 35 = 39

VQ/P = 6.24 m/s

(sin θ) /7 = (sin 60o)/6.24

θ = 76.3o

Hence velocity of Q relative to P is 6.24 m/sec in the direction 166.3o.

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ii) Q

the time taken for Q to be due east of P

cos 13.7o = 30/QQ´

QQ´ = 30 / cos 13.7o 13.7o

QQ´ = 30.88 m

The time taken for Q to be due east of P = 30.88/6.24

≈ 5 sec

P Q´

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30o VP = 25 km/h

VQ = 10 km/h

VP/Q θ

x

QUESTION NO. 2

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.

SOLUTION:

In the velocity diagram, if the vectors representing VP and VQ are produced, the angle between them will be 30o.

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.

VP/Q2 = 252+102 - 2(25)(10) cos30o

VP/Q2 = 625 + 100 – 500(0.866)

VP/Q2 = 292

VP/Q = 17.1 km/h

(sin θ) /10 = (sin 30o)/17.1

θ = 17o so x = 30o – 17.0o = 13.0o

Hence velocity of P relative to Q is 17.1 km/h with an angle of 13.0o 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 13.0o =d/5 d

d = 5tan 13.0o = 1.154 km = 1154 m 13.0o

Hence distance apart, when P is due north of Q = 1154 m. Q 5km P

VQ = 10 km/h VP = 25 km/h

120o 30o

Q 5 km P

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Q

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.

SOLUTION: N

P VP = 3m/s VP = 3m/s

30o

VQ/P

2 km θ VQ = 4m/s

VQ = 4m/s

60o Velocity diagram

Q

In the velocity diagram, if the vectors representing VP and VQ are produced, the angle between them will be 30o.

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/P2 = 32+42 - 2(3)(4) cos30o

VQ/P2 = 25 – 24(0.866)

VQ/P = 2.05 m/s

(sin θ) /3 = (sin 30o)/2.05

θ = 47.0o so direction of VQ/P = 60o – 47.0o = 013.0o

Hence velocity of Q relative to P is 2.05 m/s in the direction 013.0o.

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 13o = PQ´/2

PQ´ = 2 tan 13o = 0.462 km = 462 m 2 km

Hence the distance apart, in meters, when Q is due east of P = 462 m 13o

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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 225o from B.

SOLUTION:

A 5 km B

VB = 8 km/h θ

VA = 5 km/h VA = 5 km/h VA/B

VA = VA/ B + VB VB = 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 = √52 + 82

VA/ B = 9.43 km/h

tan θ = 8/5

θ = 58.0o

Hence velocity of A relative to B is 9.43 km/h in the direction 122.0o.

ii) To find the distance the two ships when A is on the bearing of 225o 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

32.0o 45o 225o

x

A´

Hence the distance between the two ships when A is on the bearing of 225o from B = 2.72 km.

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 = 180o – (32o + 45o) = 103o

(sin 103o) /5 = (sin 32o)/BA´

BA´ = 2.72 km

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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 120o

at 300 km/h. Find

a) the velocity of B relative to A

b) the time when B is due north of A.

SOLUTION:

A 50 km B

30o 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, 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 = 3002+4502 -2(300)(450) cos150o

VB/A 2 = 90000 + 202500 + 233826.9

VB/A = 725.5 km/h

(sin θ) /300 = (sin 150o)/725.5

θ = 11.9o

Hence the velocity of B relative to A is 725.5 km/h in the direction 281.9o i.e. (270o+11.9o)

b) B´

d

A θ=11.9o B

50 km

cos 11.9o = 50/BB´

BB´= 50 /cos 11.9o

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

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QUESTION NO. 6

VP= 8m/s VQ = 10m/s

P 30o 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.

SOLUTION:

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.

α VP= 8m/s VQ = 10m/s

30o 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 30o)/10

θ = 23.6o

b) α = 180o – (30o +23.6o) = 126.4o

(sin 126.4o) / VQ/P = (sin 30o)/10

VQ/P = 16.1

Time that elapses before the collision = 50,000/16.1 = 3106 sec

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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.

SOLUTION:

VA = 6m/s VH = 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 = 30o

VA = 6m/s VH = 12m/s

b) VA/H = √(122 - 62)

VA/H = 10.4 m/s θ

VA/H

Time that elapses before the interception = 500/10.4 ≈ 48 sec

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N

75o N

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 075o and B is travelling at V m/sec in a direction 015o.

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.

SOLUTION (a) When V = 20 m/sec

VA = 10 m/s 60o

75o VA 60o

105o VB/W VB = 20 m/sec

50 m

15o VB = V =20 m/sec 15o

SOLUTION:

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.

N

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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 180o- (105o + 15o) = 60o. 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:


VB/A2 = 202+102 -2(20)(10) cos60o

VB/A = 17.3m/s

The angle between VB/A and VB will be taken as θ. A part of θ is 15o which is the original

direction of VB.

sin θ/10 = sin 60o/17.3

θ = 30o

So the direction of VB/A will be 360o – 15o = 345o.

Hence the velocity of B relative to A is 17.3 m/sec in the direction 345o.

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.

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cos 15o = 50/BB´ B´ A

BB´ = 50 / cos15o

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 15o

= 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.

VA = 10 m/s

105o

VA = 10 m/s

105o VB/A VB = V

50 m VB = V

15o 15o

Original diagram Velocity Diagram

(i) VB = VB/A + VA

sin 105o/V = sin 15o/10

V = 37.3 m/sec

N

N

26


ii) sin 105o/37.3 = sin 60o/VB/A

VB/A = 33.4 m/sec

time taken for B to collide with A = 50/33.4 = 1.5 sec

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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.

SOLUTION:

i) VP = 20 i +10 j = �2010�

VQ = –10 i+30 j = ��10 30�

Distance travelled by P in t hours = v×t = t �2010� = �20

10 �

Distance travelled by Q in t hours = v×t = t ��10 30� = ��10

30 �

Position vectors of P at time t hours after 1200 hours= OP = � 050� � �20

10 � = � 20 50 � 10 �

Position vectors of Q at time t hours after 1200 hours= OQ = �8020� � ��10

30 � = �80 � 10 20 � 30 �

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ii) at 1400 hrs i.e. at t = 2

Position vectors of P = OP = � 20 250 � 10 2� = �40

70�

Position vectors of Q = OQ = �80 � 10 220 � 30 2� = �60

80�

PQ = OQ – OP = �6080� - �40

70� = �2010�

Distance between P and Q at 1400 hrs = |PQ| = √202 + 102 = √500 = 22.4 m

iii) If P and Q meet then for some value of t OP = OQ

� 20 50 � 10 � = �80 � 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


45o

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.

SOLUTION: N

VL = 36 km/h

36×1.5 = 54 km

90 km d

45o

315o

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.

d2 = 542+902 -2(54) (90) cos45o

d2 = 2916 + 8100 – 6873

d2 = 4143

d = 64.4

Speed of the lifeboat = 64.4 /1.5 = 42.9 km/h

30


RELATIVE VELOCITY IN VECTOR FORMAT


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]

SOLUTION:

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�40� + � 50

�70� = � 330�110� = 330i - 110j

The true velocity of the plane i.e. VP is directed from P to Q

N

330

P

VP = 330i - 110j -110

tan x = 110/330 so x = tan-1(110/330) = 18.4o Q

Bearing of Q from P = 90o + 18.4o = 108.4o

Speed of the plane = √3302 +1102= √121000 = 347.9 km/h

Time of flight = 273/347.9 = 0.785 hrs ≈ 47 min

x

31



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.

SOLUTION:

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

45o VW/C 45o VW/C

Vc = 7m/s E 135o E

Vc = 7m/s VW = 12m/s

sin x/7 = sin 135o/12 θ Velocity Diagram

sin x = 0.412

x = sin-1(0.412)

x = 24.4o

θ = 180o – (135o + 24.4o) = 20.6o

Hence the wind is blowing from the direction 020.6o.

32


α

A


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


the velocity of the water. 150m B

θ

VW = 3m/s

200 m

VB/W = 6m/s VB

x

tan θ = 200/150 so θ = tan-1(200/150)

θ = 53.1o

sin x/3 = sin 53.1o/6

x = 23.6o

α = 180o - (23.6o + 53.1o) = 103.3o

sin 103.3o/VB = sin 53.1o/6

VB = 7.30 m/s

Distance to be covered = √2002 + 1502 = 250 m

time taken to travel from A to B = 250/7.30 ≈ 34 sec

θ

33



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 150o and there is a constant wind of 120 km/h blowing towards the west.

Find the time taken for the flight.

SOLUTION:

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 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 = 0.3464

θ = 20.3o

x = 180o – (120o + 20.3o) = 39.7o

sin 39.7o/VP = sin 120o/300

VP = 221.3 km

Time taken for the flight = 720/221.3 = 3.25 hrs

34



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.

SOLUTION:

Initial Position vector of F = OF = i+12j = � 112�

Initial Position vector of S = OS = 85i + 5j = �855 �

VF = 3i +2j = �32� and VS = – 5i + kj = ��5

��

Let the spider catches the fly after “t” seconds.

Distance travelled by F in t sec = v×t = t �32� = �3

2�

Distance travelled by S in t sec = v×t = t ��5 � � = ��5

� �

Position vectors of F after t sec = OF´ = � 112� � �3

2 � = � 1 � 312 � 2 �

Position vectors of S after t sec = OS´ = �855 � � ��5

� � = �85 � 5 5 � � �

If the spider catches the fly then for some value of t, OF´ = OS´

� 1 � 312 � 2 � = �85 � 5

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


VW =150km/h


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.

SOLUTION:

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

45o VW = 150 km/h

VP = 500 km/h VP = 500 km/h

VP/W

θ

VP/W2 = 5002+1502 -2(500) (150) cos45o

VP/W2 = 250000+22500 - (150000) cos45o

VP/W = 408 km/h

ii) sin θ/150 = sin 45o/408

θ = 15.1o

Hence the plane must be steered in the direction 015.1o.

36


x


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.

SOLUTION:

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


VW = 2 m/s

90 m

VF/W VF = 1.5 m/s

x


(VF/W) 2 = (VW)2 + (VF)2 = 4 + 2.25

VF/W = 2.5 m/s

Now tan x = 1.5/2 => x = 36.9o

Hence the ferry must be steered by making an angle of 36.9o with the bank upstream.

37



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.

SOLUTION:

Initial Position vector of spacecraft = OA = 12j = � 012�

Initial Position vector of missile = OB = 46i = �460 �

VS = 40i +15j = �4015� 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 �4015� = �40

15�

Distance travelled by missile in (t – 0.5) sec = v×t = (t - 0.5 � �30� = �� 0.5�

30 � 15 �

Position vectors of spacecraft after t sec = OS = � 012� � �40

15� = � 4012 � 15�

Position vectors of missile after t sec = OM = �460 � � �� 0.5�

30 � 15 � = �46 � � � 0.5�30 � 15 �

38


i) As the missile hits the spacecraft so for some value of t, OS= OM

� 4012 � 15� = �46 � � � 0.5�

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



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.

SOLUTION:

PQ = 960i +400 j = �960400�

VW = –60i +60j = ��60 60�

As the plane takes 4 hours to travel from P to Q so

VP = ¼ PQ = ¼ �960400� = �240

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 = �240100� - ��60

60� = �300 40� = 300i + 40j

ii) N

VP/W = 300i + 40j

θ 40

300

tan x = 40/300

x = tan-1(40/300)

x = 7.60o

The bearing, to the nearest degree, on which the plane must be directed = θ = 082o

x

40



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.

SOLUTION:

i) The component of velocity along the direction due east is Vx and towards north, Vy.

North- east direction means, an angle of 45o between

North and East.

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) Initial Position vector of the ship = OP = 2i + 3j = �23�

Velocity of the ship = VS = 15i + 15j = �1515�

Position vectors of ship after 1.5 hrs (from 0900 to 1030) = OS = �23� � 1.5 �15

15� = �24.525.5�

= 24.5 i + 25.5 j

41


iii) Position vectors of ship t hrs after leaving P = �23� � �15

15� =�2 � 153 � 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 = � 025�

VS/SUB = VS - VSUB = �1515� - � 0

25� = � 15�10� = 15i – 10j

v) Position vectors of ship t hrs after leaving P = �23� � �15

15� =�2 � 153 � 15 �

Position vectors of submarine t hrs after leaving Q = � 47�27� � � 0

25� = � 47�27 � 25�

If the submarine meets the ship then for some value of t

�2 � 153 � 15 � = � 47

�27 � 25�

2 + 15t = 47

t = 3

Hence position vector of the point where submarine meets the ship = � 47�27 � 25 3�

= �4748� = 47i + 48j

42


x


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.

SOLUTION:

VB = 48/10 = 4.8 m/s

VB = VB/W + VW



VW = 1.4 m/s

48 m

VB/W VB = 4.8 m/s

x


(VB/W) 2 = (VW)2 + (VB)2 = 4.82 + 1.42

VB/W = 5 m/s

Now tan x = 4.8/1.4 => x = 73.7o

Hence the ferry must be steered by making an angle of 73.7o with the bank upstream.

43



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.

SOLUTION:

i) The component of velocity along the direction due east is Vx and towards north, Vy.

North- east direction means, an angle of 45o between

North and East.

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) Initial Position vector of the ship P = OP = - 4i + 8j = ��4 8�

Velocity of the ship P = VP = 10i + 10j = �1010�

Position vectors of ship P after 2 hrs (from 1000 to 1200) = OP´ = ��4 8� � 2 �10

10� = �1628�

= 16 i + 28 j

44


iii) Initial Position vector of the ship Q = OQ = 19i + 34j = �19 34�

Velocity of the ship Q = VQ = 8i + 6j = �86�

VP/Q = VP - VQ = �1010� - �8

6� = �24� = 2i + 4j

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 � 108 � 10 �

Position vectors of ship Q, t hrs after leaving B = �19 34� � � � 2� �8

6� = �19 � 8 � 1634 � 6 � 12�

If the ship P meets the ship Q then for some value of t

��4 � 108 � 10 � = �19 � 8 � 16

34 � 6 � 12�

-4 + 10t = 19 + 8t -16

2t = 7 => t = 3.5 hrs

Hence the ship P will meet ship Q at 1330.

Position vector of the point where the ship P meets the ship Q = ��4 � 10 3.5�8 � 10 3.5 �

= �3143� = 31i + 43j

45


MAY JUNE 2010 PAPER 1 (4037/21/M/J/10)

QUESTION NO. 10.

In this question, �10�is a unit vector due east and �0

1�is a unit vector due north

A lighthouse has position vector �2748� km relative to an origin O. A boat moves in such a way

that its position vector is given by � 4 � 8t12 � 6t� km, where t is the time, in hours, after 1200.

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.

SOLUTION:

i) Position vector of the boat at 1400 hrs =OB = � 4 � 8 212 � 6 2� = �20

24�

Position vector of the lighthouse =OL = �2748�

BL = OL – OB = �2748� - �20

24� =� 724�

|BL|= √72 + 242 = 25

Hence at 1400, the boat is 25 km from the lighthouse.

ii) Position vector of the lighthouse =OL = �2748�

Position vector of the boat = OB = � 4 � 8t12 � 6t�

BL = OL – OB = �2748� - � 4 � 8t

12 � 6t� = �23 � 8t36 � 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 + 64t2+ 1296 – 432t + 36t2 < 625

100t2 – 800t +1200 < 0

46


t2 – 8t +12 < 0

t2 – 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


60o

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 060o. There is a constant wind of 80 km/h blowing from the south. Find, to the

nearest minute, the time taken for the flight.

SOLUTION:

VW = 80

N N θ

VP VP

60o VW = 80 km/h 60o VP/W = 250

x

Velocity Diagram

sin 60o/250 = sin xo/80

sin x = 0.277

x = 16.1o

θ = 180o – (60o + 16.1o) = 103.9o

sin 103.9o/VP = sin 16.1o/80

VP = 280.0 km/h

Time of flight = 500/280.0

= 1.79 hrs = 107 min

B

A

Documents

RELATIVE VELOCITY - WELCOME IGCSE · PDF fileHence 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, we will