Solving A-B-C Ship Journey with Motorboat

[jiboom]

a ship A is traveling due east at a speed of 8m/s and a ship B is traveling due south at 10m/s. At an instant when A is 3km from B in a direction 060 a motor boat leaves A and travels in a straight line to B with speed 14m/s

1) show it reacges B in 500 seconds
2)if upon reaching B the motor boat immediately returns back to A in a straight line,again with speed 14,find the time taken for the return journey.

i need a diagram here to get started. book I am working through is no help :(

im thinking:

distance triangle: if C is point where motorboat meets B then have

AB=3
angle ABC=X say


velocity triangle:

Vb=velocity of B
Vm=velocity of motorboat

then
i have Vb pointing down, -vm at an angle up from bottom of vb and vb-vs makes other side


then angle between vb and vb-vs=180-x


but now not sure how to proceed so thinking my pictures are wrong or I am totally missing something. i seem to need another angle or length?

 

[jiboom]

any takers?

im sitting here with loads of trianlgles on various pieces of paper! I am guessing i get another angle or length from the set up from the question but I am not getting it!

 

[jiboom]

a ship A is traveling due east at a speed of 8m/s and a ship B is traveling due south at 10m/s. At an instant when A is 3km from B in a direction 060 a motor boat leaves A and travels in a straight line to B with speed 14m/s

1) show it reacges B in 500 seconds
2)if upon reaching B the motor boat immediately returns back to A in a straight line,again with speed 14,find the time taken for the return journey.

i need a diagram here to get started. book I am working through is no help :(

im thinking:

distance triangle: if C is point where motorboat meets B then have

AB=3
angle ABC=X say


velocity triangle:

Vb=velocity of B
Vm=velocity of motorboat

then
i have Vb pointing down, -vm at an angle up from bottom of vb and vb-vs makes other side


then angle between vb and vb-vs=180-x


but now not sure how to proceed so thinking my pictures are wrong or I am totally missing something. i seem to need another angle or length?

ok,have made some progress.

it appears what i was attempting was wrong.

i drew a distance diagram showing the point B and M(otorboat) met and then said relative velocity needed to be parallel to the line.

it turns out the relative velocity needs to be parallel to the initial situation, ie at 60 to north, which gives me the extra info i needed to do part (a)

can anyone run this by me though? why is this the way to go


for the second part:


i have tried to find the bearing of A from B and the distance by:

i know boats been moving for 500 seconds so A has moved 4000m right,B 5000m down.

i know when they were 3000 m apart the angle was 60 so i can use this to fill in the top left corner of the right angled triangle.

from this triangle i can find the distance AB and the bearing (roughly 9500,45)


now following the first part, i say relative velocity vector goes at 45 to north, andf in my triangle i have

-va along the base, vm from left hand side of -va and (vm-va) from rhs of -va

and angle between va and (vm-va) =90+45=135

but this does not get me the book answer. I get about 1250 to their 1790.

am i going about this the correct way?/

 

[jiboom]

still no joy matching book answer.

im sure the distance between the 2 is 9500ish number so taking book answer the relative velocity must be 5.2

does this make sense? the relative velocity for the motorboat and B was 6,but B was moving at 10 m.s so should the relative velocity for motorboat and A be greater as A is only moving at 8?

im thinking along lines of:
3 cars on motorway:
car A going 30
car b going 50
car c going 60

relative to A car C is going 30
relative to b car c is going 10
so although B is going quicker its relative velocity is slower

 

[jiboom]

anyone with any thoughts? i have whole exercise of these so can't make any progress til i get this,the first question,sorted. thanks