have no idea how solve this?

• May 15th 2010, 09:38 PM
MickG86
have no idea how to solve this?
Hi,
I'm tying to solve the following question since for a hour and still have no clue how to....(Headbang)

A ship is heading out to sea at a bearing of 028° at a constant speed. At 10:04 hours it is 5.2 km due west of Port Linseed and at 10:40 hours it is due north of Port Linseed. Radio Contact with Port Linseed can be maintaned up to distances of 32 km.

Assuming that the ship continues its course unchanged, determine its bearing from Port Linseed at 11:10 hours and the time at which radio contact with Port Linseed will be lost.

Thanks for the help (Bow)
• May 16th 2010, 08:33 AM
Ashley911
hope this helps
Don't know if this helps but a few ideas might help you

Since due west of port linseed and then due north port linseed we have a right angle triangle with two sides the same length 5.2 and therfore the other two angles will be both 45 degrees as 90+45+45 = 180 and 180-45 will be the other angle, we know the times 10:04 and 10:40 and distance between these two points is x squared = 5.2 squared + 5.2 squared so can get the distance and work out the speed from that, then we know the time from 11:10 to 10:40 and use this same speed to get the length of the side of new triangle.

maybe this will spark you with some fresh ideas how to solve this
• May 16th 2010, 09:48 AM
obesechicken13
I don't think that my explanation will be very useful without my drawing a picture. Sry too lazy . Did you draw one? If you did, it should be simple. However for the first part of your question. the answer should be 6.67km, 49.4 degrees north of east at 11:10.

Then finding out when the distance is equal to PL is 32 km, should be 187 minutes after 10:40.
• May 18th 2010, 07:14 PM
MickG86
Thank you both of you,

anyway in the end i figured it out myself and when i did a drawing it all made sense (Clapping).

Just had to use Trigonemtry and simple Math and I got the same answer as u did obesechicken13 (Cool)

Thank You again