A ship moves so that its position vector,in metres,relative2a lighthouse at time,t is r=(80-0.4t)i +(2t-80)j?

Question

where i and j are unit vectors directed east and north respectively.

a) Find the distance of the ship fromthe lighthouse when t = 60?

b) Find the times when the ship is
i) Due north of the lighthouse?
ii) North east of the lighthouse?

Answer 1

I've seen your other questions. You need to learn about vectors. If a vector is pointing north or south (Such as, the ship is travelling north - velocity vector, the ship is north of the lighthouse - position vector etc.) Then the east west component is 0. So everything beside "i" will = 0.
80-.4(t)=0
North of lighthouse at t=20
Likewise, if something is due east or heading due east etc, the j component will be zero.
2t-80=0
East of lighthouse at t=40

As for a), its easy to see that r=(80-0.4(60))i + (2(60)-80)j
r=56i+40j so the ship is 56 east of the lighthouse and 40 north of the lighthouse at t=60.

If you want this as a straight distance, as with all vectors (again, learn it!) Get the square root of the i component squared plus the j component squared. This is just the pythagoras theorem. So r=68.8

Now go draw diagrams of these questions and figure them out or you'll never pass an exam on the subject.

Answer 2

Man, you're in serious trouble with three days left to answer and you about to run aground on those rocks beneath the lighthouse. Good luck!

Answer 3

You sad soul.
This is not a quiz.I am happy to suggest sensible answers,
But - If you are lost at sea, call the coastgard.
Twit

Answer 4

I answered your other two questions, now you should be able to answer this one on your own.