a patrol boat, travelling at 40km/h on a brearing of 030 degree, sees a suspect craft travelling at 15 km/h due north. It intercepts the craft one hour later. What was the position of the craft relative to the patrol boat when it was first sighted?
I'm really confused about this question, hope anyone here can solve it.
hi
Vp= velocity of patrol boat , Vc=velocity of craft
cVp=velocity of c relative to P
cVp=Vc-Vp
=15 j -20 i -20\sqrt{3} j
=-20 i +(15-20\sqrt{3})j
s=vt , since t=1 , s=v
cRp= displacement of craft relative to patrol boat
= |cVp |
the direction given in the answer is measured from south .
There are a number of different ways to do this problem. Since you made no attempt to solve it yourself, we don't know which would be appropriate for you. If you are given a problem like this, you must have had some instruction in them. Do you know how to "solve triangles", using the sine and cosine laws? Do you know how to separate a vector into its components?