Answer:
The answer is 40.25 [nmi]
Explanation:
1) We use the equation for linear movement with constant velocity
- [tex]x_{f} =x_{0} +vt[/tex]
2) Let´s focus on ship 1 first
- [tex]x_{1f} =x_{1o} +v_{1} t\\x_{1f} =0+12(1.5)\\x_{1f} =18[nmi][/tex]
3) Now ship 2
- [tex]x_{2f} =x_{2o} +v_{2} t\\x_{2f} =0+24(1.5)\\x_{2f} =36[nmi][/tex]
4) Now we know how far apart are the ships from the port, if we draw a line from Ship 1 to Ship 2, we can see we form a triangle rectangle, since ship 1 left on a bearing of 42° and ship 2 on a bearing of 132°C, we know they left on a bearing of 90° from each other, since 132-42 = 90
5) We have a triangle rectangle, we know two of its sides, we can use the pythagoras therom to solve for the distance between the two ships:
- [tex]distance = \sqrt{18^{2}+36^{2} } \\distance = 40.25 [nmi][/tex]
