Question

Two ships leave a port at the same time. The first ship sails on a bearing of 42 degrees at 12 knots​ (nautical miles per​ hour) and the second on a bearing of 132 degrees at 24 knots. How far apart are they after 1.5 ​hours? (Neglect the curvature of the​ earth.)

Answer 1

Answer:

The answer is 40.25 [nmi]

Explanation:

1) We use the equation for linear movement with constant velocity

• $x_{f} =x_{0} +vt$

2) Let´s focus on ship 1 first

• $x_{1f} =x_{1o} +v_{1} t\\x_{1f} =0+12(1.5)\\x_{1f} =18[nmi]$

3) Now ship 2

• $x_{2f} =x_{2o} +v_{2} t\\x_{2f} =0+24(1.5)\\x_{2f} =36[nmi]$

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:

• $distance = \sqrt{18^{2}+36^{2}  } \\distance = 40.25 [nmi]$