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

The answer is 40.25 [nmi]

Step-by-step 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)  } \\<strong>distance = 40.25 [nmi]`