Question

A ship leaves port at noon and travels at a bearing of 214°. The ship’s average rate of speed is 15 miles per hour.

Answer 1

Answer: (-31.09 mi, -20.97 mi)

Step-by-step explanation:

If the noon is our initial time, then we have that 2:30 pm is t = 2.5 hours.

Now, the position (x, y) of the ship can be writen as:

(velocity*time*cos(angle), velocity*time*sin(angle))

or:

(15mph*t¨*cos(214°), 15mph*t*sin(214°))

then the position at t = 2.5 hours is:

(15mph*2.5h*cos(214°), 15mph*2.5h*sin(214°))

(-31.09 mi, -20.97 mi)