a) The ship left the port and sailed for 2 hours on a course of 75O, at an average speed of 2.5 nautical miles per hour.
b)
North
It changed its course to 165O and travelled for 3 hours, at an average speed of 4 nautical miles per hour.
Your team is tasked to lead the rescue. You will be needing the following:
1. distance of the ship from the port
2. bearing from the port to the ship
Assuming those directions are 75° and 165°.
and ignoring the
b)
North , whatever that means
you have a very simple problem.
I made a sketch with O the port, P as its first position and Q as its final position.
With some simple geometry you should see that angle P is a right angle, OP = 5 and PQ=12
so OQ^2 = 5^2+12^2
OQ = 13
in triangle OPQ,
tan O = 12/5
angle O = appr 67.4°
so the bearing from port i = (75+67.4)° = 142.4°
To calculate the distance of the ship from the port and the bearing from the port to the ship, we can use trigonometry and vector addition. Here's how we can find the answers:
a) The ship first sailed on a course of 75 degrees for 2 hours at an average speed of 2.5 nautical miles per hour. We can start by finding the distance covered in this leg of the journey.
Distance = Speed * Time
Distance = 2.5 nautical miles/hour * 2 hours
Distance = 5 nautical miles
b) After that, the ship changed its course to 165 degrees and traveled for 3 hours at an average speed of 4 nautical miles per hour. We can find the distance covered in this leg as well.
Distance = Speed * Time
Distance = 4 nautical miles/hour * 3 hours
Distance = 12 nautical miles
To calculate the total distance of the ship from the port, we need to find the magnitude of the resulting vector. We can use the Pythagorean theorem as follows:
Total Distance = sqrt(distance^2 + distance^2)
Total Distance = sqrt(5^2 + 12^2)
Total Distance = sqrt(25 + 144)
Total Distance = sqrt(169)
Total Distance = 13 nautical miles
Now, to find the bearing (direction) from the port to the ship, we can use trigonometry. We'll need to find the angles of the two vector components (the lengths of each leg of the journey) relative to the positive x-axis and add them together.
Angle = arctan(distance2 / distance1)
Angle = arctan(12 / 5)
Angle ≈ 67.38 degrees
However, since the ship changed its course from north, we need to subtract the angle from 90 degrees to get the final bearing.
Bearing = 90 degrees - Angle
Bearing = 90 degrees - 67.38 degrees
Bearing ≈ 22.62 degrees
Therefore, the distance of the ship from the port is approximately 13 nautical miles, and the bearing from the port to the ship is approximately 22.62 degrees.