A ship regularly travels between two ports that are 350 miles apart. When traveling directly from A to B, the ship sails on a bearing of . However, to avoid a storm, one day the ship leaves A on a bearing of and travels for 6 hours at 17mph. When the bad weather has passed, it turns and heads for B.How far is the ship from B when it makes the turn? What is the bearing for the second part of the trip?
Ships travel on headings, not bearings.
From the ship's perspective, other objects have bearings.
Other than that, supply the missing numbers and maybe we can get somewhere.
Ships travel on "headings" not "bearings". Math teachers are not seamen :)
You need to know what the heading is for the first part of the trip to do the problem.
one day the ship leaves A on a HEADING of ##### WHAT #### for 102 miles (6*17)
To find the distance from point A to point B when the ship makes the turn, we can use the concept of vector addition.
First, let's establish a coordinate system where point A is the origin (0,0). From point A, the ship travels on a bearing of θ (angle measured clockwise from the positive x-axis).
Given that the ship travels at a speed of 17 mph for 6 hours, we can calculate the displacement vector for the first part of the trip. The magnitude of the displacement vector is given by the formula:
Magnitude = Speed × Time
Magnitude = 17 mph × 6 hours = 102 miles
Now, using trigonometry, we can find the x and y components of the displacement vector. The x-component can be found using the cosine function:
x-component = Magnitude × cos(θ)
Similarly, the y-component can be found using the sine function:
y-component = Magnitude × sin(θ)
Since the angle is not given in the question, we cannot directly calculate the x and y components of the displacement vector. Therefore, we need to know the value of θ to find the distance from point A to B when the ship makes the turn.