How Do You Calculate the Distance and Relative Speed Between Two Ships?

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To calculate the distance between ships A and B after two hours, the correct approach involves breaking down their velocities into x and y components using trigonometry. Ship A travels at 35.0 mph at an angle of 25.0 degrees west of north, while ship B moves at 30.0 mph at 80.0 degrees east of north. The distance can be found by determining the resultant position of both ships after two hours and applying the distance formula. For the relative speed of ship A as seen by ship B, vector addition or subtraction is necessary, considering their respective directions and speeds. Accurate component calculations, including both x and y components, are crucial for obtaining the correct results.
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Ships A and B leave port together. For the next two hours, ship A travels at 35.0 mph in a direction 25.0 degrees west of north while the ship B travels 80.0 degrees east of north at 30.0 mph.
What is the distance between the two ships two hours after they depart?
What is the speed of ship A as seen by ship B?

basically i tried using trig to find the x component of both vectors and then added them. it wanted distance at t = 2 so i did 70 * sin (25) + 60 * sin(80) but it was incorrect. i tried this method with the example in the book and it was fairly close. so I am guessing i did it incorrectly but got an answer that was close by coincidence? and for the second part I am really unsure where to begin, i just know reference frames have something to do with it and maybe some sort of vector addition/subtraction. any suggestions will be much appreciated.
 
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anyone help please? I am stuck
 
try using

v = \frac{d}{t}
 
i got it using the law of cosines. thanks for your suggestion though.
 
When you're calculating the length of a component-wise vector it's
\sqrt{x^2+y^2}
from the way you describe what you did, you might have forgotten the y component.

The second part looks like straightforward vector addition as well.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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