Question about 2 ships - vectors?

[Star Forger]

Homework Statement

Ships A and B leave port together. For the next two hours, ship A travels at 35 mph in a direction 20° west of north while the ship B travels 10° east of north at 40 mph.

(a) What is the distance between the two ships two hours after they depart (in miles)?


(b) What is the speed of ship A as seen by ship B (in mph)?

Homework Equations

The Attempt at a Solution

I have no idea how to approach this problem. I believe it is vectors, however as I said I'm not sure. Any help would be greatly appreciated.

 

[iRaid]

Start by drawing 2 vectors from one point, let's say (0,0)
-One vector will go at a 20° angle to the west (of north) and the other will go 10° east (of north)
-Since it's 2 hours you can make the vector length double (70 and 80).
-You will end up have 1 angle, 2 sides
-You can now use the law of cosines to solveI believe this is correct, if not someone will correct me :p

Edit: that's part A.

 

[Star Forger]

Thank you, that made a lot of sense and I calculated part A correct. How would I calculate part B though? What does speed as seen by the other mean?

 

[SteamKing]

Pretend you were standing on boat B watching boat A. If both boats were traveling in tje same direction at the same speed, boat A would appear stationary to someone on boat B.

 

[rwisz]

I can provide a response to this question using mathematical and scientific principles. First, let's define some variables to help us solve the problem.

Let d be the distance between the two ships after two hours, vA be the speed of ship A, and vB be the speed of ship B.

(a) To find the distance between the two ships after two hours, we can use the formula for distance: d = vt, where v is the speed and t is the time. We know that ship A travels at 35 mph for 2 hours, so dA = 35*2 = 70 miles. Similarly, ship B travels at 40 mph for 2 hours, so dB = 40*2 = 80 miles. To find the distance between the two ships, we can use the Pythagorean theorem: d = √(dA^2 + dB^2) = √(70^2 + 80^2) = √(4900 + 6400) = √11300 ≈ 106.3 miles.

(b) To find the speed of ship A as seen by ship B, we can use the concept of relative velocity. Since ship B is traveling at 40 mph, it can be thought of as a frame of reference moving at 40 mph. This means that the velocity of ship A as seen by ship B will be the sum of its actual velocity and the velocity of the frame of reference (ship B). We can use the formula vA' = vA - vB to find the velocity of ship A as seen by ship B: vA' = 35 - 40 = -5 mph. This negative sign indicates that ship A appears to be moving in the opposite direction as seen by ship B. However, to find the speed, we need to take the magnitude of this velocity, which is 5 mph. So, ship A appears to be moving at a speed of 5 mph as seen by ship B.

In conclusion, the distance between the two ships after two hours is approximately 106.3 miles, and the speed of ship A as seen by ship B is 5 mph.