At noon boat A is 100 km West of boat B. A is moving at 25 km/h South, and B is moving at 35 km/h North. At what rate is the distance between A and B changing at 4pm?

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At noon boat A is 100 km West of boat B. A is moving at 25 km/h South, and B is moving at 35 km/h North. At what rate is the distance between A and B changing at 4pm?

Mathematics
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The change on horizontal distance (y) is 0
@Ahmad-nedal thanks so much! How did you get 500?
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Oops I'm sorry. It's 240 Since (35+25)*4=240 The speeds add up since they go in opposite directions
The two boats can never be moving at a rate more than 70km/h away from one another.
I'm afraid there are other mistakes, if the idea is clear, just forget about the final numbers
Right, I don't quite understand [t(35+25)] this part.. i mean the logic behind it
Ahh okay, so using Pythag we always know the distance between the two boats...
't' being the time passed in hours. Distance^2 = 100^2 + [(35+25)t]^2
U want the distance in the equation, which is x in this case, Distance=time*speed=x From that you can also find z, the other distance
Ex: after an hour, the boats will be 100km apart on the x-axis, and 70km apart on the y-axis right? after 2hrs, the boats will still be 100km apart on the x-axis, but 140km apart on the y-axis
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Derp! YES It is 60... hah!
Ohhh right! Now I get it! I totally forgot about the formula d=vt
That makes sense now
Be careful u forgot about many things Tracy ,just like me :P Good luck
Yep! You are solving for v in this problem.
|dw:1435206409506:dw| So in fact, we are using pythag for this black triangle right?
Exactly!
Hahaha thanks @Ahmad-nedal I've got an exam tomorrow morning >.<
@gFunc Thanks so much!!! That was really really really helpful!! Appreciate it! xx
My pleasure! Good luck on the test!
Thanks! :D

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