SOLUTION: Ship A leaves port sailing north at a speed of 15 mph. A half hour later, ship B leaves the same port sailing east at a speed of 33 mph. Let t (in hours) denote the time ship B has
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-> SOLUTION: Ship A leaves port sailing north at a speed of 15 mph. A half hour later, ship B leaves the same port sailing east at a speed of 33 mph. Let t (in hours) denote the time ship B has
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Question 338372: Ship A leaves port sailing north at a speed of 15 mph. A half hour later, ship B leaves the same port sailing east at a speed of 33 mph. Let t (in hours) denote the time ship B has been at sea.
Find an expression in terms of t giving the distance between the two ships.
I don't know how to start it thanks! Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Ship A leaves port sailing north at a speed of 15 mph.
A half hour later, ship B leaves the same port sailing east at a speed of 33 mph.
Let t (in hours) denote the time ship B has been at sea.
Find an expression in terms of t giving the distance between the two ships.
:
This is a right triangle problem
The path of ship A is one leg
The path of ship B is the other leg
The distance between the ships is the hypotenuse (d)
:
Dist = speed * time
:
The distance traveled by Ship A:
A = 15(t+.5),
A = (15t+7.5) (it traveled a half hour longer than B)
:
The distance traveled by Ship B:
B = 33t
:
d^2 = A^2 + B^2
which is
d^2 = (15t+7.5)^2 + (33t)^2
:
FOIL (15t+7.5)(15t+7.5); square (33t)
d^2 = 225t^2 + 225t + 56.25 + 1089t^2
:
d^2 = 1314t^2 + 225t + 56.25
:
d(t) =
:
You can see this when t = 0, Ship A will have traveled half an hour
and will have traveled .5(15) 7.5 mi, before Ship B starts
:
Using our equation for t=0 we have
d =
Which is
d = 7.5 mi as you would expect, did this make sense to you?
