SOLUTION: A motorboat went 10 miles upstream in 1 hour. The return trip took only 30 minutes. Assume that the "water speed" and the current speed were constant during both parts of the trip.

Algebra ->  Coordinate-system -> SOLUTION: A motorboat went 10 miles upstream in 1 hour. The return trip took only 30 minutes. Assume that the "water speed" and the current speed were constant during both parts of the trip.      Log On


   

Question 180585: A motorboat went 10 miles upstream in 1 hour. The return trip took only 30 minutes. Assume that the "water speed" and the current speed were constant during both parts of the trip. Find the water speed of the boat and the current speed.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let r%5Bb%5D= the speed of the boat in still water
Let r%5Bc%5D= the speed of the current
For the trip upstream:
(1) d%5Bu%5D+=+r%5Bu%5D%2At%5Bu%5D
For the trip downstream:
(2) d%5Bd%5D+=+r%5Bd%5D%2At%5Bd%5D
given:
d%5Bu%5D+=+10 mi
d%5Bd%5D+=+10 mi
t%5Bu%5D+=+1 hr
t%5Bd%5D+=+.5 hr
upstream:
r%5Bu%5D+=+r%5Bb%5D+-+r%5Bc%5D
r%5Bd%5D+=+r%5Bb%5D+%2B+r%5Bc%5D
------------------------
(1) d%5Bu%5D+=+r%5Bu%5D%2At%5Bu%5D
(1) 10+=+r%5Bu%5D%2A1
(1) 10+=+%28r%5Bb%5D+-+r%5Bc%5D%29%2A1
(1) 10+=+r%5Bb%5D+-+r%5Bc%5D
and
(2) d%5Bd%5D+=+r%5Bd%5D%2At%5Bd%5D
(2) 10+=+r%5Bd%5D%2A.5
(2) 10+=+%28r%5Bb%5D+%2B+r%5Bc%5D%29%2A.5
(2) 10+=+.5r%5Bb%5D+%2B+.5r%5Bc%5D
-------------------------------
Subtract (2) from (1)
(1) 10+=+r%5Bb%5D+-+r%5Bc%5D
(2) -10+=+-.5r%5Bb%5D+-+.5r%5Bc%5D
(3) 0+=+.5r%5Bb%5D+-+1.5r%5Bc%5D
1.5r%5Bc%5D+=+.5r%5Bb%5D
3r%5Bc%5D+=+r%5Bb%5D
Now I can rewrite (1)
(1) 10+=+r%5Bb%5D+-+r%5Bc%5D
(1) 10+=+3r%5Bc%5D+-+r%5Bc%5D
(1) 10+=+2r%5Bc%5D
r%5Bc%5D+=+5 mi/hr
and, since
(1) 10+=+r%5Bb%5D+-+r%5Bc%5D
(1) 10+=+r%5Bb%5D+-+5
(1) r%5Bb%5D+=+15 mi/hr
The speed of ther boat in still water is 15 mi/hr
The speed of the current is 5 mi/hr
check:
(2) 10+=+%28r%5Bb%5D+%2B+r%5Bc%5D%29%2A.5
10+=+%2815+%2B+5%29%2A.5
10+=+20%2A.5
10+=+10
OK