I can't seem to figure this one out! I am sure I'm making it more complicated than it is.
A motorboat can maintain a constant speed of 16 miles per hour relative to the water, and make a trip upstream in 20minutes; the return trip takes 15 minutes. What is the speed of the current?
2006-08-01 12:49:26 · 5 answers · asked by Monica S 2 in Science & Mathematics ➔ Mathematics
(16 - x)/3 = (16 + x)/4
4(16 - x) = 3(16 + x)
64 - 4x = 48 + 3x
7x = 16
x = 2.286 mph
2006-08-01 13:04:34 · answer #1 · answered by Jim R 3 · 0⤊ 0⤋
Let's say "C" is the speed of the current. The two things constant here are the distance of the trip "d" and the speed of the boat relative to the water (16 mph)
upstream trip d = (16 - C)(20 minutes or 1/3 of an hour)
downstream trip d = (16 + C)(15 minutes or 1/4 of an hour)
since both trips are the same distance d=d
therefore (16 - C)(1/3) = (16 + C)(1/4)
solve I think C = 16/7
2006-08-01 20:05:38 · answer #2 · answered by Whitman Lam 5 · 0⤊ 0⤋
To solve this problem, begin with the formula for speed:
speed = distance / time
Rearrange to find distance, which is the same going upstream as downstream:
distance = speed x time
Let s = speed of current
therefore:
distance upstream = (speed - s) x 20 which is equal to
distance downstream = (speed +s) x 15
so
(speed - s) x 20 = (speed +s) x 15
Solve for s (I get 2.2857)
Ideally, substitute this value into your equation to check.
If you are keen, you could now determine the distance of the trip.
2006-08-01 20:12:07 · answer #3 · answered by Auriga 5 · 0⤊ 0⤋
Sorry for my poor English:
Distance = x
Speed of the water = v
x/(16+V) = 1/4
x/(16-V) = 1/3
So
4x = 16 + v
3x = 16 - v
So
x = 2v
Substituting : x = 16/7
2006-08-01 20:10:28 · answer #4 · answered by Anonymous · 0⤊ 0⤋
I think lam gets the apple for this one kiddo. Sorry I let you down but I went back to work for a bit and was late getting back to this stuff...
2006-08-02 01:42:49 · answer #5 · answered by Walter J 3 · 0⤊ 0⤋