Help...More Math Help!!!...Please!
posted by Jessica .
A kayaker paddled 2 hours with a 6 mph current in a river. The return trip against the same current took 3 hours. Find the speed the kayaker would make in still water.
The speed is 6 mph slower going upstream and 6 mph faster going downstream. Distance traveled = time x speed. The distance for both is equal. Let X = the speed in still water.
With the above information, you should be able to develop the equation and solve it for X.
I hope this helps. Thanks for asking.
Let y = speed of the kayaker.
distance = rate x time.
For the trip down stream, his rate is y+6 and his rate back up stream is y6.
The distance traveled is the same; therefore,
upstream rate x time = downstream rate x time
(y6)*3hrs=(y+6)*2hrs
Solve for y.
Post your work if you get stuck.
You just give names to the variables you don't know and write down the equations.
The speed the kayaker in still water? Don't know? Let's call it v.
Then the speed when travelling with the current is
v + 6 mph.
The distance traveled in the two hours is:
(v + 6 mph)* 2 hours
let's call this d, so we put:
d = (v + 6 mph)* 2 hours
The speed on the return trip is
v  6 mph
The time mneeded for that is given to be 3 hours, but it is also equal to the distance divided by the speed. So, you have that:
d/(v6 mph) = 3 hours >
(v + 6 mph)* 2 hours/(v6 mph) = 3 hours >
(x+6)/(x6) = 3/2
where x = v/(mph)
x + 6 = 3/2 x  9 >
1/2 x = 15 >
x = 30 >
v/(mph) = 30 >
v = 30 mph
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