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Help...More Math Help!!!...Please!

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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 y-6.
The distance traveled is the same; therefore,
upstream rate x time = downstream rate x time
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/(v-6 mph) = 3 hours -->

(v + 6 mph)* 2 hours/(v-6 mph) = 3 hours --->

(x+6)/(x-6) = 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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