A kayaker paddled 2 hours with a 6 mph current in a river. The return trip against the same current took 3 hours. How do you find the speed the kayaker would make in still water?

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Answer 1 · 2018-08-05T02:55:32

$\frac{6}{5} m p h$ $6/5 mph$

D = RT
As the Distance is the same

$R_{1} \times T_{1} = R_{2} \times T_{2}$ $R_1 xx T_1 = R_2 xx T_2$

Let K equal the rate of the Kayaker in still water

$(6 + K) \times 2 = (6 - K) \times 3$ $( 6 + K) xx 2 = ( 6 -K ) xx 3$

Multiplying across the parenthesis gives

$12 + 2 K = 18 - 3 K$ $12 + 2K = 18 - 3K$

Adding the opposite to both sides of the equation gives

$12 - 12 + 2 K + 3 K = 18 - 12 - 3 K + 3 K$ $12 - 12 + 2K + 3K = 18 - 12 - 3K + 3K$

The result is

$5 K = 6$ $5K = 6$

Dividing both sides by 5

$5 \frac{K}{5} = \frac{6}{5}$ $5K/5 = 6/5$

$K = \frac{6}{5}$ $K = 6/5$