Two friends canoe upstream for 3 hours, only to realize that their campsite is downstream.

algebra-precalculusphysics

They turn around and paddle downstream for 4 hours. The next morning, they pack up and get back on the river to canoe to their original starting point 28 miles upstream and arrive at 2 pm. Assume the river always flows at a constant rate of 2 miles per hour, and the two friends always paddle at a constant rate. What time did they leave for their return trip?


My approach:

When going up they have the stream against them, so they go at their rate $(r) – 2m/h$. On the other hand, when going downstream they gain 2m/h.

$D=r*t$

$(r+2)*4-(r-2)*3 = 28$

$r=14$

The original point is upstream so they will paddle at $12m/h$

$28 = 12*t$

$t= \frac{7}{3}$ $\rightarrow$ $t=2h 20m$. Since they arrived at $2$ pm, they left at $11:40$ am.

What do you think?

Best Answer

Your math is correct. Here's an illustration I made to go with it:

enter image description here

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