Question

A motorboat maintained a constant speed of 10 miles per hour relative to the water in going 19 miles upstream and then returning. The total time for the trip was 20.0

hours. Use this information to find the speed of the current.

Answer 1

I was trying to make head or tails of "A motorboat maintained a constant speed of 10 miles per hour relative to the water in going 19 miles upstream and then returning"

well... all it's really saying, " the boat on its way up and down, kept the speedometer at 10mph"

that simply means the boat's "still water speed" is 10mph

however, going upstream, if say the boat has a rate of 10 in this case, and the stream has a rate of "r", the boat is not really going 10mph fast, is going "10 - r" fast, because the stream's rate is eroding some from those 10mph

now, on its way down, is not going 10mph fast either, is going faster, is going "10+r", because is going with the stream and thus the stream's rate is adding to it

notice, it went up 19miles, and then returned, so is 19miles up, and 19miles down

took 20hrs total, so if it went up in "t" hours, the way down takes the slack from 20, or "20 - t"

now, recall your d = rt, or distance = rate * time


`\bf \begin{array}{lccclll} &distance&rate&time\\ &-----&-----&-----\\ upstream&19&10-r&t\\ downstream&19&10+r&20-t \end{array}\\\\ -------------------------------\\\\ \begin{cases} 19=(10-r)t\implies \cfrac{19}{10-r}=t\\\\ 19=(10+r)(20-t)\\ ----------\\ 19=(10+r)\left(20- (19)/(10-r) \right) \end{cases}`


`\bf 19=(10+r)\left( \cfrac{200-20r-19}{10-r} \right)\implies 19=(10+r)\left( \cfrac{181-20r}{10-r} \right) \\\\\\ 19(10-r)=(10+r)(181-20r) \\\\\\ 190-19r=1810-200r+181r-20r^2 \\\\\\ 190-19r=1810-19r-20r^2\implies 20r^2=1810-190 \\\\\\ 20r^2=1620\implies r^2=\cfrac{1620}{20}\implies r^2=81 \\\\\\ r=√(81)\implies \boxed{r=9}`