Asked by Sean charles
On Monday afternoon, Andrew rowed his boat with current for 4.5 hours and covered 27 miles, stopping in the evening at a campground. On Tuesday morning he returned to his starting point against the current in 6.75 hours. Find the speed of the current and the rate at which Andrew rowed in still water. (I have problems solving word problems)
Answers
Answered by
Damon
speed down = (v+c)
time down =4.5 = 27/(v+c)
speed up = (v-c)
time up = 6.75 = 27/(v-c)
so
6.75(v-c) = 4.5(v+c)
6.75 v -4.5 v = 4.5 c + 6.75 c
2.25 v = 11.25 c
v = 5 c
now back
6.75 = 27/(5c-c) = 27/4c
27 = 27c
c = 1
time down =4.5 = 27/(v+c)
speed up = (v-c)
time up = 6.75 = 27/(v-c)
so
6.75(v-c) = 4.5(v+c)
6.75 v -4.5 v = 4.5 c + 6.75 c
2.25 v = 11.25 c
v = 5 c
now back
6.75 = 27/(5c-c) = 27/4c
27 = 27c
c = 1
Answered by
Megan Brown
What is V+C
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