SOLUTION: the tugboat moves at a rate of 10mph in still water. it travels 24 mi upstream and 24 mi downstream in a total time of 5 hr. what is the speed of the current?
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Question 137672This question is from textbook
: the tugboat moves at a rate of 10mph in still water. it travels 24 mi upstream and 24 mi downstream in a total time of 5 hr. what is the speed of the current?
This question is from textbook
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Let r=speed of the current
tugboat's speed upstream=10-r
tugboat's speed downstream=10+r
Time for tugboat to travel upstream=24/(10-r)
Time for tugboat to travel downstream=24/(10+r)
Now we are told that the sum of the above two times equals 5 hours, so our equation to solve is:
24/(10-r)+24/(10+r)=5 multiply each term by (10-r)(10+r)
24(10+r)+24(10-r)=5(10-r)(10+r) get rid of parens
240+24r+240-24r=500-5r^2 subtract 500 from and add 5r^2 to both sides
240+24r+240-24r-500+5r^2=500-500-5r^2+5r^2 collect like terms
-20+5r^2=0 rearranging:
5r^2-20=0 divide each side by 5
r^2-4=0-------------------------------quadratic in standard form and it can be factored
(r-2)(r+2)=0
r=2 mph--------------------speed of the current
and
r=-2 mph----------------------DISREGARD--- CURRENT SPEED IS POSITIVE
ck
24/(10-2)+24/(10+2)=5
3+2=5
5=5
Hope this helps----ptaylor
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