A boat is traveling in a river with a current that has a speed of 1.5 km/h. In one hour, the boat can travel twice the distance downstream as it can travel upstream. What is the boat's speed in still water?
Help!!!
i hate word problems
but i am sure we can do this using distance equals rate times time
Me and my mom are having real hard problems with this. In class we are using the elimination method. I just can't figure it out
We are also using d=rt
we need a variable for the rate is still water lets call it \(x\) and see what equations we can come up with
going downstream the rate will be \(1.5\) faster so \(x+1.5\) downstream similarly upstream the rate will be \(x-1.5\)
b+1.5=2(the upstream) b-1.5=the uptream
? Idk
then since \(T=1\) we have \[x+1.5=2D\\ x-1.5=D\] and we can solve this one
you can solve for \(x\) using a number of methods. one way it to double the second one and get \[x+1.5=2D\\ 2x-3=2D\] then solve \[2x-3=x+1.5\]
Wouldn't it be b=the speed of the boat in still water c=the speed of the current 1.5=c downstream=2(upstream) So… b+c=2(upstream) b-c=upstream Or... b+1.5=2(upstream) b-1.5=upstream ??
calulate the speed im upstream then calculate in down strea (as its given twice divide by 2) then take the difference in down and up stream
i think instead of using the word "upstream" make that word "distance" or just D
but they are equal so when you wrote \[b+1.5=2D\\ b-1.5=D\] that is really the same double the second one get \[b+1.5=2D\\ 2b-3=2D\] then set them equal to get \[b+1.5=2b-3\]
Then substitute... Upstream=b-1.5 2(upstream)=b+1.5 2(b-1.5)=b+1.5 2b-3=b+1.5?
you had it right here b+1.5=2(upstream) b-1.5=upstream so again double the second one and set them equal
yup looks good to me solve that for \(b\) in two steps and you are done
@satellite73 but my method was correct, right?
yes
since one is twice as fast as the other we probably could have gone straight to \[2x-3=x+1.5\]
Thank you!!!
yw
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