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Mathematics 11 Online
OpenStudy (mokeira):

Solving a distance, rate, time problem using a system of linear equations

OpenStudy (mokeira):

OpenStudy (mokeira):

@TylerD

OpenStudy (mokeira):

@ziqbal103 @aryandecoolest

OpenStudy (tylerd):

so it travels at 25mph upstream and 37mph downstream

OpenStudy (mokeira):

yes it does

OpenStudy (tylerd):

i think you can just take an average of the 2 to find it in still water

OpenStudy (tylerd):

im not sure how id do this with a system of linear equations though

OpenStudy (mokeira):

but why do you do so? that is the part i dont understand

OpenStudy (tylerd):

so i get 31 mph in still water with a water current at 6

OpenStudy (mokeira):

@TylerD

OpenStudy (anonymous):

It easy calculate speed upstream and downstream and then substitute in this: \[U=1/2(a+b)\] \[V=1/2(a-b)\] where a and b are speed downstream and upstream and U and V are speed of boat in still water and in stream respectively...

OpenStudy (mokeira):

ok..i was confused somewhere but now i get it. Thank u guys!!!!

OpenStudy (tylerd):

i just know from physics half of Velocity final + velocity initial = avg velocity

OpenStudy (mokeira):

oooh...that could wok too!!! there seem to be so many ways to do this

OpenStudy (anonymous):

Let the speed of the boat in still water be x and the speed of the current be y. While going upwards, speed of boat = x-y time = 4 hours distance = 100 miles speed = dist/time x-y = 100/4 = 25

OpenStudy (mokeira):

continue... @Brainybeauty

OpenStudy (anonymous):

Okay So while going downwards, speed of boat = x+y time = 4 hours distance = 148 miles speed = dist/time x+y = 148/4 = 37

OpenStudy (mokeira):

so then i find the value of x and the value of y?

OpenStudy (anonymous):

So you can find x and y now?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

You basically have to make two equations from the given information and then solve them to get the answer. It is as simple as that.

OpenStudy (mokeira):

amazing! thank you soooo much

OpenStudy (anonymous):

U R welcome! :)

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