Question

I'm stuck on this one algebra question which I was suppose to create, yet I don't know how to solve it.

Sally drove a Jet Ski for 4 hours upstream against a water current flowing at 10 MPH. When returning back home with the water current it was a 2 hour ride. Calculate how fast Sally traveling in still water?

Thanks in advance for any help! :)

Answer 1

hmmm,

speed = distance/time

speed = current +- jetski.

assuming the distance is constant, and the speeds of the river and jet are constant ....

Answer 2

(c-j) = d/4

4(c-j) = d

2(c+j) = d

4(c-j) = 2(c+j)

4(10-j) = 2(10+j)

solve for j?

Answer 3

i seem to have deleted my (c+j) = d/2 :)

Answer 4

4(10-j) = 2(10+j)

2(10-j) = 10+j

20-2j = 10+j

10 = 3j

10/3 = j

Answer 5

does that make sense? and did i think it thru correctly?

Answer 6

I seem to understand how you solved it but got lost in the beginning when you were setting up the equation, but I got that the speed of the Jet Ski, or the speed of it in "still" water, which is the same (from my understanding) is 30 not 10/3...

when solving I figured that both traveled at different times, so I would multiply each one by their respective time traveled.

4(j-10) = 2(j+10)

4j-40 = 2j+20

2j=60

j = 30

Is this wrong? How did you set up your equation? Still a bit confused... :/

Answer 7

hmmm, i do tend to get these things mixed around at times :)

your set up with the jet +- current seems better.

Answer 8

in fact, to get upstream he would have to be going faster than the current :)

Answer 9

So just to recap, the speed of the Jet ski is 30 MPH? and thanks for your input!

Answer 10

yep, thats would be correct.