anonymous
  • anonymous
PLEASE HELP The current in a nearby river is 5 mph. John wants to be able to travel up the river for a distance of 30 miles and return to his starting point in a total of 8 hours. How fast must John be able to row his boat in order to accomplish this? (Assume that the speed that John can row refers to how fast he can row in still water.)
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
mathstudent55
  • mathstudent55
He travels a total time of 8 hours. If he travels up the river for t time, how much time will he travel down the river?
anonymous
  • anonymous
i have no idea!! please explain

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anonymous
  • anonymous
mathstudent55
  • mathstudent55
Ok. The total trip is 8 hours. time going up + time going down = 8 hours If you let the time going up be called just t, then you have t + time going down = 8 hours Ok so far?
anonymous
  • anonymous
ok
mathstudent55
  • mathstudent55
Now let's see if we can get an expression for the time going down. t + time going down = 8 Subtract t from both sides: time going down = 8 - t Now we have: Time going up = t Time going down = 8 - t Ok?
anonymous
  • anonymous
ok
mathstudent55
  • mathstudent55
Ok. I put the info we have in the table below. Now we need the speed going up and the speed going down, and the distances going up and down. |dw:1436391156182:dw|
mathstudent55
  • mathstudent55
Let's do the distances first bec they are very easy.
mathstudent55
  • mathstudent55
How far does he row each way?
anonymous
  • anonymous
like how am i supposed to know im so confused on this question
mathstudent55
  • mathstudent55
Have you read the problem? It's written right there. There is no calculation needed.
mathstudent55
  • mathstudent55
Does this help? \(\sf John ~wants ~to ~be ~able ~to ~travel ~up ~the ~river ~for ~a ~distance ~of ~\huge \color{red}{30 ~miles}\) \( \sf and ~return ~to ~his ~starting ~point ~in ~a ~total ~of ~8 ~hours. \)
anonymous
  • anonymous
so for up its 30 ?
mathstudent55
  • mathstudent55
Right, Each part of the trip is 30 miles. We can add the distances to our table.
anonymous
  • anonymous
ok
mathstudent55
  • mathstudent55
|dw:1436391741905:dw|
mathstudent55
  • mathstudent55
Now we need to work on the speed going up and the speed going down. His speed in still water is an unknown. Let's call it s. When he goes up the river, does the 5 mph speed of the river speed him up or slow him down?
anonymous
  • anonymous
slows him down?
mathstudent55
  • mathstudent55
Right. That means it takes away from his speed. If the speed in still water is s, then going up the river against the 5-mph current, his speed will be s - 5 By the same token, going down the river, the 5-mph current helps him, so his speed will be s + 5 down the river. Now we add the speeds to our table.
mathstudent55
  • mathstudent55
|dw:1436392136839:dw|
anonymous
  • anonymous
ok now that we have the info how would i answer the question?
mathstudent55
  • mathstudent55
Now that we have all the info we need, we can solve the problem. speed = distance/time That means distance = speed * time
mathstudent55
  • mathstudent55
We write an equation for going up the river and an equation for going down the river.
mathstudent55
  • mathstudent55
Up the river distance = speed * time 30 = (s - 5)t Down the river distance = speed * time 30 = (s + 5)(8 - t)
anonymous
  • anonymous
s-5 = 30 /t for up
anonymous
  • anonymous
oh nvm ur right
mathstudent55
  • mathstudent55
Now we have a system of equations that we need to solve: 30 = (s - 5)t 30 = (s + 5)(8 - t)
mathstudent55
  • mathstudent55
Solve the first equation for t and substitute in the second one: \(t = \dfrac{30}{s - 5} \) \(30 = (s + 5)(8 - \dfrac{30}{s - 5}) \) Now we have an equation in only s, so we can solve for s.
mathstudent55
  • mathstudent55
Multiply both sides by s - 5 to get rid of the denominator: \(30(s - 5)= (s + 5)(s - 5)(8 - \dfrac{30}{s - 5}) \) \(30s - 150 = (s + 5)(8s - 40 - 30)\) \(30s - 150 = (s + 5)(8s - 70) \) \(30s - 150 = 8s^2 - 70s + 40s - 350\) \(30s - 150 = 8s^2 - 30s - 350 \) \(8s^2 - 60s - 200 = 0\) \(2s^2 - 15s - 50 = 0\) \((2s + 5)(s - 10) = 0\) \(s = -\dfrac{5}{2} \) or \(s = 10\) We discard the negative speed, -5/2, and the answer is the speed is 10 mph.
mathstudent55
  • mathstudent55
Let's see if the answer makes sense. The speed in still water is 10 mph The speed going up is 5 mph The speed going down is 15 mph The time going up is 30/5 = 6, 6 hours The time going down is 30/15 = 2, 2 hours 6 hours + 2 hours = 8 hours. Yes, 10 mph is the correct speed he needs to row in still water.
anonymous
  • anonymous
thank you so much!
mathstudent55
  • mathstudent55
You're welcome.
anonymous
  • anonymous
could you help me with this as well?
anonymous
  • anonymous

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