- 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.)

- schrodinger

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- anonymous

- 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

i have no idea!! please explain

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## More answers

- anonymous

- 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

ok

- 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

ok

- 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

Let's do the distances first bec they are very easy.

- mathstudent55

How far does he row each way?

- anonymous

like how am i supposed to know im so confused on this question

- mathstudent55

Have you read the problem?
It's written right there.
There is no calculation needed.

- 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

so for up its 30 ?

- mathstudent55

Right, Each part of the trip is 30 miles.
We can add the distances to our table.

- anonymous

ok

- mathstudent55

|dw:1436391741905:dw|

- 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

slows him down?

- 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

|dw:1436392136839:dw|

- anonymous

ok now that we have the info how would i answer the question?

- mathstudent55

Now that we have all the info we need, we can solve the problem.
speed = distance/time
That means
distance = speed * time

- mathstudent55

We write an equation for going up the river and an equation for going down the river.

- mathstudent55

Up the river
distance = speed * time
30 = (s - 5)t
Down the river
distance = speed * time
30 = (s + 5)(8 - t)

- anonymous

s-5 = 30 /t for up

- anonymous

oh nvm ur right

- mathstudent55

Now we have a system of equations that we need to solve:
30 = (s - 5)t
30 = (s + 5)(8 - t)

- 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

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

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

thank you so much!

- mathstudent55

You're welcome.

- anonymous

could you help me with this as well?

- anonymous

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