# Another Algebra Word Problem - Speed, Distance

• Apr 9th 2010, 08:18 AM
guest84
Another Algebra Word Problem - Speed, Distance
Am stuck again. Anyone have any ideas?

In a stream whose current flows at the speed of 3 miles/hr, a man rowed upstream and returned to his starting point in 6.25 hr. If the man rows at a speed of 5 mi/hr, in still water, how long did he row upstream.

Tup + Tdown = 6.25 Hrs.

Vup = (Vman - 3)
Vdown = (Vman + 3)

Dup = Ddown = D , Tup (Vb-3) = Tdown (Vb+3)

D / (Vman - 3) + D / (Vman + 3) = 6.25 Hrs
• Apr 9th 2010, 09:08 AM
qmech
You're almost there. They told you Vman=5, so Vup=2, Vdown=8, etc.
• Apr 9th 2010, 09:34 AM
Wilmer
OR go this way:
speeds are (5-3) = 2 and (5+3) = 8
let h = time @ 2mph; then time at 8mph = 25/4 - h
SO:
2h = 8(25/4 - h)
2h = 50 - 8h
10h = 50
h = 5

Wrap 'er up !
• Apr 9th 2010, 09:39 AM
mathemagister
Quote:

Originally Posted by guest84
Am stuck again. Anyone have any ideas?

In a stream whose current flows at the speed of 3 miles/hr, a man rowed upstream and returned to his starting point in 6.25 hr. If the man rows at a speed of 5 mi/hr, in still water, how long did he row upstream.

Tup + Tdown = 6.25 Hrs.

Vup = (Vman - 3)
Vdown = (Vman + 3)

Dup = Ddown = D , Tup (Vb-3) = Tdown (Vb+3)

D / (Vman - 3) + D / (Vman + 3) = 6.25 Hrs

Where exactly are you stuck? In the last bit of algebra?

You know how to solve
$\displaystyle \frac{D}{2} + \frac{D}{8} = 6.25$, right?

Thereafter, you know everything you need to solve for T_up.

Good luck :)

Mathemagister
• Apr 9th 2010, 09:59 AM
guest84
5 mi/hr in still water no?
I could be misinterpreting the question...I thought the 5 mi/hr does not apply for the current.

The way I interpreted it was that you need to determine the distance D from down and back with the current.

Then...

Using the velocity known in STILL water, determine how long it took him to go the the same distance encountered with the current.

In a stream whose current flows at the speed of 3 miles/hr, a man rowed upstream and returned to his starting point in 6.25 hr. If the man rows at a speed of 5 mi/hr, in still water, how long did he row upstream.
• Apr 9th 2010, 12:56 PM
Wilmer
Quote:

Originally Posted by guest84
I could be misinterpreting the question...I thought the 5 mi/hr does not apply for the current.

Correct: it does not, 5mph is the speed if NO current;
BUT with 3mph current, speed is 2mph if against, 8mph if with.
Stop your paralysis by analysis (Smirk)
• Apr 10th 2010, 04:26 AM
mathemagister
Quote:

Originally Posted by guest84
I could be misinterpreting the question...I thought the 5 mi/hr does not apply for the current.

The way I interpreted it was that you need to determine the distance D from down and back with the current.

Then...

Using the velocity known in STILL water, determine how long it took him to go the the same distance encountered with the current.

In a stream whose current flows at the speed of 3 miles/hr, a man rowed upstream and returned to his starting point in 6.25 hr. If the man rows at a speed of 5 mi/hr, in still water, how long did he row upstream.

That way, you are just adding more work. The 5mi/hr does apply for the current in that it is the "base number" that the current adds to. So a current of 3mi/hr downstream will mean that the water is moving at 5mi/hr + 3mi/hr = 8mi/hr.

Hope you are less confused now :)

Mathemagister