algebra help please?

Question

A boat takes 1(1/2) hours to ttravel 6 miles downstrem and the same distance back. If the boat travels at 9 mph in still water, what is the speed of the current? Show work and give eplanations for each step. Use one variable.

how do i solve thi sproblem?
thanks

Answer 1

A. The simple formula for distance traveled is d = rt where r is the rate of travel and t = the time traveled, and d = the distance traveled.

B. Solving for t we get t = d/r

C. d is given as 6 miles each way.

If c is the rate of the current,
the rate of travel for the first six miles (down stream... with the current) was 9 + c. So, using B above,
t for the first six miles was 6/(9+c).

Similarly, the rate of travel for the last six miles was 9 -c.
t for the last six miles was 6/(9-c).

And we know that the total of those two times was 1.5 hours.
So we have 6/(9+c) + 6/(9-c) = 3/2

The common denominator is 81 - c² ... we get that by multiplying (9+c)(9-c)
6/(9+c) = 6(9-c)/(81 - c²), and 6/(9-c) = 6(9+c)/(81 - c²)

Adding we get
[(54 - 6c) + (54 + 6 c)]/(81 - c²) = 3/2
or 108/(81 - c²) = 3/2

We can factor out a 3 here and get
36/(81 - c²) = 1/2

To get rid of the denominators, multiply both sides by 2(81 - c²) and we get
72 = 81 - c²

Using a little algebra we get
c² = 81-72 = 9... making c = 3

So the speed of the current is 3 mph.

Checking...
t (downstream) = 6/(9+3) = 1/2 hour
t (upstream) 6/(9-3) = 1 hour
Total time 1.5 hours... Works!!!

Answer 2

Let's call the current speed "c". The boat travels 9mph, and for six miles. Distance = rate * time (or time = distance/rate):

the downstream time is distance/rate, where distance=6 and rate=9+c, because the speed of the current (relative to the shore) is added to the speed of the boat. t1 = 6/(9+c)

the upstream time is distance/rate, where distance=6 and rate=9-c, because the speed of the current (relative to the shore) is subtracted from the speed of the boat. t2 = 6/(9-c)

The sum of those times (t1+t2) is 1.5 hours:

6/(9+c) + 6/(9-c) = 3/2

Multiply by 2(9+c)(9-c), or 2(81 - c^2), to eliminate everything in the denominators, and you get:

12(9-c) + 12(9+c) = 3(81 - c^2)

108 - 12c + 108 + 12c = 3(81 - c^2)

216 = 3(81 - c^2)
72 = 81 - c^2
c^2 = 9

c = 3. (c could also be -3, but a negative current speed makes no sense, so we can eliminate that possibility).

Checking the answer, that would be 9+3 (12 mph) downstream, and 9-3 (6 mph) upstream. That would take a half-hour (6mi/12mph) to make the 6-mile trip downstream, and a full hour (6mi/6mph) upstream... so the total of 1.5 hours works out.

Answer 3

let s = speed of boat in still water. Let w =speed of current. Note that he could go well over 9 miles with no current - so the current is working against him. He's going up stream. His ground speed is s-w. Now use d=rt formula 6=(9-w)*3/2 or 12=27-3w
3w=27-12=15
w=5

Answer 4

Let's call the current speed "c". If the boat travels 9mph, and for six miles...

the downstream time is 6/(9+c)
the upstream time is 6/(9-c)

I just copied the answer above