heres a fun little problem?

Question

a boat can travel 5 miles uptream and 5 miles downstream in a total of 6 hour. If the speed of the current is 2mph, what is the speed of the boat in still water?

Answer 1

Use b for boat's speed.

time = distance/rate, and the sum of the times for upstream (b-2) and downstream (b+2) for 5 miles each way is 6 hours:

5/(b-2) + 5/(b+2) = 6

Multiply by (b-2)(b+2) to eliminate denominators:

5/(b-2) + 5/(b+2) = 6
5(b+2) + 5(b-2) = 6(b+2)(b-2)
5b + 5b + 10 - 10 = 6(b^2 - 4)
10b = 6b^2 - 24
6b^2 -10b - 24 = 0
3b^2 - 5b - 12 = 0

Now use the quadratic:

b = (-(-5) +/- sqrt((-5)^2 - 4*3*(-12)))) / (2*3)
b = (5 +/- sqrt(25 + 144)) / 6
b = (5 +/- sqrt(169)) / 6
b = (5 +/- 13) / 6

Two solutions:

b = (5 + 13)/ 6 = 18/6 = 3 mph
b = (5 - 13)/6 = -8/6 mph

The negative answer can be ignored. The answer is 3 mph.

Checking the answer: if the boat's speed is 3mph, then the upstream speed is 3-2 = 1 mph. It would take the boat 5 hours to go 5 miles upstream. And the downstream speed is 3+2 = 5 mph. It would take the boat 1 hour to go 5 miles downstream. 5 + 1 = 6 hours, which is correct.

Answer 2

Let S be the speed in still water. Then going upstream the speed is s - 2, and downstream it's s + 2.

Speed is distance over time (s = d/t), so time is distance over speed (t = d/s). The time upstream was 5/(s-2) and downstream it was 5/(s+2). These add up to be 6 hours, so:
5/(s-2) + 5/(s+2) = 6.

Solve this for s (hint: multiply both sides by (s-2) and (s+2) first to clear the fractions).

Answer 3

It doesn't matter what the speed of the current is if the boat is in still water. It's speed will "still" be zero.

Answer 4

[ 5 / (x-2) ] + [ 5 / (x+2) ] = 6

5(x+2) + 5(x-2) = 6(x+2)(x-2)

10x = 6x^2 - 24

6x^2 - 10x - 24 = 0

3x^2 - 5x - 12 = 0

(3x + 4)(x - 3) = 0

x = 3
x = -4/3

3 miles per hour