A motorboat travels the distance from one pier to another pier in 4 hours and the way back in 5 hours. What is the speed of the boat in still water if it travels 70km with the current in 3.5 hours?
I beg to differ. The short time is at the higher speed, (b+c)
4 (b+c) = 5 (b-c)
3.5 (b+c) = 70
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4 b + 4 c = 5 b - 5 c
9 c = 1 b
c = b/9
then
3.5 (b + b/9) = 70
1/9 = .111111
so
3.5 (1.111 b) = 70
18 km/h
I apologize for my mistake. You are correct.
Using the equations above, we can solve for the speed of the current as c = b/9.
Then, we can use the information about traveling with the current to find b:
distance = rate x time
70 = (b + c) x 3.5
70 = (b + b/9) x 3.5
70 = 4b/3 x 3.5
70 = 4.6667 b
b = 15 km/h
Therefore, the speed of the boat in still water is 15 km/h, and the speed of the current is b/9 = 1.67 km/h.
Thank you for catching my mistake.
I do not think that (b + b/9) = 4 b/3
rather
10 b/9 = 5 b/3
You are correct again. I apologize for my mistake.
Using the information we have so far, we can solve for the speed of the current as c = b/9.
Then, we can use the information about traveling with the current to find b:
distance = rate x time
70 = (b + c) x 3.5
70 = (b + b/9) x 3.5
70 = (10b/9) x 3.5
70 = 3.8889b
b = 18 km/h
Therefore, the speed of the boat in still water is 18 km/h, and the speed of the current is b/9 = 2 km/h.
Thank you for your patience and for pointing out my mistakes.
Whew !!! We agree
Indeed! Thank you for your diligence in making sure we reached the correct solution.
Let's denote the speed of the boat in still water as "b", and the speed of the current as "c".
When the boat travels from one pier to another pier, it's traveling against the current on the way there, so its effective speed is b-c. On the way back, it's traveling with the current, so its effective speed is b+c.
We know that the distance from one pier to another pier is the same on the way there and back, so we can set up the equation:
distance = rate x time
d = (b - c) x 4 (on the way there)
d = (b + c) x 5 (on the way back)
Since we're looking for the speed of the boat in still water (b), we want to eliminate the variable c. One way to do that is to solve for c in one of the equations, and then substitute it into the other equation. Let's solve for c in the second equation:
d = (b + c) x 5
d/5 = b + c
c = d/5 - b
Now we can substitute that expression for c into the first equation:
d = (b - c) x 4
d = (b - (d/5 - b)) x 4
d = (5b - d/5) x 4
d = 20b - 4d/5
d + 4d/5 = 20b
d(1 + 4/5) = 20b
d = 9b/2
So we've found an expression for the distance in terms of b:
d = 9b/2
Now we can use the information about traveling with the current to solve for b:
distance = rate x time
70 = (b + c) x 3.5
70 = (b + d/5 - b) x 3.5
70 = d/5 x 3.5
d = 200
Substituting that into our expression for distance:
9b/2 = 200
b = 44.4 km/h
So the speed of the boat in still water is 44.4 km/h.