A boat travels down the river in 2.9 hours. The return trip up river takes 3.6 hours. A one way trip down or up river is a distance of 30 miles. If the boat is traveling at maximum speed, what is the boats speed measured in waters with no current? You need to use systems to solve this problem. Round your speed to the tenth of a mile per hour.
2.9Vb + 2.9Vc = 30
0.6Vb - 0.6Vc = 5
1.74Vb + 1.74Vc = 18
1.74Vb - 1.74Vc = 14.5
3.48Vb = 32.5
Vb = 9.3 mi/h = Velocity of boat.
To solve this problem, we can set up a system of equations using the concept of relative velocity.
Let's define the speed of the boat as "b" and the speed of the river current as "r". Then, the speed of the boat moving downstream (with the current) would be "b + r", and the speed of the boat moving upstream (against the current) would be "b - r".
We can use the formula time = distance / speed to set up the following equations:
Downstream trip: 2.9 = 30 / (b + r)
Upstream trip: 3.6 = 30 / (b - r)
We can simplify the equations by cross-multiplying:
2.9(b + r) = 30
3.6(b - r) = 30
Now, we can solve this system of equations.
Multiplying the first equation by 10 to eliminate the decimal:
29(b + r) = 300
Expanding the equation:
29b + 29r = 300
Likewise, multiplying the second equation by 10:
36(b - r) = 300
Expanding the equation:
36b - 36r = 300
Now we have a system of equations:
29b + 29r = 300
36b - 36r = 300
We can solve this system of linear equations by using the method of elimination.
First, let's multiply the first equation by 36 and the second equation by 29 to eliminate the "r" variable:
(36)(29b + 29r) = (36)(300)
(29)(36b - 36r) = (29)(300)
Expanding:
1044b + 1044r = 10800
1044b - 1044r = 8700
Adding the equations together, the "r" term cancels out:
2088b = 19500
Dividing both sides by 2088:
b = 9.32
Therefore, the boat's speed in still water, without any current, is approximately 9.3 miles per hour (rounded to the tenth of a mile per hour).