A new piece of laboratory equipment costing $10,000 promises to save $4000 per year in materials and overtime pay. If the cost o
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By solving a system of equations we will find that the rate of current in the stream is S = 2 mi/h.
When the motorboat travels downstream, the total velocity will be the velocity of the motorboat in still water plus the velocity of the stream, while if the motorboat travels upstream, we have the velocity of the stream subtracted.
So upstream the speed is:
(14 mi/h - S)
Downstream the speed is:
(14 mi/h + S)
Where S is the rate of current in the stream.
We know that downstream it takes 15 minutes more to travel 12 miles, then we can write the system of equations:
(14 mi/h + S)*T = 12 mi
(14 mi/h - S)*(T - 15 min) = 12 mi
To solve this, we need to isolate one of the variables in one of the equations, I will isolate T in the first one:
T = (12 mi)/(14 mi/h + S)
Replacing that in the other equation we will get:
(14 mi/h - S)*((12 mi)/(14 mi/h + S) - 15 min) = 12 mi
Now we can solve this for S. Now we can multiply both sides by (14 mi/h + S).
(14 mi/h - S)*12 mi - (14 mi/h + S)*(14 mi/h - S)*(- 15 min) = 12 mi*(14 mi/h + S)
Also notice that the speeds are in hours, so we can rewrite:
- 15 min = -0.25 h
(14 mi/h - S)*12 mi - (14 mi/h + S)*(14 mi/h - S)*(- 0.25 h) = 12 mi*(14 mi/h + S)
168 mi^2/h - 12mi*S + 49mi^2/h + 0.25h*S^2 = 168mi^2/h + 12mi*S
- 12mi*S + 49mi^2/h - 0.25h*S^2 = 12mi*S
-24mi*S - 0.25h*S^2 + 49mi^2/h = 0
This is a quadratic equation, the solutions are:
We only take the positive solution, so we get:
S = (24 mi - 25 mi)/(-0.5 mi) = 2 mi/h
The rate of current in the stream is 2 mi/h.
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