Car A travels 120 miles in the same time that car B travels 150 miles. If car B averages 10 mph faster than car A, what is the s
1 answer:
Answer:
Speed of car A = 40 mph
Speed of car B = 50 mph
Step-by-step explanation:
Given:
Distance travelled by car A = 120 miles
Distance travelled by car B = 120 miles
To Find:
speed of each car = ?
Solution:
Let the speed of car A be x
then speed of car B is (x +10)
The Time taken for each car is same
Time taken for car A = Time taken for car B
We know that time =
Time taken for car A
=> ---------------------------(1)
Similarly
Time taken for car B
=> -----------------------(2)
Equating (1) and (2), we get
=
120x + 1200 = 150x
1200 = 150x-120 x
1200 = 30x
x= 40
Speed of car A = 40mph
Speed of car B = (x+10) = (x+40) = 50mph
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Answer:
7.85 km
Step-by-step explanation:
Let x represent the distance from the refinery to point P. Then the distance under the river from point P to the storage tanks is ...
√(2² +(9 -x)²) = √(x² -18x +85)
In units of $200,000, the cost of the pipeline will be ...
c = x + 2√(x² -18x +85)
The derivative with respect to x is ...
dc/dx = 1 +(2x -18)/√(x² -18x +85)
When we set this to zero, we can get the equation ...
√(x² -18x +85) +(2x -18) = 0
And this can be rewritten as ...
3x^2 -54x +239 = 0
which has solution
x = 9 -(2/3)√3 ≈ 7.8453
Point P should be located about 7.85 km from the refinery.
_____
It is interesting to note that the angle the pipe makes with the riverbank is given by arccos(1/2), where the 1/2 is the ratio of overland to under-river costs. That angle is 60°, so the distance along the riverbank from the storage facility to point P is 2cot(60°) = (2/3)√3 ≈ 1.1547 km. You will recognize this as the value subtracted from 9 km in the solution above.
This "angle solution" is the generic solution to this sort of problem where the route costs are different and part of the route is along the edge of the higher-cost path.
