SOLUTION: A man in a rowboat that is a = 4 miles from the nearest point A on a straight shoreline wishes to reach a house located at a point B that is b = 10 miles farther down the shoreline
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-> SOLUTION: A man in a rowboat that is a = 4 miles from the nearest point A on a straight shoreline wishes to reach a house located at a point B that is b = 10 miles farther down the shoreline
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Question 1145867: A man in a rowboat that is a = 4 miles from the nearest point A on a straight shoreline wishes to reach a house located at a point B that is b = 10 miles farther down the shoreline. He plans to row to a point P that is between A and B and is x miles from the house, and then he will walk the remainder of the distance.
Suppose he can row at a rate of 3 mi/hr and can walk at a rate of 5 mi/hr. If T is the total time required to reach the house, express T as a function of x.
the rowboat is at point R.
the shorted distance point on the shore if point A.
the hours is at point B.
the distance from the rowboat to point A is 4 miles which is the length of line RA.
the distance from point A to the house at point B is 10 miles which is the length of line AB
the distance from point P to the house at point B is x miles which is the length of line PB.
this makes the distance from point A to point P equal to (10 - x).
the distance from the rowboat to point P is equal to the hypotenuse of the right triangle RAP which is the line RP.
that makes the distance equal to sqrt(4^2 + (10-x)^2).
simplify that to get distance from point A to P equal to sqrt(x^2-20x+116).
the basic formula to use is r * t = d
r is the rate.
t is the time.
d is the distance.
the man rows the boat at 3 miles per hour.
the man walks at 5 miles per hour.
let T1 equal the time it takes to get from point R to point P rowing the boat.
let D1 equal the distance from point R to point P.
you get r * t = d becomes 3 * T1 = D1
since D1 = sqrt(x^2-20x+116), you get 3 * T1 = sqrt(x^2-20x+116)
solve for T1 to get T1 = sqrt(x^2-20x+116)/3
let T2 equal the time it takes to get from poinnt P to point B walking.
let D2 equal the distance from point P to point B.
you get r * t = d becomes 5 * T2 = D2
since D2 = x, you get 5 * T2 = x
solve for T2 to get T2 = x/5
let the total time it takes to get from point R to point B through point P equal to T.
you get T = T1 + T2 = sqrt(x^2-20x+116)/3 + x/5
that's your solution.
you can combine that equation into one common denominator to get:
T = (5*sqrt(x^2-20x+116) + 3x)/15
the two equations of T, shown above, are equivalent.
they produce the same value for T.
the domain of the T function is 0 < x < 10, where x represents the distance from point P to point B.
x cannot be 0 because then point P would be the same point as point B.
x cannot be 10 because then point P would be the same point as point A.
that's because the problem statement says that point P is between point A and point B and can therefore not be at either of them.
either of the equations involving T will be your solution, depending on how you want to show them.
