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he rows 6 miles downstream in 1 hour.
the bomb is timed to go off in 3 hours.
he just barely makes it back upstream 6 miles to defuse the bomb.
if it takes him 3 hours to get back upstream, he will be there just in the nick of time to defuse the bomb.
that's the assumption here.
it doesn't take into account the time to get out of the boat, find the bomb, and defuse the bomb.
if that had to be taken into account, then the problem would have to make assumptions that were not presented to you.
since each person doing the problem might make different assumptions, then each person's answer would be different.
i will assume that 0 time is required to defuse the bomb once he gets back upstream to the starting point.
under that assumption, this is what happens.
it took him 1 hour to get downstream.
it takes him 3 hours to get upstream.
obviously the stream is going in the downstream direction.
his rate of speed going downstream is:
x + y
his rate of speed going upstream is:
x - y
x is the speed of the boat.
y is the speed of the current.
the equation to use is rate * time = distance
x + y = the rate going downstream.
time = 1 hour going downstream.
distance = 6 miles going downstream.
x - y = the rate going upstream.
time = 3 hours going upstream.
distance = 6 miles going upstream (same distance as going downstream).
the 2 equations we have to work with are:
(x + y) * 1 = 6
(x - y) * 3 = 6
since both equations are equal to 6, we can set them equal to each other to get:
(x + y) * 1 = (x - y) * 3
we remove parentheses to get:
x + y = 3x - 3y
we subtract x and y from both sides of the equation to get:
0 = 2x - 4y
we add 4y to both sides of the equation to get:
4y = 2x
we divide both sides of the equation by 2 to get:
2y = x
x = speed of the boat.
y = speed of the current.
the speed of the boat is 2 times the speed of the current.
we go back to our original equations and replace x with 2y.
(x + y) * 1 = 6 becomes:
(2y + y) * 1 = 6 which becomes:
3y * 1 = 6 which becomes:
3y = 6 which becomes:
y = 2.
(x - y) * 3 = 6 becomes:
(2y - y) * 3 = 6 which becomes:
y * 3 = 6 which becomes:
y = 2
we get y = 2 miles per hour in each equation.
if y = 2, and x = 2y, then x has to be equal to 4 miles per hour.
we plug those values for x and y in the original equations again to get:
(x + y) * 1 = 6 becomes (4 + 2) * 1 = 6 which becomes 6 = 6
(x - y) * 3 = 6 becomes (4 - 2) * 3 = 6 which becomes 2 * 3 = 6 which becomes 6 = 6.
the values for x and y are good because they satisfy the requirements of both equations simultaneously.
the answer to the problem is:
the speed of the boat in still water is 4 miles per hour.
the speed of the current is 2 miles per hour.
zero amount of time is allotted for getting out of the boat and defusing the bomb. the assumption is that defusing the bomb occurs instantly as soon as he gets back to the original location where he started from.