Question 454466
with the current, which is 2 mph, i can travel 48 miles in my boat.
Against the current, it takes 2 more hours to travel the same 48 miles.
what is the speed of my boat in still water?
:
Let s = boat speed in still water
then
(s+2) = effective speed with the current
and
(s-2) = effective speed against the current
:
Write a time equation time = distance/speed
:
Time downstr + 2 hrs = time upstr
{{{48/((s+2))}}} + 2 = {{{48/((s-2))}}}
:
multiply by (s+2)(s-2), results
48(s-2) + 2(s+2)(s-2) = 48(s+2)
48s - 96 + 2(s^2-4) = 48s + 96
:
Subtract 48s from both sides
2s^2 - 8 = 96 + 96
2s^2 = 192 + 8
2s^2 = 200
s^2 = {{{200/2}}}
s^2 = 100
s = {{{sqrt(100)}}}
s = 10 mph is speed in still water
:
:
Check solution, find the times for each trip
48/8 = 6 hr
48/12= 4 hrs
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diff: 2 hrs