Asked by Scott
The boat travels 60 miles to an island and the same 60 miles coming back. Changes in the wind and tide made the average speed on the return trip 3 mph slower than the speed on the way out. If the total time of the trip took 9 hours, find the speed going to the island and the speed on the return trip.
Answers
Answered by
Damon
return speed = v - 3
time out = 60/v
time back = 60/(v-3)
so
60/v + 60/(v-3) = 9
60(v-3) + 60 v = 9 v(v-3)
9 v^2 -27v = 120 v - 180
9 v^2 - 147 v + 180 = 0
3 v^2 - 49 v + 60 = 0
(3 v -4)(v-15) = 0
v = 15 or v = 4/3
4/3 will not work because v - 3 is negative
try v = 15
15 and 15-3 = 12
then
60/v = 4
60/12 = 5
sure enough 9 hours
time out = 60/v
time back = 60/(v-3)
so
60/v + 60/(v-3) = 9
60(v-3) + 60 v = 9 v(v-3)
9 v^2 -27v = 120 v - 180
9 v^2 - 147 v + 180 = 0
3 v^2 - 49 v + 60 = 0
(3 v -4)(v-15) = 0
v = 15 or v = 4/3
4/3 will not work because v - 3 is negative
try v = 15
15 and 15-3 = 12
then
60/v = 4
60/12 = 5
sure enough 9 hours
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