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dURING THE FIRST PART OF A TRIP, A CONOEIST TRAVELS53 MILES AT A CERTAIN SPEED . The canoeist travels 9 miles on the second par...Asked by Anonymous
during the first part of a trip a canoeist travels 64 miles at a certain speed. the canoeist travels 11 miles on the second part of the trip at a 5 mph slower. the total time for the trip is 2 hrs. what speed on each part of the trip
Answers
Answered by
Terry
Given:
d1 = 64
d2 = 11
t(total) = 2
s1 = ?
s1 = s2 + 5
Ask yourself, what t really is?
t(total) = d1/s1 + d2/s2
substituting s1 as s2 + 5:
2 = 64/(s2+5) + 11/(s2)
Common denominator: multiply s(s+5)/s(s+5)and you eventually get:
2s^2 - 65s - 55 = 0
Same old quadratic formula:
s2 = 33.325 mph
s1 = s2 + 5 = 38.325 mph
Plug the # back in the original equation... 2 = 2, checked!
d1 = 64
d2 = 11
t(total) = 2
s1 = ?
s1 = s2 + 5
Ask yourself, what t really is?
t(total) = d1/s1 + d2/s2
substituting s1 as s2 + 5:
2 = 64/(s2+5) + 11/(s2)
Common denominator: multiply s(s+5)/s(s+5)and you eventually get:
2s^2 - 65s - 55 = 0
Same old quadratic formula:
s2 = 33.325 mph
s1 = s2 + 5 = 38.325 mph
Plug the # back in the original equation... 2 = 2, checked!
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