Asked by Jonah
Jessie paddled her canoe 20 miles upstream, then paddled back. If the speed of the current was
3 mph and the total trip took 7 hours, what was Jessie’s speed?
3 mph and the total trip took 7 hours, what was Jessie’s speed?
Answers
Answered by
scott
[20 / (j - 3)] + [20 / (j + 3)] = 7
20 j + 60 + 20 j - 60 = 7(j - 3)(j + 3) = 7 j^2 - 63
0 = 7 j^2 - 40 j - 63
use quadratic formula to find j
20 j + 60 + 20 j - 60 = 7(j - 3)(j + 3) = 7 j^2 - 63
0 = 7 j^2 - 40 j - 63
use quadratic formula to find j
Answered by
Henry
d = V*T = 40.
V*7 = 40,
V = 5.71 mi/h.
V*7 = 40,
V = 5.71 mi/h.
Answered by
Henry
Do you mean "What was Jessica's speed in still air"? If so:
Upstream: (V-3)*T1 = 40.
Eq1: T1V - 3T1 = 40.
Downstream: (V+3)*T2 = 40.
Eq2: T2V + 3T2 = 40.
Add Eq1 and Eq2:
T1V - 3T1 = 40
T2V + 3T2 = 40
Sum: T1V + T2V = 80.
V(T1+T2) = 80.
T1 + T2 = 7 Hours.
7V = 80,
V = 11.43 mi/h = Velocity in still air.
Upstream: (V-3)*T1 = 40.
Eq1: T1V - 3T1 = 40.
Downstream: (V+3)*T2 = 40.
Eq2: T2V + 3T2 = 40.
Add Eq1 and Eq2:
T1V - 3T1 = 40
T2V + 3T2 = 40
Sum: T1V + T2V = 80.
V(T1+T2) = 80.
T1 + T2 = 7 Hours.
7V = 80,
V = 11.43 mi/h = Velocity in still air.
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