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?
To find Jessie's speed, we can set up a distance equation based on the given information.
Let's say Jessie's speed is represented by "S mph". Since she paddled 20 miles upstream, against the current, her effective speed would be (S - 3) mph. Similarly, when she paddled back downstream, her effective speed would be (S + 3) mph.
Now, we can plug these values into the distance equation:
Distance = Speed x Time
The distance covered upstream is 20 miles, and the time taken is (20 / (S - 3)) hours.
The distance covered downstream is also 20 miles, and the time taken is (20 / (S + 3)) hours.
The total trip took 7 hours, so we can create the equation:
(20 / (S - 3)) + (20 / (S + 3)) = 7
Now, let's solve this equation to find Jessie's speed (S).
[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
d = V*T = 40.
V*7 = 40,
V = 5.71 mi/h.
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.