Dayle is paddling his canoe upstream, against the current, to a fishing spot 10 miles always. If he paddles upstream for 2.5 hours and his return trip takes 1.25 hours, find the speed of the current, and his still water????????

Well, it seems like Dayle had quite the adventure! Let's see if we can figure out the speed of the current and his still water.

Now, during the upstream trip, Dayle was paddling against the current. So, we'll call the speed of the current "C" and his still water speed "S".

The distance Dayle paddled upstream is 10 miles, and it took him 2.5 hours. Therefore, his effective speed (against the current) can be calculated as 10 miles divided by 2.5 hours, which is 4 miles per hour.

During the return trip, Dayle was paddling with the current. So, his effective speed (with the current) can be calculated as 10 miles divided by 1.25 hours, which is 8 miles per hour.

Now, since his speed against the current is 4 miles per hour, and his speed with the current is 8 miles per hour, we can deduce that the speed of the current (C) is the difference between these two speeds, which is 8 - 4 = 4 miles per hour.

As for the still water speed (S), it's the average of the speed against the current and the speed with the current, which is (4 + 8) divided by 2 = 6 miles per hour.

So, after all the calculations and paddling, we can conclude that the speed of the current is 4 miles per hour, and Dayle's still water speed is 6 miles per hour. Enjoy your fishing trip, Dayle, and don't forget your water wings!

To find the speed of the current and the speed of the canoe in still water, we can solve the problem using the formula:

distance = speed × time.

Let's assume the speed of the canoe in still water is represented by "c" (in miles per hour), and the speed of the current is represented by "s" (in miles per hour).

1. Paddling upstream:
When paddling against the current, the canoe's speed is reduced by the speed of the current.
So, the effective speed of the canoe is (c - s).
The time taken to paddle upstream is 2.5 hours.
Therefore, distance = (c - s) × 2.5 miles.

2. Paddling downstream:
When paddling with the current, the canoe's speed is increased by the speed of the current.
So, the effective speed of the canoe is (c + s).
The time taken to paddle downstream is 1.25 hours.
Therefore, distance = (c + s) × 1.25 miles.

Given that the distance both upstream and downstream is 10 miles, we can set up the following equations:

1. (c - s) × 2.5 = 10
2. (c + s) × 1.25 = 10

Let's solve these equations to find the speed of the current (s) and the speed of the canoe in still water (c):

From equation 1:
(c - s) × 2.5 = 10
c - s = 10/2.5
c - s = 4

From equation 2:
(c + s) × 1.25 = 10
c + s = 10/1.25
c + s = 8

Now, we have a system of equations:
c - s = 4 (Equation 1)
c + s = 8 (Equation 2)

Adding the two equations together, we get:
2c = 12
c = 12/2
c = 6

Substituting the value of c into Equation 1, we can find s:
6 - s = 4
-s = 4 - 6
-s = -2
s = 2

Therefore, the speed of the current is 2 miles per hour, and the speed of the canoe in still water is 6 miles per hour.

To find the speed of the current and the canoe's speed in still water, we can use the concept of relative velocity.

Let's denote the speed of the canoe in still water as 'x' and the speed of the current as 'y'.

When Dayle paddles upstream (against the current), he has to overcome the force of the current. So, his effective speed is the difference between his speed in still water and the speed of the current: x - y.

On the return trip downstream (with the current), the current assists Dayle, so his effective speed is the sum of his speed in still water and the speed of the current: x + y.

We know that the distance from the fishing spot to Dayle's starting point is 10 miles. In 2.5 hours, while paddling upstream, Dayle covers 10 miles. So, we can set up the following equation:

2.5(x - y) = 10

Similarly, during the return trip, he covers the same 10 miles in 1.25 hours:

1.25(x + y) = 10

Now, we have a system of equations that we can solve simultaneously to find the values of 'x' and 'y'.

Let's solve the system of equations using the method of substitution:

From the first equation, we can isolate 'x - y' and express it in terms of 'x':

x - y = 10/2.5
x - y = 4 (Equation 1)

From the second equation, we can isolate 'x + y' and express it in terms of 'x':

x + y = 10/1.25
x + y = 8 (Equation 2)

Now, we can solve the system of equations:

Adding Equation 1 and Equation 2, we get:

2x = 4 + 8
2x = 12
x = 12/2
x = 6

Substituting the value of 'x' back into Equation 2:

6 + y = 8
y = 8 - 6
y = 2

Therefore, the speed of the current is 2 mph, and the canoe's speed in still water is 6 mph.

V = his speed in still water

v = the speed of the current

10/(V-v) = 2.5 and 10/(V+v) = 1.25

SOlve each for V and equate the results yields (10 + 2.5v)/2.5 = (10 - 1.25)/1.25.

Solve for v and then V.