Chinook salmon are able to move through water especially fast by jumping out of the water periodically. This behavior is called porpoising. Suppose a salmon swimming in still water jumps out of the water with velocity 6.41 m/s at 44.1° above the horizontal, sails through the air a distance L before returning to the water, and then swims the same distance L underwater in a straight, horizontal line with velocity 3.91 m/s before jumping out again.
(a) Determine the average velocity of the fish for the entire process of jumping and swimming underwater.
(b) Consider the time interval required to travel the entire distance of 2L. By what percentage is this time interval reduced by the jumping/swimming process compared with simply swimming underwater at 3.91 m/s?
For (a), I got 5.16 m/s.
I have no idea how to do (b).
uhyi
To solve part (b), we need to compare the time it takes for the salmon to cover a distance of 2L using only swimming underwater at a speed of 3.91 m/s with the time it takes when including the jumps and swimming underwater.
Let's break down the process:
1. Swimming in the air:
The salmon jumps out of the water with a velocity of 6.41 m/s. It travels a distance L before returning to the water. We can find the time it takes for this air travel using the equation of motion:
L = Vx * t_air ...(1)
where Vx is the horizontal component of the velocity (Vx = V * cosθ) and t_air is the time in the air.
Plugging in the values, we have:
L = (6.41 m/s) * cos(44.1°) * t_air
2. Swimming underwater:
After it returns to the water, the salmon swims underwater with a velocity of 3.91 m/s for the same distance L. The time it takes for this underwater travel is given by:
L = Vx * t_underwater ...(2)
where Vx is the horizontal component of the velocity (Vx = 3.91 m/s) and t_underwater is the time underwater.
From equation (2), we can solve for t_underwater:
t_underwater = L / Vx
3. Total time for 2L:
The total time for the salmon to cover a distance of 2L, including both air and underwater travel, is given by:
t_total = t_air + t_underwater + t_air ...(3)
Substituting equations (1) and (2) into equation (3), we get:
t_total = (L / (V * cosθ)) + (L / Vx) + (L / (V * cosθ))
Simplifying further, we have:
t_total = (2L / (V * cosθ)) + (L / Vx)
Now, let's compare the percentage reduction in time using the jumping/swimming process compared to simply swimming underwater:
t_underwater_only = (2L) / (3.91 m/s)
The percentage reduction in time is given by:
Percentage reduction = ((t_underwater_only - t_total) / t_underwater_only) * 100
Plug in the values to calculate the percentage reduction.