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Chinook Salmon Swimming Pattern

  • Thread starter DatAshe
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  • #1
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This question has me puzzled, I only have a little bit of a clue of how to do this. Any and all help is greatly appreciated:)

Homework Statement


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 5.87 m/s at 46.5° 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 3.67 m/s before jumping out again.
(a) Determine the average velocity of the fish for the entire process of jumping and swimming underwater.
___________m/s

(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 3.67 m/s?
_________%


Homework Equations



V_avg =( x_1+x_2 )/ 2 , ΔX=V_x*t, Δy=V_oy+(1/2)a_y(t^2)

The Attempt at a Solution


A) I got 5.86 m/s by finding the velocities of both parts (8.08m/s for the jump, 3.67m/s for the underwater part), adding them together and dividing by two, to get an average.
B) I got 45.5% by finding.....looking at my notes I have no clue how I got that. I apologize, but I have no clue what I did, and no clue how to do it correctly. Sorry for being a bum on this part:(
 

Answers and Replies

  • #2
PeterO
Homework Helper
2,425
46
This question has me puzzled, I only have a little bit of a clue of how to do this. Any and all help is greatly appreciated:)

Homework Statement


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 5.87 m/s at 46.5° 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 3.67 m/s before jumping out again.
(a) Determine the average velocity of the fish for the entire process of jumping and swimming underwater.
___________m/s

(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 3.67 m/s?
_________%


Homework Equations



V_avg =( x_1+x_2 )/ 2 , ΔX=V_x*t, Δy=V_oy+(1/2)a_y(t^2)

The Attempt at a Solution


A) I got 5.86 m/s by finding the velocities of both parts (8.08m/s for the jump, 3.67m/s for the underwater part), adding them together and dividing by two, to get an average.
B) I got 45.5% by finding.....looking at my notes I have no clue how I got that. I apologize, but I have no clue what I did, and no clue how to do it correctly. Sorry for being a bum on this part:(
Given that it left the water with a speed of 5.67 m/s, I am not sure how you calculated a velocity of 8.08 m/s ????

Also, the way you averaged the velocities is concerning.

SUppose the "jump phase" took 1 second, while the swim phase took 1 minute, the average velocity would be much closed to the swim velocity.

Average velocity is total displacement divided by total time. You would need to calculate the displacement and time for each part of the journey.
 

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