Relative Velocities: Dolphin Swims Home - (a) _______° N of W, (b) ________min

  • Thread starter Jtappan
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In summary: The angle between \vec{v_w} and \vec{v_c} is 45 degrees, so the dolphin needs to swim in the direction of \vec{v_w} and away from the current.
  • #1
Jtappan
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Homework Statement



A dolphin wants to swim directly back to its home bay, which is 0.79 km due west. It can swim at a speed of 4.14 m/s relative to the water, but a uniform water current flows with speed 2.93 m/s in the southeast direction.

(a) What direction should the dolphin head?
_______° N of W
(b) How long does it take the dolphin to swim the 0.79-km distance home?
________min


Homework Equations



kinematics equations

The Attempt at a Solution

 
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  • #2
Jtappan said:

Homework Statement



A dolphin wants to swim directly back to its home bay, which is 0.79 km due west. It can swim at a speed of 4.14 m/s relative to the water, but a uniform water current flows with speed 2.93 m/s in the southeast direction.

(a) What direction should the dolphin head?
_______° N of W
(b) How long does it take the dolphin to swim the 0.79-km distance home?
________min


Homework Equations



kinematics equations

The Attempt at a Solution



Welcome to the forums.

This sum has to be done by vectors or relative velocity.

And as you are new to the forums i would like to remind you that every homework thread should also carry the attempts made by the person in solving that problem. We are here to help you in your homework not to do your homework.
 
  • #3
Sorry

First I have been trying to find all of my variables in both directions. The angle should be 45 degrees for the 2.93m/s in the south east directions, I am pretty sure. Then I am just completely lost after that. If it is a 45 degree angle then you do 2.93sin45 and get 2.07 roughly. This is where it loses me because there are two different velocities for the Dolphin relative to the water.
 
  • #4
Be frank

Jtappan said:
First I have been trying to find all of my variables in both directions. The angle should be 45 degrees for the 2.93m/s in the south east directions, I am pretty sure. Then I am just completely lost after that. If it is a 45 degree angle then you do 2.93sin45 and get 2.07 roughly. This is where it loses me because there are two different velocities for the Dolphin relative to the water.

Try do it with relative velocities.Here

[tex]\vec{v}_{dg}[/tex] = [tex]\vec{v}_{dr}[/tex] - [tex]\vec{v}_{gr}[/tex]

Hence [tex]\vec{v}_{dg}[/tex] = [tex]\vec{v}_{dr}[/tex] + [tex]\vec{v}_{rg}[/tex]

Now here [tex]\vec{v}_{dg}[/tex] is the velocity of the dolphin wrt the ground.

[tex]\vec{v}_{dr}[/tex] is the velocity of the dolphin wrt the river.

[tex]\vec{v}_{rg}[/tex] is the velocity of the river wrt the ground.
 
  • #5
A question

i have a question :) can u solve it :)


find the equation of circle which passes through the point (-2,-4) and has the same center as the circle whos equation is x^2+y^2-4x-6y-23=0 ??
 
  • #6
worldisonline said:
i have a question :) can u solve it :)


find the equation of circle which passes through the point (-2,-4) and has the same center as the circle whos equation is x^2+y^2-4x-6y-23=0 ??

worldisonline, you should start your own thread in the Precalculus section of the Homework forums. Also, show us how you approached the problem, and where you got stuck.
 
  • #7
I think that you are a bit weak at relative velocities.

Lets start it again, I have inserted the diagram have a look at it.Here first consider the x components and you ewill that

Vrgcos(45) + Vdrcos(90) = Vdgcos(A+45)
Hence Vrgcos45=Vdgcos(A+45)

Now consider the y components

Vrgsin45+Vdrsin90 = Vdgsin(45+A)

Vrg=2.93 and Vdr=4.14 m/s.

Now solve the two.

Sorry learning and doc but i had to give the whole method. I was getting PM from the OP and believe me he was completely at sea.
 
  • #8
No prob Fedex. But I'm confused by your method... Why are you using 90 degrees and A+45?
 
  • #9
learningphysics said:
No prob Fedex. But I'm confused by your method... Why are you using 90 degrees and A+45?

Forgot to attach the diagram.Sorry.
 

Attachments

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  • #10
The resultant motion of the dolphin needs to be to the west, let's call it [tex]\vec{v_w}[/tex]. This motion is the combined motion of the current, let's call it [tex]\vec{v_c}[/tex], and the direction in which the dolphin needs to swim, let's call it [tex]\vec{v_d}[/tex]. These three vectors will form a closed triangle with the resultant of the other two being [tex]\vec{v_w}[/tex].
 

Attachments

  • dolphin.gif
    dolphin.gif
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Last edited:

What is the concept of relative velocities?

The concept of relative velocities is based on the idea that the velocity of an object or a point is relative to another object or point. It refers to the motion of one object in relation to another object, taking into account their respective velocities and directions.

How do you calculate relative velocities?

To calculate relative velocities, you need to first determine the velocities of the two objects or points in question. Then, you can use vector subtraction to find the difference in their velocities. This will give you the relative velocity of one object with respect to the other.

What are the units of relative velocity?

The units of relative velocity are the same as that of regular velocity, which is distance per unit time. This can be in meters per second, miles per hour, or any other unit of length divided by time.

In which direction is the relative velocity of the dolphin swimming home?

The direction of the relative velocity of the dolphin swimming home can be determined by looking at the direction of the dolphin's velocity and the direction it is swimming in relation to due west. For example, if the dolphin's velocity is pointing towards the northwest and it is swimming towards the west, the relative velocity can be described as (a) 45° N of W.

How long will it take the dolphin to swim home?

The time it takes for the dolphin to swim home can be calculated by dividing the distance between the dolphin's current position and its home by its velocity. This will give you the time in seconds, which can be converted to minutes if needed. Keep in mind that this calculation assumes a constant velocity and does not take into account factors such as fatigue or changes in direction.

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