Solving Boat Speed & Distance with Drag Force

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The discussion centers on deriving the speed v(t) of a boat experiencing drag force F=-αeβv. The user expresses confusion about the relationship between instantaneous velocity and the drag force, which complicates their calculations. They initially found v(t) and calculated the time and distance until the boat stops but later realized a critical error in their interpretation of the drag equation. The importance of correctly rearranging the equation to separate variables for integration is emphasized, as well as the need for clarity in mathematical representations. The user acknowledges the necessity to correct their calculations due to misinterpretation of the drag force's exponential term.
Macykc2
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Homework Statement


A boat with initial speed vo is launched on a lake. The Boat is slowed by the water by a force F=-αeβv. Find an expression for the speed v(t), and find the time and distance for the boat to stop.

Homework Equations


Drag force F=-αeβv
alpha and beta are not specified as to what they represent.

The Attempt at a Solution


I have an attached attempted solution, I'm just kind of confused on my results, I feel like it makes sense but at the same time it doesn't. My confusion lies in that I try to imagine using my derived v(t) equation but the drag force is constantly changed based on the instantaneous velocity at any moment, the v in the force equation is what throws me off, and makes me think I need to use a different approach.
Once I found my v(t) I found time when v(t)=0, then I found velocity as a function of position to find distance when v=0.
But again, now that I look at it, the v that is apart of e should probably be 0 as well, then the positions and times won't be affected by the drag force...
 

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Macykc2 said:

Homework Statement


A boat with initial speed vo is launched on a lake. The Boat is slowed by the water by a force F=-αeβv. Find an expression for the speed v(t), and find the time and distance for the boat to stop.

Homework Equations


Drag force F=-αeβv
alpha and beta are not specified as to what they represent.

The Attempt at a Solution


I have an attached attempted solution, I'm just kind of confused on my results, I feel like it makes sense but at the same time it doesn't. My confusion lies in that I try to imagine using my derived v(t) equation but the drag force is constantly changed based on the instantaneous velocity at any moment, the v in the force equation is what throws me off, and makes me think I need to use a different approach.
Once I found my v(t) I found time when v(t)=0, then I found velocity as a function of position to find distance when v=0.
But again, now that I look at it, the v that is apart of e should probably be 0 as well, then the positions and times won't be affected by the drag force...
Posting working as an image makes it hard to comment on individual lines. (It can also be hard to read, but yours is ok.)
Better to take the trouble to type the equations in, but if you must post images please number all the equations.

You cannot integrate a function of v just by multiplying it by t. v is a variable.
Rearrange the equation so that all the references to v are on one side and t on the other (including dv, dt).
 
I figured that's what i had to do, I took a look at it after I posted and tried some other things including that, and although I do get results it's once again hard to see if it makes sense, as they do not specify what alpha and beta are.

Also I will try in the future to just use this site instead of a picture, I'm just not good with that math function as I've never used it.

I ended up with v(t)=##\frac{1}{β}##ln(eβvo - ##\frac{αβt}{m})## but as I said, it's hard to know if what I did is right with the lack of info.
 
I also just noticed a very crucial mistake, this whole time I've been writing eβv as e-βv, so I need to re-do everything...
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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