Rectilinear motion, not uniform accleration

In summary, the conversation discusses a problem involving differentiating and integrating the equation F=-b exp(αv) to express v(t) or a(t) as a function of time. The conversation also discusses using the formula vdv/dx=-b/m e^αv to solve the problem and suggests dividing by e^αv and integrating with respect to t. The conversation concludes with the user successfully solving the problem.
  • #1
Terrycho
20
2
Homework Statement
A boat of mass m with initial velocity v_o is slowed by a frictional force
F=-bexp(αv), α and b are positive constants .
Relevant Equations
F=-bexp(αv)
241315


The photo above is the problem I'm struggling now and the photo below is what I have done so far.

241316


By differentiating and integrating the given equation, which is F=-b exp(αv), I tried to express v(t) or a(t) as a function related to t, which is time.

But the more I did, the more complicated it got...

Force is related to the velocity so when you differentiate it, the acceleration function about t comes out so you cannot just express its motion.

How was I supposed to approach this problem to?

Look forward to hearing from you guys.
Thanks a lot!

If this posts makes any inconvenience or you offended, I will delete it right away.

No matter how long comments are, any comments would be appreciated!
 
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  • #2
Hello terycho, ##\qquad## :welcome: ##\qquad## !

Do us a favor and read the guidelines on posting photos. If you promise to type instead of snap I'll try to help:

Don't understand your first step: you multiply with ##e^{-v(0)}\ ## on one side only: no can do.

Your second step doesn't help you, as you conclude yourself.

If you rewrite ##\ a(t) = - {b\over m} e^{\alpha v} \ ## as ##\ {dv\over dt} = - {b\over m} e^{\alpha v} \ ## does it look a bit more familiar ?
Do you know how to solve this kind of differential equation ?
(hit: think chain rule for differentiation)
 
  • #3
BvU said:
Hello terycho, ##\qquad## :welcome: ##\qquad## !

Do us a favor and read the guidelines on posting photos. If you promise to type instead of snap I'll try to help:

Don't understand your first step: you multiply with ##e^{-v(0)}\ ## on one side only: no can do.

Your second step doesn't help you, as you conclude yourself.

If you rewrite ##\ a(t) = - {b\over m} e^{\alpha v} \ ## as ##\ {dv\over dt} = - {b\over m} e^{\alpha v} \ ## does it look a bit more familiar ?
Do you know how to solve this kind of differential equation ?
(hit: think chain rule for differentiation)
I modified the post and read the guide! Sorry for breaking the rules.. My bad

And I know how to solve this kind of equations. We need to use this formula(attached the photo below). I read the formula you rewrote, but isn’t v also related to time? If I move dv to the right and do integration , it gets much more complexed since v is also the function related to time
 

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  • #4
Terrycho said:
I modified the post and read the guide! Sorry for breaking the rules.. My bad

And I know how to solve this kind of equations. We need to use this formula(attached the photo below). I read the formula you rewrote, but isn’t v also related to time? If I move dv to the right and do integration , it gets much more complexed since v is also the function related to time
You don't "move" anything, especially you can't move ## dv ##.

Use the form of the equation suggested by @BvU . ##\displaystyle \quad \ {dv\over dt} = - {b\over m} e^{\alpha v} ##

Divide both sides by ## e^{\alpha v} ## and integrate with respect to ##t## .

Note that ##\displaystyle \int f(v(t)) \frac{dv}{dt} dt = \int f(v) dv \ \ \ ## (Integration by substitution)
 
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  • #5
SammyS said:
You don't "move" anything, especially you can't move ## dv ##.

Use the form of the equation suggested by @BvU . ##\displaystyle \quad \ {dv\over dt} = - {b\over m} e^{\alpha v} ##

Divide both sides by ## e^{\alpha v} ## and integrate with respect to ##t## .

Note that ##\displaystyle \int f(v(t)) \frac{dv}{dt} dt = \int f(v) dv \ \ \ ## (Integration by substitution)
I didn’t think of dividing both sides by e^(av). I solved this problem! Thanks a lot!
 
  • #6
Terrycho said:
I modified the post and read the guide! Sorry for breaking the rules.. My bad

And I know how to solve this kind of equations. We need to use this formula(attached the photo below). I read the formula you rewrote, but isn’t v also related to time? If I move dv to the right and do integration , it gets much more complexed since v is also the function related to time
I solved it! Thanks!
 
  • #7
Terrycho said:
...
And I know how to solve this kind of equations. We need to use this formula(attached the photo below). I read the formula you rewrote, but isn’t v also related to time?
Your image →
0b93a057-a8df-4fcc-a523-6adaf7b1feba-jpeg.jpg


This formula is useful in the case where acceleration is given as some function of position or velocity, and in some cases both.

In your case it can be used to to find velocity as a function of position.

The equation ##\
\displaystyle \quad \ {dv\over dt} = - {b\over m} e^{\alpha v} \
## can be written as:

## \displaystyle \quad \ v{dv\over dx} = - {b\over m} e^{\alpha v} ##​

Dividing by ##\ e^{\alpha v} \ ## and integrating with respect to position, ##x## give the following,

## \displaystyle \int v \ e^{-\alpha v} \left({dv\over dx}\right) dx = - \int {b\over m} dx ##
 

1. What is rectilinear motion?

Rectilinear motion is a type of motion where an object moves along a straight line path.

2. What is the difference between rectilinear motion and uniform acceleration?

Rectilinear motion refers to the path of an object, while uniform acceleration refers to the rate at which the object's velocity changes. In rectilinear motion, the object's velocity can remain constant or change at irregular intervals, while in uniform acceleration, the object's velocity changes at a constant rate.

3. How is rectilinear motion represented on a position-time graph?

Rectilinear motion is represented on a position-time graph by a straight line with a constant slope. The slope of the line represents the object's velocity, and the line's y-intercept represents the object's initial position.

4. How is rectilinear motion related to Newton's first law of motion?

Rectilinear motion is related to Newton's first law of motion, also known as the law of inertia, which states that an object at rest will remain at rest and an object in motion will remain in motion with constant velocity unless acted upon by an external force. In rectilinear motion, the object will continue to move along a straight line path unless a force acts upon it to change its velocity.

5. What are some real-life examples of rectilinear motion?

Some real-life examples of rectilinear motion include a car driving along a straight road, a person walking in a straight line, or a ball rolling down a ramp. Any motion that occurs along a straight line path can be considered rectilinear motion.

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