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Expressing velocity as a function of the distance (not time elapsed)

  • Thread starter Beez
  • Start date
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Hi, I encountered the great difficulty solving the following question.
(The reason why its is hard is because I have to express the velocity as a function of the distance of the object has traveled in stead of as a function of the time elapsed.)

Starting from rest at t=0, an 8 lb. object is pulled along a surface with a force that is equal to twice the distance (in feet) of the object from its starting point x=0. The coefficient of sliding friction is 1/4.
The air resistance is numerically equal to one-eight the square of the velocity (in feet/second).
What is the velocity formula of the object as a function of the distance of the object has traveled?

So, weight = 8lb.
[tex] m=\frac{8}{32}=\frac{1}{4}[/tex]
[tex]Air Resistance = \frac{1}{8}v^2[/tex]
[tex]Friction = \frac{-1}{4}8 = -2[/tex]
Pulling force = 2x


First I tried to write the velocity formula as a function of the time elapsed. The I got

[tex]\frac{1}{4}\frac{dv}{dt}=2x - 2-\frac{1}{8}v^2[/tex]
[tex]2\frac{dv}{dt}=16x - 16-v^2[/tex]
[tex]\frac{dv}{v^2+16}=(-8x)dt[/tex]
Integrate, I had
[tex]-0.004363tan^-^1(0.25v)=8xt+C[/tex]
Applying the I.C., v(0)=0
c = 0
so, [tex]tan^-^1(0.25v)=\frac{-8xt}{0.004363}[/tex]
I don't know how to solve this equation for v from here.

Besides, the answer provided for this problem was [tex]v=\sqrt{16x-32+32e^-^x}[/tex]
which gives me the idea that my approach is off and should be a way to obtain the desired formula without obtaining the formula as a function of time elapsed first...

What is the best way to attack this problem?
 
88
0
Beez said:
Hi, I encountered the great difficulty solving the following question.
(The reason why its is hard is because I have to express the velocity as a function of the distance of the object has traveled in stead of as a function of the time elapsed.)

Starting from rest at t=0, an 8 lb. object is pulled along a surface with a force that is equal to twice the distance (in feet) of the object from its starting point x=0. The coefficient of sliding friction is 1/4.
The air resistance is numerically equal to one-eight the square of the velocity (in feet/second).
What is the velocity formula of the object as a function of the distance of the object has traveled?

So, weight = 8lb.
[tex] m=\frac{8}{32}=\frac{1}{4}[/tex]
[tex]Air Resistance = \frac{1}{8}v^2[/tex]
[tex]Friction = \frac{-1}{4}8 = -2[/tex]
Pulling force = 2x


First I tried to write the velocity formula as a function of the time elapsed. The I got

[tex]\frac{1}{4}\frac{dv}{dt}=2x - 2-\frac{1}{8}v^2[/tex]
[tex]2\frac{dv}{dt}=16x - 16-v^2[/tex]
[tex]\frac{dv}{v^2+16}=(-8x)dt[/tex]
Integrate, I had
[tex]-0.004363tan^-^1(0.25v)=8xt+C[/tex]
Applying the I.C., v(0)=0
c = 0
so, [tex]tan^-^1(0.25v)=\frac{-8xt}{0.004363}[/tex]
I don't know how to solve this equation for v from here.

Besides, the answer provided for this problem was [tex]v=\sqrt{16x-32+32e^-^x}[/tex]
which gives me the idea that my approach is off and should be a way to obtain the desired formula without obtaining the formula as a function of time elapsed first...

What is the best way to attack this problem?
begin with your correct equation:

[tex]2\frac{dv}{dt} \ = \ 16x - 16-v^2[/tex]

rearrange terms:

[tex]\frac{dv}{dt} + \frac{v^2}{2} \ = \ 8x - 8 [/tex]

then use:

[tex]\frac{dv}{dt} \ = \ \frac{dv}{dx} \cdot \frac{dx}{dt} \ = \ \frac{dv}{dx} \cdot v [/tex]

so that placing this result into the diff eq:

[tex]v\frac{dv}{dx} + \frac{v^2}{2} \ = \ 8x - 8 [/tex]

or:

[tex]2v\frac{dv}{dx} + v^2 \ = \ 16x - 16 [/tex]

[tex]\frac{d(v^{2})}{dx} + v^2 \ = \ 16x - 16 [/tex]

now let y=v2 and solve for y(x) using standard techniques.
 
Last edited:
3,761
8
geosonel said:
begin with your correct equation:

[tex]2\frac{dv}{dt} \ = \ 16x - 16-v^2[/tex]

rearrange terms:

[tex]\frac{dv}{dt} + \frac{v^2}{2} \ = \ 8x - 8 [/tex]

then use:

[tex]\frac{dv}{dt} \ = \ \frac{dv}{dx} \cdot \frac{dx}{dt} \ = \ \frac{dv}{dx} \cdot v [/tex]

so that placing this result into the diff eq:

[tex]v\frac{dv}{dx} + \frac{v^2}{2} \ = \ 8x - 8 [/tex]

or:

[tex]2v\frac{dv}{dx} + v^2 \ = \ 16x - 16 [/tex]

[tex]\frac{d(v^{2})}{dx} + v^2 \ = \ 16x - 16 [/tex]

now let y=v2 and solve for y(x) using standard techniques.
this is correct

marlon
 
32
0
Thanks for helping me again geosonel. I got it!
 

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