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

[Beez]

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.
$$m=\frac{8}{32}=\frac{1}{4}$$
$$Air Resistance = \frac{1}{8}v^2$$
$$Friction = \frac{-1}{4}8 = -2$$
Pulling force = 2x


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

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

Besides, the answer provided for this problem was $$v=\sqrt{16x-32+32e^-^x}$$
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?

 

[geosonel]

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.
$$m=\frac{8}{32}=\frac{1}{4}$$
$$Air Resistance = \frac{1}{8}v^2$$
$$Friction = \frac{-1}{4}8 = -2$$
Pulling force = 2x


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

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

Besides, the answer provided for this problem was $$v=\sqrt{16x-32+32e^-^x}$$
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:

$$2\frac{dv}{dt} \ = \ 16x - 16-v^2$$

rearrange terms:

$$\frac{dv}{dt} + \frac{v^2}{2} \ = \ 8x - 8$$

then use:

$$\frac{dv}{dt} \ = \ \frac{dv}{dx} \cdot \frac{dx}{dt} \ = \ \frac{dv}{dx} \cdot v$$

so that placing this result into the diff eq:

$$v\frac{dv}{dx} + \frac{v^2}{2} \ = \ 8x - 8$$

or:

$$2v\frac{dv}{dx} + v^2 \ = \ 16x - 16$$

$$\frac{d(v^{2})}{dx} + v^2 \ = \ 16x - 16$$

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

 

[marlon]

begin with your correct equation:

$$2\frac{dv}{dt} \ = \ 16x - 16-v^2$$

rearrange terms:

$$\frac{dv}{dt} + \frac{v^2}{2} \ = \ 8x - 8$$

then use:

$$\frac{dv}{dt} \ = \ \frac{dv}{dx} \cdot \frac{dx}{dt} \ = \ \frac{dv}{dx} \cdot v$$

so that placing this result into the diff eq:

$$v\frac{dv}{dx} + \frac{v^2}{2} \ = \ 8x - 8$$

or:

$$2v\frac{dv}{dx} + v^2 \ = \ 16x - 16$$

$$\frac{d(v^{2})}{dx} + v^2 \ = \ 16x - 16$$

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

this is correct

marlon

 

[Beez]

Thanks for helping me again geosonel. I got it!