- #1

Beez

- 32

- 0

(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?