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

In summary, the conversation is about solving a question involving expressing velocity as a function of distance instead of time. The problem involves an 8 lb. object being pulled along a surface with a force equal to twice the distance, with air resistance and friction taken into account. The discussion includes an attempted solution and the correct method of solving the equation.
  • #1
Beez
32
0
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?
 
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  • #2
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
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
 
  • #4
Thanks for helping me again geosonel. I got it!
 

1. How is velocity expressed as a function of distance?

Velocity can be expressed as a function of distance by dividing the change in distance by the change in time. This is known as the average velocity formula and is represented by V = Δd/Δt.

2. Can velocity be calculated without considering time?

No, velocity is a measure of how fast an object is moving in a certain direction over a period of time. Therefore, time is an essential factor in calculating velocity and cannot be disregarded.

3. What is the difference between speed and velocity in terms of distance?

Speed is a measure of how fast an object is moving without considering its direction, while velocity takes into account both speed and direction. Therefore, velocity as a function of distance will include the direction of the object's movement.

4. How does acceleration affect velocity as a function of distance?

Acceleration is the rate of change of velocity over time. As an object accelerates, its velocity as a function of distance will also change. The steeper the slope of the distance-time graph, the greater the acceleration and the faster the object is moving.

5. Can velocity as a function of distance be represented graphically?

Yes, velocity as a function of distance can be represented on a graph by plotting the distance on the x-axis and the velocity on the y-axis. This will result in a velocity-time graph, where the slope of the line represents the velocity as a function of distance.

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