Expressing velocity as a function of the distance

[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?

 

[Beez]

Woops

Sorry, somehow I posted the same question twice. Please disregard this and leave your advice in the other one with the same title with (not time elapsed) added.
Thanks.

 

[thepatientmental]

It is understandable that you are facing difficulty in solving this question as it requires expressing velocity as a function of distance rather than time. However, with a systematic approach, we can find the solution. Let's break down the problem into smaller steps.

Step 1: Understanding the given information
Before we start solving the problem, it is important to understand the given information. We have an object with a weight of 8lb, which means its mass is 1/4. The object is pulled with a force that is twice the distance it has traveled from its starting point. The coefficient of sliding friction is 1/4 and air resistance is 1/8 of the square of the velocity.

Step 2: Writing the force equation
We know that force equals mass times acceleration (F=ma). In this case, the force acting on the object is the pulling force, friction, and air resistance. So, we can write the equation as:

F = ma = 2x - \frac{1}{4}v - \frac{1}{8}v^2

Step 3: Finding the acceleration
To find the acceleration, we need to use Newton's second law, which states that the net force acting on an object is equal to its mass times acceleration (F=ma). So, we can rearrange the equation to find the acceleration:

a = \frac{F}{m} = \frac{2x - \frac{1}{4}v - \frac{1}{8}v^2}{\frac{1}{4}} = 8x - 2v - \frac{1}{2}v^2

Step 4: Using kinematic equations
We know that velocity is the derivative of displacement with respect to time (v=\frac{dx}{dt}). However, in this case, we need to express velocity as a function of distance, not time. To do this, we can use the kinematic equations of motion, specifically the one that relates velocity, acceleration, and displacement:

v^2 = u^2 + 2as

Where v is the final velocity, u is the initial velocity (in this case, it is 0 as the object starts from rest), a is the acceleration, and s is the displacement.

Step 5: Substituting values and simplifying
Substituting the values from step 3 into the kinematic equation, we get: