Need HELP equation of motion with variable acceleration

AI Thread Summary
The discussion revolves around solving the equation of motion for a particle dropped into oil, where the acceleration is defined as a = g - kv. Participants clarify that the problem can be approached as a first-order differential equation regarding velocity, rather than a second-order equation. One user expresses frustration over their calculations leading to a zero result, seeking insight into potential errors. The conversation emphasizes the importance of correctly interpreting the differential equations involved. Ultimately, understanding the correct order of the differential equation is crucial for solving the problem effectively.
Behroz
Messages
5
Reaction score
0

Homework Statement


A particle is dropped from rest, at the surface, into a tank containing oil
The acceleration of the particle in the oil is a = g – kv
where g is the gravitational acceleration and –kv being denoted by
the resistance put on the particle by the oil.
Solve for x as a function of time!


The Attempt at a Solution


I'm attaching an image file containing my calculations.
As can be seen I get a differential equation of the second order
but as I proceed to solve this equation and try to determine the
constants the whole thing turns into ZERO?!? What am I doing wrong??
 

Attachments

  • en.jpg
    en.jpg
    31.4 KB · Views: 573
Physics news on Phys.org
You don't need to consider this as a 2. order diff. equation. You can view the equation as

\ddot{x} = g - k\dot{x}
or
\dot{v} = g - kv

The second equation is just a 1. order diff. equation with respect to the velocity v.
 
P3X-018 said:
You don't need to consider this as a 2. order diff. equation. You can view the equation as

\ddot{x} = g - k\dot{x}
or
\dot{v} = g - kv

The second equation is just a 1. order diff. equation with respect to the velocity v.

Yeah, you're right.. but one SHOULD be able solve it by puting it up as a second order right? I just want to know what I'm doing wrong, it really bugs me.
 
what have you got? Kind of hard to know what's wrong when we just have a generic list of eqns...
John S
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top