Integration (displacement or position)

AI Thread Summary
The discussion focuses on the integration challenge in a physics problem involving thrust, mass, and acceleration. The equation mg - uk - m*a is set up with known constants and variables for velocity and acceleration. The user attempts to manipulate the equation to prepare for integration but encounters difficulties, particularly with the left side of the equation. They correctly identify the integration of the right side as yielding -1/3000s + c, while seeking guidance on properly integrating the left side. The goal is to determine the displacement (s) when the velocity (v) reaches zero, using initial conditions provided.
kieran1black2
Messages
23
Reaction score
0

Homework Statement


the trouble is in the integration on the left side (go straight o my answer down the bottom of the page)... please help me...

mg-u-kv=m*a

where (all constant)
m= mass= 3000
g=gravity=10
u=thrust=172000
k=30

v=velocity=variable
a=acceleration=variable


Homework Equations



accel=v*dv/ds=dv/dt=a

The Attempt at a Solution



30000-172000-30v = 3000*v*dv/ds
-142000-30v = 3000*v*dv/ds
-(142000+30v) = 3000*v*dv/ds
142000+30v = -3000*v*dv/ds
(142000+30v)/v*dv = -3000*ds
v/(142000+30v)*dv = -1/3000*ds

and then integrate both sides, this is where my calculus stops

i know the right side (-1/3000*ds) becomes -1/3000s + c
where
s= displacement or position and
c= constant
i have 2 scenarios

when s=0, v=1000
and what i need to find is when v=0 what is s?


so once i integrate the left side i sub those values in.
 
Physics news on Phys.org
kieran1black2 said:
(142000+30v)/v*dv = -3000*ds
v/(142000+30v)*dv = -1/3000*ds

Are you sure about this transition?
 
if you think you can do it without that transition... go for it... that's just how i interpreted the question
 
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