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mechanics (1 Viewer)

shuning

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a particle of mass m falls from rest under gravity in a medium whose resistance is mkv. where k is a constant. taking the initial postion as the orgion and the x axis downwards. prove that:

velocity v at time t is given by

the distance x fallen in time t is given by or

and the terminal velocity is given by
 
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GUSSSSSSSSSSSSS

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wudnt it just be easier to use: F = ma = mg - mkv

a = g - kv .......replace acelleration by dv/dt
dv/dt = g - kv
dv/(g-kv) = dt
t = Sdv/(g-kv) = (-1/k)ln(g-kv) + c
**when t=0, v=0** .......it was initially at rest
0 = (-1/k)ln(g) + c ---> c = (1/k)ln(g)
therefore: t = (-1/k)ln(g-kv) + (1/k)ln(g)
t = (1/k)ln(g/(g-kv))
kt = ln(g/(g-kv))
**inverse**
g/(g-kv) = e^kt
g-kv = g/(e^kt)
kv = g - ge^-kt
v = (g/k)(1 - e^-kt)

part 2

replace v by dx/dt

dx/dt = (g/k)(1 - e^-kt)
dx = (g/k)(1 - e^-kt)dt
x = (g/k)S(1 - e^-kt)dt = (g/k)(t + (1/k)e^-kt) + c
**when t=0, x=0**
0 = (g/k)(0 + 1/k) + c --> c = -g/k^2
therefore: x = (g/k)(t + (1/k)e^-kt) - g/k^2
x = (g/k)(t + (1/k)e^-kt - (1/k))


part 3

terminal velocity happens when the resistive force is equal to the force goin down (lol gud wording gus xD)

ie: mg = mkv
g = kv
v = g/k
 

GUSSSSSSSSSSSSS

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Re: 回复: Re: mechanics

hahaha........well this friday i MIGHT be in chatswood not sure......so if u got more q's i can do em then xD
 

Aerath

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part 3

terminal velocity happens when the resistive force is equal to the force goin down (lol gud wording gus xD)

ie: mg = mkv
g = kv
v = g/k
You could just say let x'' = 0. :p
Or let x go towards infinity. or let t go towards infinity. =P
 

GUSSSSSSSSSSSSS

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You could just say let x'' = 0. :p
Or let x go towards infinity. or let t go towards infinity. =P
yea there r a number of ways to say it xD

i just picked one which i failed at explaining .... eloquently ....
=P
 

Aerath

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I just love how you said: "force going down".
 

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