integration from mechanics (1 Viewer)

[bboyelement]

i cant get this for some reason ... please help

dt/dv = 1/(g-kv^2)

 

[pLuvia]

After using partial fractions
1=A[sqrt{g}+v.sqrt{k}]+B[sqrt{g}-v.sqrt{k}]
A=1/2sqrt{g}
B=1/2sqrt{g}
t=1/2sqrt{g}[ln(sqrt{g}+v.sqrt{k})]+1/2sqrt[{g}ln(sqrt{g}-v.sqrt{k})]

 

[bboyelement]

There's also the question of what you want it in respect to...

Cambridge book

exercise 7.2 question 2

a particle is moving vertically downward in a medium which exerts a resistance to the motion which is proportional to the square of the speed of the particle. it is released form rest at O and its terminal velocity is V. Find the distance it has fallen below O and the time taken when its velocity is one-half of its terminal velocity.

 

[STx]

hey is it possible to use the (1/2a).ln[(x-a)/(x+a)] + C 'form' directly

 

[bboyelement]

Cambridge book

exercise 7.2 question 2

a particle is moving vertically downward in a medium which exerts a resistance to the motion which is proportional to the square of the speed of the particle. it is released form rest at O and its terminal velocity is V. Find the distance it has fallen below O and the time taken when its velocity is one-half of its terminal velocity.

help? like i could get the distance but cant get the time ...

 

[STx]

well find the (terminal velocity)/2 and use dv/dt to find and expression for t in terms of v, then sub in (terminal velocity)/2 for v. hope that helps

 

[Mountain.Dew]

BIG BIG BIG HINT:

Spoiler

use dv/dt = 1/2d(v^2)/dx

 

[bboyelement]

BIG BIG BIG HINT:

Spoiler

use dv/dt = 1/2d(v^2)/dx

but im trying to find an expression for t so i can find the time ... i stated above that i found the distance (x) but cant find time (t)

 

[STx]

umm flip dv/dt to dt/dv, for both sides, integrate w.r.t 'v'

 

[bboyelement]

After using partial fractions
1=A[sqrt{g}+v.sqrt{k}]+B[sqrt{g}-v.sqrt{k}]
A=1/2sqrt{g}
B=1/2sqrt{g}
t=1/2sqrt{g}[ln(sqrt{g}+v.sqrt{k})]+1/2sqrt[{g}ln(sqrt{g}-v.sqrt{k})]

thanks

 

[Wackedupwacko]

hmm that looks so messy tho i just use

1/ g - kv²
= 1/k ( 1/ [g/k - v²])
= 1/k (1/ [c² - v²]) where c² = g/k
= 1/2ck (1/[c+v] + 1/[c-v])
integral of that is just
1/2ck (ln | (c+v) / (c-v)| + c

sub bak the g/k if u want but i dunno it looks much less of a mess if i use c² = g/k

btw my answer is

1/2sqrt(kg) ( ln| {sqrt [k/g] +v} /{sqrt[k/g] -v}|) + c