mechanics question! (1 Viewer)

[haboozin]

hey, can't do the first question in 7.2 cambridge, eeekk.

full explanation would be appreciated.

<i>A particle is moving vertically downwards in a medium which exerts a resistance to the motion which is proportional to the speed of the particle. The particle is released from rest at O, and at time t its position is at a distance x below O and its speed is v. if the terminal velocity is V, show that gx +Vv=Vgt</i>

no weird ways of solving it please, strictly 4unit syllabus. Let this not be another Pi and e debate haha.

well i have an idea, but i better let people start from scatch 'cause i want to see how you would do it...
thanks.

 

[no_arg]

ma=mg-kmv
a=g-kv

Term vel when a=o ie V=g/k

Now a=g(1-(k/g)v)=g(1-v/V) implying

Va=g(V-v)
Va+vg=Vg

integrating wrt t yields

Vv+xg=Vgt (C=0)


g may of course depend upon a varying G

 

[haboozin]

ma=mg-kmv
a=g-kv

Term vel when a=o ie V=g/k

<b>Now a=g(1-(k/g)v)</b>=g(1-v/V) implying

Va=g(V-v)
Va+vg=Vg

<b>integrating wrt t yields</b>

Vv+xg=Vgt (C=0)


g may of course depend upon a varying G

i dont know how u got the bolded parts.

 

[no_arg]

The first bold is just rewriting a=g-kv (tease out the g)

for the second remember that the int of acc wrt to t is vel
and the integral of vel wrt t is posn!

Vg is just a constant so the int of Vg wrt t is Vgt.

 

[who_loves_maths]

I thought this question looked familiar lol.

Trev has asked this question quite a while ago:

Mechanics Q

- the simpler solution is the latter one in that thread.