Mechanics q. (1 Viewer)

[Trev]

It's an easy one (first question of 7.2 in Cambridge) I just don't know where to start... So full working isn't needed.

A partice is moving vertically downward 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 it's position is at a distance x below O and it's speed is v. If the terminal velocity is V show that gx + Vv = Vgt.
Thanx in advance!

 

[who_loves_maths]

i haven't done Mechanics 4u yet, so i'm not aware of any easier way out of this other than just finding the expresions for 'x', and 'v' in terms of 't':

a = (g -kv) ; where, 'a' is acceleration, and 'k' is a constant ---> dt/dv = 1/(g -kv)

integrate and account for the constant 'c' to get: v = (g/k)(1 - e^(-kt))

but terminal velocity 'V' occurs when a = 0 ---> V = g/k ;

hence, v = V(1 -e^(-kt)) ---> dx/dt = V(1 -e^(-kt)); also, Vv = (V^2)(1 -e^(-kt))

integrate that and you get: x = Vt + (V/k)(e^(-kt)) - V/k ; but k = g/V ,

ie. x = Vt + (V^2/g)(e^(-kt)) - (V^2/g) ---> gx = Vgt + (V^2)(e^(-kt) - 1)

therefore, gx + Vv = Vgt


this is the systematic way of doing it, there's probably a simpler method to it... but like i said, i haven't done Mechanics 4u so i can't help you with the 'easier' method(s), sry. hope this helps anyway

 

[Trev]

Ah yes, I remember now. I had v² instead of v; thanx!

 

[who_loves_maths]

okay, well i just remembered something too Trev, i fiddled around a bit with the question and i think this is the 'easier' way, i hope it's not too late:

a = (g -kv) , and V = g/k when a =0.

now, re-arrange it: kv = g -a ---> v = (1/k)(g -a) = dx/dt

hence, integrating: x = (1/k)Int[(g -a)dt] = (1/k)(gt - v) , since Int[a dt] = v

therefore, x = g/k*t - v/k ---> gx = g^2/k*t - g/k*v = Vgt - Vv ;

ie. {after re-arranging} gx + Vv = Vgt


hope this helps now

 

[Trev]

okay, well i just remembered something too Trev, i fiddled around a bit with the question and i think this is the 'easier' way, i hope it's not too late:

a = (g -kv) , and V = g/k when a =0.

now, re-arrange it: kv = g -a ---> v = (1/k)(g -a) = dx/dt

hence, integrating: x = (1/k)Int[(g -a)dt] = (1/k)(gt - v) , since Int[a dt] = v

therefore, x = g/k*t - v/k ---> gx = g^2/k*t - g/k*v = Vgt - Vv ;

ie. {after re-arranging} gx + Vv = Vgt


hope this helps now

I'm not sure if you purposely disregarded the constant when integrating as it becomes zero, or you just failed to include it... Thanx for the help!

 

[who_loves_maths]

I'm not sure if you purposely disregarded the constant when integrating as it becomes zero

yeah i didn't include it cause i thought it was obvious that it was 0 since t = 0 ---> v = 0 ---> x = 0, so c = 0 as well...

but i should've made some sort of reference to it in my last post so sorry about that... of course if you were doing it in an exam, then you must include the constant.