Resisted Motion Question (1 Viewer)

[mel4dan]

Hey everyone,

I have a question for you from Fitzpatrick. 35b, Q12:

A particle has an initial velocity U and after travelling a distance d in time T along a straight path its velocity is V. The retardation of the particle is proportional to its velocity at any time. Show that

U = Ve^(T(U-V)/d)​

If someone could show me how to do it that would be great...

 

[Roobs]

i cant remember mechanics.... but ive got worked solutions to that chapter, if some random smart doesent post a timely solution email me and i'll do you a crude "take a photo of page with digital camera" job.....

 

[Roobs]

crap..rubendownunder@hotmail.com

 

[mel4dan]

Ok i will email...

 

[hyparzero]

This is just a basic tutorial on it...

Retardation = kv

Therefore:
ΣF = ma = - mkV

Hence

a = -kV
=> dV/dT = -kV
=> -dV/kV = dT

Thus, -∫dV/kV = ∫dT

=> -[ln(kV)]k = T + C ........(1)

When T= 0, V = U
Hence C = -[ln(kU)]k

(1)... => T= 1/k [ln(U/V)] ......... (2)

Again,
ΣF = ma = - mkV

Hence
a = -kV
=> 1/2 dV²/dD = -kV

Rearrange, Integrate and Solve, note here that you will be integrating V with respect to V², so simply treat V as √(V²)

After integration, remember to let V= U when D= 0 to find C

You should get:
V= U - kD
=> k = (U - V)/D ...... Subsitute into (2) to eliminate k

Hence
(1)...... T= 1/k [ln(U/V)]
=> T= [ln(U/V)] / [(U - V)/D]

Simplify, take exponential of both sides....

U = Ve^(T(U-V)/d) .... as required...

 

[mel4dan]

yeah i got it out