Physics Help Forum Falling in a viscous liquid problem

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

Jul 29th 2015, 12:36 PM   #11
Physics Team

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 Originally Posted by jackthehat the correct equation should be .. v(t) = v_t (1 - e to the power -(k/m)t ) ..
Yes!

 Originally Posted by jackthehat My problem with this originally would have still been the same by the way .. that the term for time t forms part of exponent's power .. and how to get a value for t from that equation .. but now I am thinking (after going away and refreshing myself with some basic maths .. as I left full-time education 40 years ago !) that if you take the log (logarithm function) of each side of the above equation you can take t down from the power as I believe ... log(e to the power x) = x ... that's right isn't it ?
Yes!

 Originally Posted by jackthehat my only problem now (I think) is that the equation would then become .. log (v(t)) = v_t(1 - (-k/m)t)
Not quite. It's easiest to start by getting the e^(-kt/m) term by itself:

e^(-kt/m) = 1 - v(t)/v_t

Now take the log of both sides:

-kt/m = ln(1-v(t)/v_t)

Rearrange:

t = (-m/k)ln(1-(v(t)/v_t))

Now set v(t)/v_t = 0.9, and solve for t.

 Jul 29th 2015, 01:12 PM #12 Member   Join Date: Jan 2015 Posts: 96 Hi ChipB .. Yes I see it now .. that's great. Thank you very much for bearing with me and for all your help. Regards JacktheHat

 Tags falling, liquid, problem, viscous