Question

Lets begin with the assumption that R=kv. What are the units of the coefficient k in terms of kilograms, metere, and/or seconds?

dv 6. The equation of motion now becomes mº mg - Kv. dt This equation is also separable because all the terms involving v can be brought to the left side of the dv equation: 1 = 1. dig-(K/m) As before, we integrate both sides of the equation with respect to t and use a change of variables. The resulting equation is s [dt . g-(K/m)v dy K dt, where a mg a-v2 к Evaluate the integrals on both sides of this equation and use the initial condition (O) = 0 to determine the arbitrary constant. Show that the velocity function is given in either of the two forms dy It's easiest to write the equation as I am not jdt, mg K mg (1-e K 1+

Answer 1

Inch كوا my k dt kg. m Iky. ky. m 167 m² g & [1]m2 = Kg.m vuit of K = Hug m K ar a² uz f dt t where a ²= my K m du = klat fa-rotato) k dt + I 7 za + atu J av m Gov -lua-u) t lu lato t + 6

= - I lulete t to قاسم q m vol=0 ; at t=o ato = t.co) + 6 -la 4-0 I le 1) = w oto Za o=ote : co Hence I. t 2 => za la [ pm en que ] = reaks of => qa.K. t el tu e are &act atua a-6) e za KE => ofteen) --utite

alct V64) = ate -17 26ict +17 이 Substitute १८ mg C mg kt - 2- al => (4) E F e 6 가 ) mg t (é 키 25 kg/mk e = 이터 mg 2 e 2 rica/nit th which i also equals to 2. Icg/mit UH C mg If e2