Question

5. In certain circumstances, we can model the velocity of a falling mass subject to air resistance as - dv m7 = mg – kv?, where v (t) is the velocity of the object, m is the mass of the object, g is acceleration due to gravity, and k is a constant of proportionality. Assume the positive direction is downward. (a) Solve this equation subect to the initial condition v (0) = vo. (b) What is the terminal velocity of the object (when dx = 0)? (c) Let s = 0 be the height at which the mass s released above the ground, and assume di = v (t). Determine and expression for s (t) if we are given that s (0) = 0.

Answer 1

Given differ emtial equahion Zo) Thus v(Ut e(t)(-) m mg-ku2 (11) (Vot ) exp (2 t)+ (-) Sepoah ng vaniablen in (), we have = dt dv dt Using d m du m in 0), we haue =O mg-ku du dt Km-2) mg-ku-o where a= mg id the termiral velocity a2-u2 Sntegrating dv his ue hawe Using ds 1O = ds K(dt C, wRere Cisan bitrany cndon thus we have dt 2 Jak. t n a+u -u) e kt a-u 2a m (Vot ) e en atu a-u () with rA= Vot m, 2ak t+ 2ac taking sac A Sas= (Ae Bt J A eBE+C dt +D, e Bt AeBtyCdt c(dt Aebt+c= Z eBt Z-C exp (aakt+A) applying comadando 1+ exp(tA) -1t emp (akat+A) 1+ exp( 2ak t +A) exp ( t+A)- g. a+u a-13 G-+9)+ (a-9) B AJ Cg-U D arbitrory conotord D Adividando at19-ca-u) Aeeie a AB ebt dt dz dt= dz ABEBE HB(z) 14-0xp-(az+ A) OT EX dz dz Using v(o) = Vo exp A- EXPA+I Vo= a ( enp.A-I) (expA+1) aepA-a= voep A +U dz +D BJZ Cz-C) ln Z a dz +1 Z-c exp A= Vo+a a-Vo Thus +D=en/z exp (kat). exA) exp (2kat).exp(A)+ t a.Vota) exp(3akt)- (0-o) ota) exp(2akt )+ (a-v) explatb) =expa.exp(b St= +D= A eBt Using Sto) o en(A)tD st= ene nei7 4D In (Atc)2 D here a ma 1ak k m (AtC)2 87 oxet