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Unformatted text preview: Solutions 2.9
1. We have the differential equation v = g  which has the solution v(t) =  mg mg kt + exp( ). k k m k v, m v(0) = 0, The velocity is half of the terminal velocity when exp( i.e. t = m ln 2. k 2.a) We have mv = kv, hence kt ). m We are told that v(0) = 220 mph, t = 4 sec, v(t) = 50 mph, and mg = 3000 lb. Since g = 32 ft/sec2 , we find m = 93.75 lb sec2 /ft. Finally v(t) = v(0) exp( k= m ln(v(0)/v(t)) = 34.73 lb sec/ft. t kt 1 )= , m 2 b) The distance traveled is
4 0 v(t) dt = v(0) m 4k (1  exp( )) = 0.1274 miles. k m (Do not forget the conversion from hours to seconds here!) 4. We have mv = mg  kv, v(0) = v0 . The solution of the differential equation is v(t) = mg mg kt + (v0 + ) exp( ). k k m 1 The maximum height is reached when v(t) = 0, i.e. t= 11. We obtain m Integration yields mv = k ln(1 + x) + C. From the initial condition, C = mv0 . The equilibrium position is given by k ln(1 + xf ) + mv0 = 0, i.e. xf = 1 + emv0 /k . 12. We have m i.e. m Consequently, v(x) = v0 ekx/m . dv = kv 2 , dt dv = kv. dx m kv0 + mg ln . k mg dv k = . dx 1+x 2 ...
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This note was uploaded on 10/16/2009 for the course MATH 2214 at Virginia Tech.
 '06
 EDeSturler
 Equations

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