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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 e-kx/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.