problem04_56

University Physics with Modern Physics with Mastering Physics (11th Edition)

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4.56: a) The equation of motion, dt dv m Cv = - 2 cannot be integrated with respect to time, as the unknown function ) ( t v is part of the integrand. The equation must be separated before integration; that is, , 1 1 0 2 v v m Ct v dv dt m C + - = - = - where 0 v is the constant of integration that gives 0 v v = at 0 t = . Note that this form shows that if 0 0 v = , there is no motion. This expression may be rewritten as , 1 1 0 - + = = m Ct v dt dx v which may be integrated to obtain . 1 ln 0 0 + = - m Ctv C m x x To obtain x as a function of v , the time t must be eliminated in favor of
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Unformatted text preview: v ; from the expression obtained after the first integration, 1-= v v m Ctv , so . ln =-v v C m x x b) By the chain rule, , v dx dv dt dv dx dv dt dv = = and using the given expression for the net force, m dx dv v Cv =-2 v dv dx m C =- =--ln ) ( v v x x m C . ln =-v v C m x x...
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