p0326 - ma = D = a = r g v 2 L Using the fact that a =...

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240 CHAPTER 3. FORCE AND ACCELERATION 3.26 Chapter 3, Problem 26 Problem: The sum of the aerodynamic drag force and the road-friction force on a Jaguar XF coasting along a horizontal surface is D = m D μ r g + v 2 /L i ,wh e r e m is the car’s mass, v is speed, μ r is rolling friction coefficient, g is gravitational acceleration and L is a characteristic length. Assuming the Jaguar’s initial speed is v o when it starts coasting, find the distance, d , that it coasts until it comes to rest. Express your answer as a function of g , L , μ r and v o . Also, find the time, t f , as a function of g , L , μ r and v o that it takes the Jaguar to come to rest. F = D i v = v i x . . . . . . ............................................................. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .......
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Unformatted text preview: ma = D = a = r g v 2 L Using the fact that a = vdv/dx , we have v dv dx = r g v 2 L = Lvdv r gL + v 2 = dx Thus, we find L 8 v o vdv r gL + v 2 = 8 d dx = L 2 f n D r gL + v 2 ie e v =0 v = v o = d Evaluating the left hand side yields L 2 f n w r gL r gL + v 2 o W = d Therefore, the total distance traveled is d = L 2 f n w 1 + v 2 o r gL W Turning to the time elapsed, we use the fact that a = dv/dt . Hence, dv dt = r g v 2 L = Ldv r gL + v 2 = dt We integrate as follows. L 8 v o dv r gL + v 2 = 8 t f dt = L r gL tan 1 v r gL r gL e e e e v =0 v = v o = t f 3.26. CHAPTER 3, PROBLEM 26 241 Simplifying and rearranging terms, we conclude that t f = L r g tan 1 w v o r gL W...
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This note was uploaded on 05/11/2011 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.

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p0326 - ma = D = a = r g v 2 L Using the fact that a =...

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