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p0326 - ma = − D = ⇒ a = − μ r g − v 2 L Using the...

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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 = − μ r g − v 2 L Using the fact that a = vdv/dx , we have v dv dx = − μ r g − v 2 L = ⇒ Lvdv μ r gL + v 2 = − dx Thus, we find L 8 v o vdv μ r gL + v 2 = − 8 d dx = ⇒ L 2 f n D μ r gL + v 2 ie e v =0 v = v o = − d Evaluating the left hand side yields L 2 f n w μ r gL μ r gL + v 2 o W = − d Therefore, the total distance traveled is d = L 2 f n w 1 + v 2 o μ r gL W Turning to the time elapsed, we use the fact that a = dv/dt . Hence, dv dt = − μ r g − v 2 L = ⇒ Ldv μ r gL + v 2 = − dt We integrate as follows. L 8 v o dv μ r gL + v 2 = − 8 t f dt = ⇒ L √ μ r gL tan − 1 v √ μ r gL μ 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 μ r g tan − 1 w v o √ μ r gL W...
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