1_8 - Solutions 1.8-Page 89 Problem 2 Suppose that a body...

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Solutions 1.8-Page 89 Problem 2 Suppose that a body moves through a resisting medium with resistance proportional to its velocity , so that . (a) Show that its velocity and position at times t are given by v kv dt dv = / kt e v t v = 0 ) ( and ( ) ) 1 ( ) ( 0 0 kt k v e x t x + = . (b) Conclude that the body travels only a finite distance, and find that distance. kv dt dv = / is a separable equation. Separating variables and integrating yields: kt e C v C kt v kdt v dv = + = = ~ ln Substituting t = 0 shows that 0 ~ v = C . Therefore, kt e v t v = 0 ) Recall that . Making this substitution in the above equation, then separating variables and integrating yields: v dt dx = / C e k v x dt e v dx e v dt dx kt kt kt + = = = 0 0 0 / The constant C is solved for by substituting 0 ) 0 ( x x = into the above position function. k v x e k v x and k v x C C k v x kt 0 0 0 0 0 0 0 + + = + = + =
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Simplifying gives ( ) ) 1 ( ) ( 0 0 kt k v e x t x + = (b) The body travels its maximum distance when v = 0. From , this occurs when . Substituting this value for time into the position function gives the maximum distance the body travels. kt e v t v = 0 ) ( = t ( ) ) 1 ( ) ( ) ( 0 0 + = k k v e x t x k v x x 0 0 max + =
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Problem 7 Suppose that a car starts from rest, its engine providing an acceleration of 10 ft/s 2 , while air resistance provides 0.1 ft/s 2 of deceleration for each foot per second of the car’s velocity. (a) Find the car’s maximum possible (limiting) velocity. (b) Find how long it takes the car to attain 90% of its limiting velocity, and how far it travels while doing so. The given acceleration information takes the following equation form: v dt dv 1 . 0 10 / = (a) Since the acceleration equation is the derivative of the velocity equation, velocity has a maxima when acceleration is zero.
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1_8 - Solutions 1.8-Page 89 Problem 2 Suppose that a body...

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