CHAPTER 2 - 22 - 88 v0 1 v0 D x =0 L v0 2 x =x v t!a t =0 v...

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88. We use a coordinate system with the origin at the initial position of the car.   The passing car’s position is given by  x 1  =  x 01  +   v 0 t  +  ! a 1 t 2  = 0 +  v 0 t  +  ! a 1 t 2 . The truck’s position is given by  x truck  =  x 0truck  +   v 0 t  =  D  +  v 0 t . The oncoming car’s position is given by  x 2  =  x 02  –   v 0 t   =  L  –  v 0 t . For the car to be safely past the truck, we must have x 1  –  x truck  = 10 m; v 0 t  +  ! a 1 t 2  – ( D  +  v 0 t ) =  ! a 1 t 2  –  D  = 10 m, which allows us to find the time required for passing: ! (1.0 m/s 2 ) t 2  – 30 m = 10 m, which gives  t  = 8.94 s. At this time the car’s location will be x 1  =  v 0 t  +  ! a 1 t 2  = (25 m/s)(8.94 s) +  ! (1.0 m/s 2 )(8.94 s) 2  = 264 m from the origin. At this time the oncoming car’s location will be x 2  =  L  –  v 0 t  = 400 m – (25 m/s)(8.94 s) = 176 m from the origin.
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