CHAPTER 2 18 - x =x v 2 02 max 2 t L =0 v 2 max 2 t max to0 0=v a t max 3 3 0=25m/s(2.0m/s)t,whichgivest =12.5s x =x v 3 03 3 max 3 3 2 t!a t 33 2

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a 2 v 1 x  = 0 d 1 d 2 d 3 d 4 d 5 d 6 x 2  =  x 02  +  v max t 2   ; L 2  = 0 +  v max t 2   . Motion 3 is the acceleration from  v max  to 0.   We find the time for this motion from 0 =  v max   +   a 3 t 3 ; 0 = 25 m/s + (– 2.0 m/s 2 ) t 3   , which gives  t 3  = 12.5 s. We find the distance for this motion from x 3  =  x 03  +  v max t  +  ! a 3 t 3 2  ; L 3  = 0 + (25 m/s)(12.5 s) +  ! (– 2.0 m/s 2 )(12.5 s) 2  = 156 m. The distance for Motion 2 is L 2  = 800 m –  L 1  –   L 3  = 800 m – 284 m – 156 m = 360 m, so the time for Motion 2 is t 2  =  L 2 / v max  = (360 m)/(25 m/s) = 14.4 s. Thus the total time for the 45 segments and 44 stops is T  = 45( t 1  +  t 2  +  t 3 ) + 44(20 s) = 45(22.7 s + 14.4 s + 12.5 s) + 44(20 s) = 3112 s =        52 min . ( b ) There are (36 km)/3.0 km) = 12 trip segments, which means 13 stations (with 11 intermediate 
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This note was uploaded on 03/08/2010 for the course PHYSICS 7A/7B taught by Professor All during the Fall '08 term at University of California, Berkeley.

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CHAPTER 2 18 - x =x v 2 02 max 2 t L =0 v 2 max 2 t max to0 0=v a t max 3 3 0=25m/s(2.0m/s)t,whichgivest =12.5s x =x v 3 03 3 max 3 3 2 t!a t 33 2

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