CHAPTER 2 - 15

CHAPTER 2 - 15 - 0=(100km/h(3.6ks/h...

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0 1 2 3 4 5 6 t  (s) B C D A E 0 10 20 30 25 15 5 x  (m) 0 = [(100 km/h)/(3.6 ks/h)] 2  + 2(– 30)(9.80 m/s 2 )( x  – 0), which gives  x  =        1.3 m .   72. If the lap distance is  D , the time for the first 9 laps is t 1  = 9 D /(199 km/h),  the time for the last lap is  t 2  =  D / æ ,  and the time for the entire trial is T  = 10 D /(200 km/h). Thus we have T  =  t 1  +  t 2  ; 10 D /(200 km/h) = 9 D /(199 km/h) +  D / æ , which gives         æ  = 209.5 km/h . 73. We use a coordinate system with the origin at the release point and down positive.  ( a ) The speed at the end of the fall is found from    v 2 v 0 2  + 2 a ( x  –  x 0 = 0 + 2 g ( H – 0), which gives  v   = (2 gH ) 1/2 .   ( b ) To achieve a speed of 50 km/h, we have v  = (2 gH ) 1/2 ;     (50 km/h)/(3.6 ks/h) = [2(9.80 m/s 2 ) H 50 ] 1/2 , which gives

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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 Berkeley.

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CHAPTER 2 - 15 - 0=(100km/h(3.6ks/h...

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