CHAPTER 2 - 19 -...

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v 2  = 0 y  = 0 H a + v 0 y 2 y 1 v 1 We convert the units for the speed limit:  (50 km/h)/(3.6 ks/h) = 13.9 m/s.   ( a ) If we assume that we are traveling at the speed limit, the time to pass through the farthest  intersection is t 1  = ( d 1  +  d 2  +  d 3  +  d 4  +  d 5  +  d 6 )/ v 1  = (10 m + 15 m + 50 m + 15 m + 70 m + 15 m)/(13.9 m/s) = 12.6 s. Because this is less than the time while the lights are green,        yes ,      you can make it through. ( b ) We find the time required for the second car to reach the speed limit: v max  =  v 02   +   a 2 t 2a ; 13.9 m/s = 0 + (2.0 m/s 2 ) t 2a   , which gives  t 2a  = 6.95 s. In this time the second car will have traveled x 2a  =  v 02 t 1  +  ! a 2 t 2a 2  = 0 +  ! (2.0 m/s 2 )(6.95 s) 2  = 48 m. The time to travel the remaining distance at constant speed is
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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 - 19 -...

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