CHAPTER 2 - 17 - H=!gt =!(9.80m/s )(3.88s) =73.9m. 1 2 2 2...

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x t t 1 Speeder Police officer H  =  ! gt 1 2  =  ! (9.80 m/s 2 )(3.88 s) 2  =        73.9 m . ( c ) We find the speeds from v 1  =  v 01  +  at 1  = 0 + (9.80 m/s 2 )(3.88 s) =        38.0 m/s v 2  =  v 02  +  at 2  = 30.0 m/s + (9.80 m/s 2 )(3.88 s – 2.00 s) =        48.4 m/s . 78. We convert the units: (110 km/h)/(3.6 ks/h) = 30.6 m/s.              ( a ) We use a coordinate system with the origin where the motorist  passes the police officer, as shown in the diagram.   ( b ) The location of the speeding motorist is given by  x m  =  x 0  +   v m t which we use to find the time required: 700 m = (30.6 m/s) t , which gives         t  = 22.9 s ( c ) The location of the police car is given by  x p  =  x 0  +   v 0p t   +  ! a p t 2  = 0 + 0 +  ! a p t 2 which we use to find the acceleration: 700 m =  ! a p (22.9 s) 2 , which gives         a p  = 2.67 m/s 2 ( d ) We find the speed of the officer from v p   v 0p   +   a p t ; = 0 + (2.67 m/s
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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 - 17 - H=!gt =!(9.80m/s )(3.88s) =73.9m. 1 2 2 2...

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