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CHAPTER 2 - 21

# CHAPTER 2 - 21 - ,asshowninthediagram...

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initial action takes place, as shown in the diagram.   The initial speed is (50 km/h)/(3.6 ks/h) = 13.9 m/s. If she decides to stop, we find the minimum stopping distance from  v 1 2  =  v 0 2  + 2 a 1 ( x 1  –  x 0 ); 0  = (13.9 m/s) 2  + 2(– 6.0 m/s 2 ) x 1   , which gives   x 1  = 16 m. Because this is less than  L 1   , the distance to the intersection, she  can safely stop in time. If she decides to increase her speed, we find the acceleration  from the time to go from 50 km/h to 70 km/h (19.4 m/s): v  =  v 0  +  a 2 t  ; 19.4 m/s = 13.9 m/s +  a 2 (6.0 s), which gives  a 2  = 0.917 m/s 2 .   We find her location when the light turns red from x 2  =  x 0  +   v 0 t 2  +  ! a 2 t 2 2  = 0 + (13.9 m/s)(2.0 s) +  ! (0.917 m/s 2 )(2.0 s) 2  = 30 m.    Because this is  L 1   , she is at the beginning of the intersection, but moving at high speed.   She should decide to stop!
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