physics  stopping distance
posted by gladys on .
when a driver brings a car to a stop by breaki.g as hard as possible the stopping can be regarded as the sum of reaction distance and breaking distance. tje following data are given, v1 m/s contains 10,20,30. reaction distance are 7.5, 15, 22.5. the breaking distance are 5.0, 20, 45. and thr last are 12.5, 35, 67.5. a) what is the reaction time around? b) what is the stopping distance if the velociry is 25 m/s?
need the ans. asap

The reaction time is t =x/v=7.5/10 = 15/20 =22.5/30 = 0.75 s.
The reaction distance for v=25 m/s is x=vt=25•0.75 = 18.75 m.
The acceleration is
a=v²/2s=10²/2•5=20²/2•20=30²/2•45=10 m/s²
The breaking distance for v=25 m/s is
s= v²/2a=25²/2•10=31.25 m
The stopping distance is x+s = 18.75+31.25=50 m