Question
police officer in a patrol car parked in a 70km/h zone observe a passing automobile travelling at a slow. constant speed. believing that of the automobile might be intoxicated the officer starts his car acceleration uniformly to 9km/h in 8s. and maintaining a constant velocity of 90km/h overtakes the motorist 42s after the automobile passed him. knowing 18s clapsed before the officer hegan pursuing the motorists determine 1. the distance the officer traveled before overtaking the motorist.2.the motorists speed. Thank's to anyone to get me an answer.
Answers
1. V = 90km/h = 90000m/3600s. = 25 m/s.
a = (V-Vo)/t = (25-0)/8 = 3.125 m/s^2.
t = 42-18 = 24 s. To catch up.
d = Vo*t + 0.5a*t^2
d = 0 + 0.5*3.125*24^2 = 900 m.
2. d = V*t = 900 m.
V * 42 = 900
V = 21.43 m/s.
a = (V-Vo)/t = (25-0)/8 = 3.125 m/s^2.
t = 42-18 = 24 s. To catch up.
d = Vo*t + 0.5a*t^2
d = 0 + 0.5*3.125*24^2 = 900 m.
2. d = V*t = 900 m.
V * 42 = 900
V = 21.43 m/s.