Vo = 95000m/h * (1h/3600s) = 26.4m/s.
d1 = 26.4m/s * 1s = 26.4m due to human reaction time.
Vf^2 = Vo^2 + 2ad,
d = (Vf^2-Vo^2) / 2a + d1,
d = ((0-(26.4)^2) / -10.4) + 26.4,
d = 67 + 26.4 = 93.4m = total stopping distance.
b Same procedure as a.
Determine the stopping distances for an automobile with an initial speed of 95km/h and human reaction time of 1.0s:
(a) for an acceleration a= -5.2m/s^2
(b) for a= -6.7m/s^2
2 answers
how the negative 10.4 fall in the bracket