Answer:
(a) for acceleration a = - 3.1 m/s^2 the stopping distance is 125.43 m.
(b) for acceleration a = - 3.1 m/s^2 the stopping distance is 70.23 m.
Step-by-step explanation:
reaction time , t = 0.5 s, u = 95 km/h = 95 x 5 / 18 = 26.38 m/s
(a) a = - 3.1 m/s^2
Let the stopping distance is s.
The distance traveled by the automobile in reaction time t is
s1 = u x t = 26.38 x 0.5 = 13.19 m
The distance traveled before stopping be s2.
v^2 = u^2 - 2 a s2
0 = 26.38^2 - 2 x 3.1 x s2
s2 = 112.24 m
Total stopping distance, s = s1 + s2 = 13.19 + 112.24 = 125.43 m
(b) a = - 6.1 m/s^2
Let the stopping distance is s.
The distance traveled by the automobile in reaction time t is
s1 = u x t = 26.38 x 0.5 = 13.19 m
The distance traveled before stopping be s2.
v^2 = u^2 - 2 a s2
0 = 26.38^2 - 2 x 6.1 x s2
s2 = 57.04 m
Total stopping distance, s = s1 + s2 = 13.19 + 57.04 = 70.23 m