Determine the stopping distances for an automobile going a constant initial speed of 95 km/h and human reaction time of 0.50 s. for an acceleration a = - 3.1 m/s^2 Determine the stopping distances fo an automobile going a constant intial speed of 95 km/h and human reaction time of 0.50 s. for an acceleration a = - 6.1 m/s^2.

(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.

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