Answer
(a) The stopping distance is 130 m.
(b) The stopping distance is 69 m.
Work Step by Step
$v = (95 ~km/h)(\frac{1000 ~m}{1 ~km})(\frac{1 ~h}{3600 ~s}) = 26.39 ~m/s$
For the first 0.40 s, the acceleration is 0 because this is the human reaction time. We can calculate the distance traveled in the first 0.40 s.
$x = v_0t = (26.39 ~m/s)(0.40 ~s) = 10.56 ~m$
We can use $x_0 = 10 ~m$ in the next parts of the question.
(a) $x = x_0 + \frac{v^2 - v_0^2}{2a} = 10.56 ~m + \frac{0 - (26.39 ~m/s)^2}{(2)(-3.0 ~m/s^2)} = 126.6 ~m$
The stopping distance is $126.6\approx130$ m.
(b) $x = x_0 + \frac{v^2 - v_0^2}{2a} = 10.56 ~m + \frac{0 - (26.39 ~m/s)^2}{(2)(-6.0 ~m/s^2)} = 68.6 ~m$
The stopping distance is $68.6\approx69$ m.
