Answer
(a) $x = \frac{v^2}{2 ~\mu_s ~g}$
(b) The minimum stopping distance is 55 meters.
(c) The minimum stopping distance is 330 meters.
Work Step by Step
(a) We can use a force equation to find the acceleration.
$ma = F_f$
$ma = mg ~\mu_s$
$a = \mu_s ~g$
We can use the deceleration to find the minimum stopping distance.
$x = \frac{0 - v^2}{2a} = \frac{-v^2}{(2)(-\mu_s ~g)} = \frac{v^2}{2 ~\mu_s ~g}$
(b) $v = (95 ~km/h)(\frac{1000 ~m}{1 ~km})(\frac{1 ~h}{3600 ~s}) = 26.4 ~m/s$
We can use this velocity to find the minimum stopping distance.
$x = \frac{v^2}{2 ~\mu_s ~g} = \frac{(26.4 ~m/s)^2}{(2)(0.65)(9.80 ~m/s^2)} = 55 ~m$
The minimum stopping distance is 55 meters.
(c) $x = \frac{v^2}{2 ~\mu_s ~(g/6)} = 6\times \frac{(26.4 ~m/s)^2}{(2)(0.65)(9.80 ~m/s^2)} = 330 ~m$
The minimum stopping distance is 330 meters.
