What is the shortest possible stopping distance for the car under such conditions?

Question

a 1300Kg car moves along a horizontal road at speed v0=25.2m/s . the road is wet so the static friction coefficient between tires and the road is only Us= .379 and the kinetic friction coefficient is even is even lower, Uk= .2653. Assume acceleration of gravity is 9.8m/s^2.

The normal force between the car and the
road follows from Newton's Laws for the ver-
tical direction of motion: Since the car moves
in the horizontal direction only, ay = 0 and
hence
Fnet
y = N - mg = 0:
Consequently, N = mg,
Fmax
F = usmg;
and since there are no horizontal forces other
than friction, maximum deceleration rate of
the car is
axmax = -1/M Ffmax =
= _USG = -3.7142 M/S^2

Answer 1

Since μk<μs, we don't allow slipping. Then deceleration a = gm*μs/m = g*μs. Then distance = at^2/2 = v^2/(2a).