you can use one of kinematic equations of motion "final velocity^2 = initial velocity^2 + 2 acc distance"
Clear[vf, vi, a, dist, t]
eq = vf^2 == vi^2 + 2*a*dist;
values = {vf -> 0, vi -> 29, dist -> 71};
(a = a /. First@Solve[eq /. values, a]) // N
And Mathematica says
-5.92254
so it is about -6 m/s. It is negative since it is deceleration. Now you can also find the time, using the other kinematic equation "final velocity=initial velocity+acc*time" (you could also use "dist = (final vel+initial vel)/2 * time)" to solve for time)
eq = vf == vi + a*t;
Solve[eq /. values, t] // N
And Mathematica says
{{t -> 4.89655}}
So it took about 5 seconds to stop.