Answer
The maximum speed of the elevator is 5.94 m/s.
Work Step by Step
Let's assume that the normal force $F_N$ on the passenger is $1.60 \times$ the passenger's weight (which is mg).
$\sum F = ma$
$F_N - mg = ma$
$1.60~mg - mg = ma$
$a = 0.6~g = 5.88~m/s^2$
We can find the speed after moving a distance of 3.0 meters.
$v^2 = v_0^2+2ay = 0 + 2ay$
$v = \sqrt{2ay} = \sqrt{(2)(5.88~m/s^2)(3.0~m)}$
$v = 5.94~m/s$
The maximum speed of the elevator is 5.94 m/s.
