#### Answer

The velocity at t = 6.0 seconds is $v = 3.18 ~m/s$.

#### Work Step by Step

Let $F_N$ be the normal force of the elevator pushing up on the person. Note that this force $F_N$ will be equal to the reading on the scale.
We can find the acceleration from t = 0 to t = 3.0 seconds.
$\sum F = ma$
$mg - F_N = ma$
$a = \frac{mg-F_N}{m}$
$a = \frac{(95~kg)(9.80~m/s^2)-830~N}{95~kg}$
$a = 1.06~m/s^2$
We can find the acceleration from t = 3.0 seconds to t = 6.0 seconds.
$\sum F = ma$
$mg - F_N = ma$
$a = \frac{mg-F_N}{m}$
$a = \frac{(95~kg)(9.80~m/s^2)-930~N}{95~kg}$
$a = 0$
For the first 3.0 seconds, the elevator accelerates downward at a rate of $1.06~m/s^2$. The elevator travels at a constant speed after that. We can find the velocity at t = 3.0 seconds.
$v = a~t = (1.06~m/s^2)(3.0~s)$
$v = 3.18~m/s$
Since the elevator does not accelerate after t = 3.0 seconds, the velocity at t = 6.0 seconds is $v = 3.18 ~m/s$