#### Answer

At t = 1.0 s, the person's apparent weight is 1040 N
At t = 5.0 s, the person's apparent weight is 735 N
At t = 9.0 s, the person's apparent weight is 585 N

#### Work Step by Step

Let $F_N$ be the normal force of the elevator floor pushing up on the person. Note that $F_N$ is equal in magnitude to the person's apparent weight.
The acceleration $a_y$ is equal to the slope of the $v_y$ versus time graph.
At t = 1.0 s:
$a_y = \frac{\Delta v_y}{\Delta t}$
$a_y = \frac{8.0~m/s-0}{2.0~s-0}$
$a_y = 4.0~m/s^2$
We can use $a_y$ to find the apparent weight.
$\sum F = ma_y$
$F_N- mg = ma_y$
$F_N = m(g+a_y)$
$F_N = (75~kg)(9.80~m/s^2+ 4.0~m/s^2)$
$F_N = 1040~N$
The person's apparent weight is 1040 N
At t = 5.0 s:
$a_y = \frac{\Delta v_y}{\Delta t}$
$a_y = \frac{8.0~m/s-8.0~m/s}{6.0~s-2.0~s}$
$a_y = 0$
We can use $a_y$ to find the apparent weight.
$\sum F = ma_y$
$F_N- mg = ma_y$
$F_N = m(g+a_y)$
$F_N = (75~kg)(9.80~m/s^2+ 0)$
$F_N = 735~N$
The person's apparent weight is 735 N
At t = 9.0 s:
$a_y = \frac{\Delta v_y}{\Delta t}$
$a_y = \frac{0-8.0~m/s}{10.0~s-6.0~s}$
$a_y = -2.0~m/s^2$
We can use $a_y$ to find the apparent weight.
$\sum F = ma_y$
$F_N- mg = ma_y$
$F_N = m(g+a_y)$
$F_N = (75~kg)(9.80~m/s^2-2.0~m/s^2)$
$F_N = 585~N$
The person's apparent weight is 585 N