Answer
$0.73\space m/s^{2}\space in\space downward\space direction$
Work Step by Step
Please see the image first.
Let's apply equation $F=ma$ to the elevator.
$\uparrow F=ma$
Let's plug known values into this equation.
$T-975g=975a$
$8.85\times10^{3}\space N-975\times9.8\space N=(975\space kg)a$
$\frac{(8850-9555)}{975}m/s^{2}=a$
$-0.73\space m/s^{2}=a$
Magnitude & the direction of the acceleration$=0.73\space m/s^{2}\space Downward$
