#### Answer

(a) The person's acceleration is $1.40~m/s^2$ downward.
(b) We can not tell if the elevator is speeding up or slowing down.

#### Work Step by Step

(a) Since the person's weight in the elevator is less than the person's weight on the ground, the person must be accelerating downward.
Let $F_N$ be the upward force exerted on the person by the elevator. Note that $F_N = \frac{120}{140}~\times weight = \frac{6mg}{7}$. We can find the person's acceleration:
$\sum F = ma$
$mg-F_N = ma$
$mg-\frac{6mg}{7} = ma$
$a = \frac{g}{7}$
$a = 1.40~m/s^2$
The person's acceleration is $1.40~m/s^2$ downward.
(b) We can not tell if the elevator is speeding up or slowing down. The elevator could be moving downward and speeding up with an acceleration of $1.40~m/s^2$, or the elevator could be moving upward and slowing down at a deceleration of $1.40~m/s^2$.