What is the passenger's apparent weight at t=1.0s?

For the complete question provided in explanation, if the elevator moves upward, then the apparent weight will be 1035 N. While for downward motion the apparent weight will be 435 N.

Explanation:

The question is incomplete. The complete question contains a velocity graph provided in the attachment. This is the velocity graph for an elevator having a passenger of 75 kg.

From the slope of graph it is clear that acceleration at t = 1 sec is given as:

Acceleration = a = (4-0)m/s / (1-0)s = 4 m/s^2

Now, there are two cases:

1- Elevator moving up

2- Elevator moving down

For upward motion:

Apparent Weight =  m(g + a)

Apparent Weight = (75 kg)(9.8 + 4)m/s^2

Apparent Weight = 1035 N

For downward motion:

Apparent Weight =  m(g - a)

Apparent Weight = (75 kg)(9.8 - 4)m/s^2

Apparent Weight = 435 N

the answer is b i think

the answer is a. hope this .