Physics Help on forces and acceleration with elevator and bathroom scale?

Question

This is a webassign question. I did all the rest but can't figure this one out.
Here it goes:
A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 847 N.

(a) As the elevator moves up, the scale reading increases to 952 N, then decreases back to 847 N. Find the acceleration of the elevator (m/s/s).

(b) As the elevator approaches the 74th floor, the scale reading drops to 782 N. What is the acceleration of the elevator?
m/s2
(c) Using your results from parts a and b, explain which change in velocity, starting or stopping, would take the longer time.
stopping
starting

(d) What changes would you expect in the scale readings on the ride back down?

Answer 1

Assuming this takes place near the surface of the Earth, we now know the mass of the student:

847 = mg

where g is the local acceleration of gravity, 9.8 m/s^2 down.

(a) Newton's Second Law

F = ma or 952 N = ma

Solve for a, which is the sum of gravity and the acceleration of the elevator. Take away g to get the acceleration of the elevator alone.

(b) Same technique as (a)

(c) Whichever had the larger acceleration has the shortest stopping time.

Answer 2

(a) upward (b) f2 - f1 = m(a2 - a1) because it truly is a linear relation f1 is at relax, so as it truly is f1 = mg the position g is the rigidity of gravity => (660 - 520)g/520 = a2 - 0 =~ 2.sixty 4 m/s^2 (c) efficient acceleration upward (faraway from gravitational rigidity). If the acceleration continues to be an similar, the speed will consistently develop.

Answer 3

How does that even help?