It takes the elevator in a skyscraper 4.5s to reach its cruising speed of 11m/s . a 70kg passenger gets aboard on the ground floor.
a- what is the passenger's weight before the elevator starts moving?
b-what is the passenger's weight while the elevator is speeding up?
c-what is the passenger's weight after the elevator reaches its cruising speed?

A and C are very simple. Since there is no acceleration in both cases, the weight of the passenger is just  (about 690 N or 150 lbs with correct sig-figs). For B, we must first find the acceleration of the elevator. Since , we can write that . Solving for a, we get that . Plugging this into Newton's second law, we get that . When calculating the apparent weight, this gets added onto the 686 N from A and C, so the final apparent weight is about 857 N, or 860 N with the correct number of sig-figs.

tl;dr:
A - 690 N
B - 860 N
C - 690 N

True will be your answer dude

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