Forces [ Finding displacement ]

AI Thread Summary
The discussion revolves around calculating the displacement of an elevator over a 20-second interval while considering varying forces acting on a student inside. The student experiences different weight readings on a scale due to changes in acceleration during the elevator's motion. Initial calculations indicate that the student correctly identified the initial conditions and mass but made errors in applying the equations for displacement during different time intervals. Specifically, the use of final velocity instead of initial velocity in the displacement formula led to incorrect results. The correct approach requires careful application of the kinematic equations to account for the changing acceleration throughout the intervals.
makemefly
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Homework Statement



Assume you are on a planet similar to Earth where the acceleration of gravity is 10m/s^2 and the positive directions for displacement velocity and acceleration are upward. At t=0sec., an elevator is at a displacement of x=0 with a velocity of v=0m/s. A student whose weight is 400N stands on a scale in an elevator and records the scale reading as a function of time. The data is shown below:

0-5s--------400N
5-10s-------500N
10-15s------600N
15-20s-----0N

g) What is the displacement of the elevator
above the starting point at the end of the
fourth 5 s interval (at 20 s)?
Answer in units of m.


I was only completely sure about a and b, for a the direction is up and for b the mass is 40kg. I obviously don't want just answers, but steps and hopefully explanations to go along with it.

Homework Equations



Vf^2 = Vi^2 + 2a X
X = ViT + 1/2 a t ^2

The Attempt at a Solution


Let
a = acceleration
X = displacement
Vi = Initial Velocity
D(X) = Displacement for time interval

Interval : 0 - 5
a = 0
v = 0
X = 0

Int : 5-10
a = 2.5
v = 12.5
X = 0 + 1/2 (2.5) (5) ^2 = 31.25 m

int : 10-15
a = 5
v = 37.5
X = (37.5)(5) +1/2*(5)(5)^2 +31.25 = 281.25 m

int : 15-20
a= -10
v = -12.5
X = (281.25) + (-12.5)(5) + (1/2)*(-10) (5)^2 = 93.75 m

I submit 93.75 as my final answer and it was wrong. Any help or ideas?
 
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hi makemefly! welcome to pf! :smile:

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in your constant acceleration equation s = vit + 1/2 at2, i think you're using vf instead of vi for c and d :redface:
 
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