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Forces [ Finding displacement ]

  • Thread starter makemefly
  • Start date
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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?
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
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welcome to pf!

hi makemefly! welcome to pf! :smile:

(try using the X2 icon just above the Reply box :wink:)

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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