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One dimensional Kinematics help

  1. Apr 7, 2015 #1
    1. The problem statement, all variables and given/known data
    2. An elevator ascends with an upward acceleration of 4.0 ft/s2. At the instant its upward speed is 8.0 ft/s, a loose bolt drops from the ceiling of the elevator 9.0 ft from the floor. Calculate:
    a. the time of flight of the bolt from ceiling to floor.
    b. the distance it has fallen relative to the elevator shaft.

    2. Relevant equations
    y=yinitial+vt+1/2at2

    3. The attempt at a solution
    assuming that elevator starts at bottom of shaft and y=0 is at bottom of shaft.

    a) y value of bolt has to equal the y value of the elevator.
    BOLT EQUATION:
    y = 9+8t-(1/2)gt^2
    FLOOR OF ELEVATOR EQUATION:
    y=8t+2t^2

    set bolt and floor equation equal to each other to get t = 1.14 seconds.

    b) delta y of bolt respect to shaft = 8t -1/2gt^2
    plug in 1.14 for t to get 2.75 meter increase?

    please tell me if i am going wrong and where....
     
  2. jcsd
  3. Apr 7, 2015 #2

    haruspex

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    Looks right so far..
    Too long. Did you try substituting this back in both equations as a check?
     
  4. Apr 7, 2015 #3
    are you sure its wrong? i double checked the arithmetic and i found it correct. i also plugged the numbers back into both equations, and they are equal.
     
  5. Apr 7, 2015 #4
    Something seems off. What value are you using for g? The problem uses imperial units but in part b) you give the answer in metric.

    The other catch is that in part a) you define the distance y=0 to be the bottom of the shaft. But clearly it must have travelled some distance first before reaching the speed of 8ft/s. So I would redefine y=0 to be the point at which the elevator reaches 8 ft/s.
     
  6. Apr 7, 2015 #5
    yeah ****. i was using 9.8. my bad...
     
  7. Apr 7, 2015 #6
    My suggestion is to get into the habit of always, always plug in the units along with the numbers. You can catch lots of mistakes this way.
     
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