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Problem on Uniform acceleration

  1. Jun 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Please can somebody help me,I have got this problem from my college but i have NO clue how to solve the problem, this is how it goes:

    A lift rises 50m in 15 seconds . During the first quarttance distance the velocity is uniformly accelerating and during the last tenth portion it is uniformly retarded. if the velocity is constant durning the centre portion of travel calculate:

    -1) the velocity in the centre portion of travel
    -2) the acceleration and the retardation values
    -3) why has question 2 got both positive and negative values



    please help me in any way you can by either solving the problems and explanation of solving
    Any help would be highly appreciated and thank you in advance.
     
  2. jcsd
  3. Jun 3, 2009 #2

    drizzle

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    this should be posted in the Introductory Physics

    start with solving 1, find the speed which is constant, first you do need to subtract [¼(15 sec) + 1/10(15 sec)] from 15, as for the last question why would the retardation be negative you think? If it isn’t, then the elevator will crash up there!!
     
  4. Jun 3, 2009 #3

    rl.bhat

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    In the problem 1/4th and 1/10th of the distance is mentioned not the time.
    Now the lift starts from rest and acquires a velocity v in quarter distance in time t1. The average velocity is v/2 and distance moved is d1. Then d1= 1/2*vt1. It moves some distance d2 with uniform velocity during time t2 which is equal to vt2. And the remaining distance d3 during time t3 which is equal to 1/2*v*t3. Add these three and solve for v.
     
  5. Jun 3, 2009 #4

    drizzle

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    gheeeeeeeeeeeeee you're right how did I miss it, still sleepy I think


    note: the third part of your solution should be negative, that is -1/2*v*t3:wink:
     
    Last edited: Jun 4, 2009
  6. Jun 3, 2009 #5

    drizzle

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    for a better understanding:

    in the quarttance distance part: d1=[(v1-v0)/2]*t1 ; v0=0

    in mid-distance: d2=[v1]*t2 ; the velocity in midway v1 is constant

    in the tenth portion part: d3=[(v3-v1)/2]*t1 ; v3=0, the lift stops

    retyping those as follows

    d1=[v1 /2]*t1 ==> d1=1/4[50 m]

    d2=[v1]*t2 ==> d2= 50-[d1+d3]

    d3=[-v1/2]*t3 ==> d3=1/10[50 m]

    v1 here is the velocity in the centre portion of travel

    what a lucky guy you are taichi2910 you got an answer :tongue:
     
    Last edited: Jun 3, 2009
  7. Jun 3, 2009 #6

    rl.bhat

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    The average velocity is positive.
     
  8. Jun 3, 2009 #7

    drizzle

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    the last velocity value SHOULD be negative so you can calculate the retardation and get a negative value:wink:
     
  9. Jun 4, 2009 #8

    rl.bhat

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    In the equation v = vo - at, vo need not be negative. In the problem the direction of the velocity remains the same from beginning to end.
     
  10. Jun 4, 2009 #9

    drizzle

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    where did you get the minus from?!! in the known kinematic equation it's plus, that is;

    v = vo + at
     
  11. Jun 4, 2009 #10

    rl.bhat

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    In the given problem, in the last portion of the motion, the lift is moving with the retardation. The above equation is for that portion of the motion. In your post you have mentioned that the velocity in that portion SHOULD be negative.
     
  12. Jun 5, 2009 #11

    drizzle

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    hey rl.bhat:smile:, first you prove it mathematically throughout your conclusions in analyzing the motion of the lift, when the (minus) sign appears then you have a physical explanation to it, which is the lift is stopping after 15 secs and it retardates.
     
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