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Movement in a straight line problem

  1. Sep 7, 2009 #1
    1. The problem statement, all variables and given/known data

    A certain elevator cab has a total run of 197 m and a maximum speed is 307 m/min, and it accelerates from rest and then back to rest at 1.21 m/s2. (a) How far does the cab move while accelerating to full speed from rest? (b) How long does it take to make the nonstop 197 m run, starting and ending at rest?

    2. Relevant equations

    i thought you could use v=vo + at
    and v= d/t

    3. The attempt at a solution

    what i did was first convert 307 m/min to 5.12 m/s
    then i said 5.12 = 0 + 1.21 (t)
    then t= 4.23 seconds

    so then i said v= d/t
    and 5.12= d/4.23, so d= 21.66 m

    then i did part b and said 5.12= 197 m/t and t= 38.48 s

    both of my answers were wrong

    please help?
     
  2. jcsd
  3. Sep 7, 2009 #2
    First you find how long it takes to accelerate to 5m/s, which is 4.23s. What you did wrong was to use 5.12 as the velocity. It isn't 5.12 though, it's velocity is only equal to 5.12m/s right at 4.23 seconds. So just use this equation.
    [tex]\Delta x = v_ot + \frac{1}{2}at^2[/tex]
     
  4. Sep 7, 2009 #3
    okay.. so, i got part a right

    but then i tried to use the equation you gave me for part b, and that was wrong :/
     
  5. Sep 7, 2009 #4
    So for (a) you got...10.8?
    Now for b you need to do this in three parts. Part 1, accelerating over distance x (10.8); part 2, traveling at a constant speed over a distance; part 3, decelerating over a distance(10.8).
     
  6. Sep 7, 2009 #5
    yes i got 10.8

    i'm not really sure what you mean by that.. what equations are these coming from?

    (i'm sorry i'm so needy, i've just never learned this stuff..)
     
  7. Sep 7, 2009 #6
    Well the elevator will speed up until it reaches 5.12m/s then it will travel at 5.12m/s until some point and then it will slow down to zero.

    Now if it takes 4.23 to sped up to 5.12m/s it will also take 4.23s to slow down to 0m/s.
    |-------|-------------------------------|-------| 197m
    ..10.8m.......................???......................10.8
     
  8. Sep 7, 2009 #7
    i accidentally entered 43.1 (which was the answer to another problem) and it turned out to be correct in this instance too, but i still can't figure out how to solve it

    i understand what you're saying, but i can't seem to make the connection between the words and the equations you would use
     
  9. Sep 7, 2009 #8
    You need to find the the total time. That would be t = t1 + t2 + t3. t1 is the time accelerating, t2 is the time traveling at a constant velocity, t3 is the time decelerating. You found t1 as 4.23s. Now you know that is the same as t3 since it's doing the exact same thing. Now to find t2 you would use t=x\v. You know v, 5.12, but you need to find x. The total length is 197 but it is only at a constant speed after it starts accelerating and before it starts decelerating. So you can find the distance by taking 197 and subtracting how far it goes for the other 2 parts.
     
  10. Sep 8, 2009 #9
    alright, i've got it now
    thank you for all of your help
     
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