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Kinematics with an elevator that has 2 acceleration constants

  1. May 9, 2006 #1
    The Problem:

    A car is to be hoisted by elevator to the fourth floor of a parking garage, which is 48ft. above the ground. If the elevator can accelerate at [itex]0.6 \frac{ft.}{s^2}[/itex], decelerate at [itex]0.3 \frac{ft.}{s^2}[/itex], and reach a maximum speed of [itex]8 \frac{ft.}{s}[/itex], determine the shortest time to make the lift, starting from rest and ending at rest.

    Work:

    [tex]s\,=\,s_0\,+\,v_0\,t\,+\,\frac{1}{2}\,a_c\,t_2[/tex]

    [tex]s_2\,=\,\left(0\right)\,+\,\left(0\right)\,\left(t_1\right)\,+\,\frac{1}{2}\,\left(0.6\,\frac{ft.}{s^2}\right)\,\left(t_1^2\right)[/tex]

    [tex]48\,ft.\,=\,\left(s_2\right)\,+\,\left(8\frac{ft.}{s}\right)\,\left(t_2\right)\,+\,\frac{1}{2}\,\left(-0.3\,\frac{ft.}{s^2}\right)\,\left(t_1^2\right)[/tex]

    [tex]s_2\,=\,0.15t_2^2\,-\,8t_2\,+\,48[/tex]

    [tex]0.3\,t_1^2\,=\,0.15\,t_2^2\,-\,8\,t_2\,+\,48[/tex]


    Here is my list of variables, starting with the final ones and going down to the initial ones:

    [tex]v_3\,=\,0\,\,&\,\,s_3\,=\,48\,ft.[/tex]
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    [tex]t_2\,=\,?[/tex]
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    [tex]v_2\,=\,8\,\frac{ft.}{s}\,\,&\,\,s_2\,=\,?[/tex]
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    [tex]t_1\,=\,?[/tex]
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    [tex]v_1\,=\,0\,\,&\,\,s_1\,=\,0[/tex]


    When I use other constant acceleration equations to solve for a position (s), I get an impossible answer like 53.3 ft. when the elevator is only supposed to go to 40 ft. at max! Please help, thanks in advance.
     
    Last edited: May 9, 2006
  2. jcsd
  3. May 9, 2006 #2

    lightgrav

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    While getting up to the top speed 8 ft/s , the average speed would be 4 ft/s;
    this would take v_max/a = t = 13.3 sec , and the car would be past the top already !
    So there's no constant speed interval, just accel time + decell time.
     
    Last edited: May 10, 2006
  4. May 10, 2006 #3
    Like this?

    [tex]0.6\,\frac{ft.}{s^2}\,\left(X\right)\,=\,8\,\frac{ft.}{s}[/tex]

    [tex]X\,=\,13.3\,s[/tex]

    So it takes 13.3 seconds for the elevator to reach maximum velocity?
     
  5. May 10, 2006 #4

    lightgrav

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    and twice as long to get back to zero speed. 40 seconds total, at average speed 4 ft/s, is 160 ft high ... the car would have to be going that far before it can reach maximum speed.

    YOUR car is only going 48 ft.
    So add the accelerating time to the decelerating time to get the total.
    you have to decelerate to zero speed, and end up at 48 ft.
     
    Last edited: May 10, 2006
  6. May 10, 2006 #5
    I got those results earlier as well, but heres the thing, the answer in the back of the book says that total time t is 21.9 seconds, not 40!

    I know it has something to do with relating the missing variables of the three points in the motion on the elevator I listed above, but which relationship?

    1) [tex]v\,=\,v_0\,+\,a_c\,t[/tex]

    2) [tex]s\,=\,s_0\,+\,v_0\,t\,+\,\frac{1}{2}\,a_c\,t^2[/tex]

    3) [tex]v^2\,=\,v_0^2\,+\,2\,a_c\,\left(s\,-\,s_0\right)[/tex]

    I think I am supposed to form a differential equation out of one of the three relationships above, right?
     
  7. May 10, 2006 #6

    Hootenanny

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    No differential equations are required. You basicially have three sections. A period when the lift is accelerating from zero to it's maximum speed. A period when it is travelling at its maximum speed and a peroid when it is deccelerating from its maximum speed back to zero. You have all the information you need. I'll get you started;

    (1)Find the distance travelled and time taken for the evlevator to accelerate from 0 to it's max speed.

    (2)Find the distance travelled and time taken for the evlevator to deccelerate from its max speed to zero.

    There's one more step you need to do. Can you go from here?

    ~H
     
  8. May 10, 2006 #7
    So the variables I listed before are wrong? The elevator does not reach it's top speed, so I have to replace 8 ft/s above with v_max here?

    [tex]v_3\,=\,0\,\,&\,\,s_3\,=\,48\,ft.[/tex]
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    [tex]t_2\,=\,?[/tex]
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    [tex]v_2\,=\,v_{max}\,\,&\,\,s_2\,=\,?[/tex]
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    [tex]t_1\,=\,?[/tex]
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    [tex]v_1\,=\,0\,\,&\,\,s_1\,=\,0[/tex]
     
    Last edited: May 10, 2006
  9. May 10, 2006 #8

    Hootenanny

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    Yeah, sorry, just replace 8 ft/s with a variable v max. You will have to use some simultaneous equations, but the process is the almost the same, except the elevator is never actually travelling at a constant speed. It will accelerate then immediatly decelerate. You will have to find the distance travelled first.

    ~H
     
  10. May 10, 2006 #9
    This is quit a routine problem once it is realised that the maximum speed possible as given in the question is never attained.

    First we have a phase where the lift is accelerating and then the deceleration.Now for the lift to start at rest and to end up at rest at the required height, there is only one possible time for the ascent which can be calculated as follows:

    Let [itex]t_1[/itex] be the time taken for acceleration.
    So
    [tex]S_{1}=\frac{1}{2}a_{1}t_{1}^2[/tex]

    Then the lift has to deccelerate a distance[itex]S_{2}[/itex] to attain 0 velocity.The lift has already attained a velocity = [itex]a_{1}t_{1}[/itex]

    Now to deccelerate back to 0 distance covered will be

    [tex]S_{2}=\frac{v^2}{2a_{2}}=\frac{{a_{1}t_{1}}^2}{2a_{2}}[/tex]
    So now
    [tex]S=48=S_{1}+S_{2}[/tex]

    And u can get [itex]t_{1}[/itex]
    Then u can easily find [itex]t_{2}[/itex] and add them together to get t which I leave to u.
    And in case you were wondering, I have obtained 21.9 sec as the answer and the method isn't half as hard as it looks.

    It would have been tougher had the const. max. velocity come into play, in which case we would have to maximise a time function with two variables.

    PS: Hi there hoot , I'm back after a long layoff in case you didn't notice.It's great to be back in PF !
     
    Last edited: May 10, 2006
  11. May 10, 2006 #10

    Curious3141

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    Actually, no, it would still be quite straightforward. You know that the lift attains the given max velocity [tex]v_{max}[/tex]. Hence you know the stopping distance required to come to rest with a retardational acceleration of [tex]a_r[/tex], given by [tex]s_r = \frac{v_{max}^2}{2a_r}[/tex]. The time taken for the retardation is simply given by [tex]t_r = \frac{v_{max}}{a_r}[/tex]

    You can find the time required to get up to max speed by [tex]t_f = \frac{v_{max}}{a_f}[/tex]. You can also find the distance from the start for the lift to have gotten up to the max speed from [tex]s_f = \frac{v_{max}^2}{2a_f}[/tex]. You have the distance for the accelerating portion and the retardational portion, so you can subtract from the total span of the lift to get the distance [tex]s_c[/tex] travelled at the constant maximal speed. From that you can find the time taken to travel that leg with [tex]t_c = \frac{s_c}{v_{max}}[/tex]

    Now just add to get the total time [tex]T = t_f + t_c + t_r[/tex] to get the requisite time. No calculus involved.
     
    Last edited: May 10, 2006
  12. May 10, 2006 #11
    Yep Curious you are right. Thanks for the correction.
    I guess I just got carried away by calculus.
     
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