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How to find the length of time for which the lift is travelling at constant speed? Watch

    • Thread Starter

    Dear TSR people,

    I am currently stuck on part iii) of the question you can see below. The answer to part ii) is 0.8ms-2. However I have no idea how to approach part iii), other then establish that the constant speed = 10ms-1 and find the total distance and total time left after reaching the constant speed.

    That does not help me get any closer to figuring out what to do next though. Could somebody please clarify this for me?

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    Thank you in advance for your time and patience,


    So you know that in total it takes 1 minute, of which 12.5s is for the first part which is the acceleration.

    You can use the speed-time graph that you have to determine the middle phase which is constant velocity, so a = 0, which will tell you the time, if you have the right equations in hand.
    • Study Helper

    (Original post by DeadManProp)
    So, you can work out the constant speed.

    Suppose it travels t seconds at the constant speed.
    How long does it decelerate for in the final stage, as a function of t?

    Then express the area under the velocity time graph as a function of t (it will be a quadratic), equate to 500 and solve for t.

    You'll get 2 values, one of which can't be true....

    Edit: Yep, you found a better method below.
    • Thread Starter

    Ahhhh, thank you, I forgot to consider the fact that the area under the graph represents the distance.

    This made me realise something obvious. The graph is a trapezium with parallel sides of lengths 60 and t, and height 10. The area under it equals 500. Therefore 500=0.5(60+t)*10. => t=40

    From there I can calculate the rest.

    Thanks a lot ;] Sometimes it's the simple fact that unlocks the way.
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