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Function of displacement

  1. Feb 4, 2015 #1
    1. The problem statement, all variables and given/known data
    A man runs at a speed of 4.0 m/s to catch a standing bus. When his distance to the door is 6.0 m, the bus starts moving forward and continues with a constant acceleration of 1.2 m/s2.

    1. How many seconds does it take for the man to reach the door ?
    2. If at the beginning he is 10.0 m from the door, will he (running at the same speed) ever catch up?

    2. Relevant equations
    S=vt
    S=1/2 at2

    3. The attempt at a solution
    1. Solution for the first question.
    Function for the man displacement is: Sman=4t+6
    Function for the bus displacement is: Sbus=1/2(1.2)t2=0.6t2
    When the man catches the bus: Sman=Sbus ↔ 4t+6 = 0.6t2 ↔ 0.6t2-4t-6 = 0 (t>0) → t≈ 7.9 s.

    2.Solution for the second question.
    If at the beginning he is 10.0 m from the door, function for his displacement is: Sman=4t+10
    Sman=Sbus ↔ 0.6t2-4t-10 = 0 (t>0) → t ≈ 8.6 s
    Therefore if at the beginning he is 10 m from the door, he will catch the bus after 8.6 s.

    I wonder if my solutions are correct? because I'm not sure whether I should add 6 meters and 10 meters or subtract them from the displacement function of the man ?
     
  2. jcsd
  3. Feb 4, 2015 #2

    BvU

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    Science Advisor
    Homework Helper
    Gold Member

    In the first question you calculate the time needed for the bus to cover the same distance as the man, when the bus starts at zero speed and the man starts 6 m ahead. That's not what you are after !

    Tip: make a graph with time horizontal and position (man, door) vertical.

    By the way, if you calculate 1.5 sec for 6 m, then alarm bells must start ringing if you find almost 8 sec as an answer to a) !

    Some humor input: Where I live, the doors close before the bus takes off., so there's no catching anyway :)

    And: don't try this in your home town, even if the doors close later :)

    Yet another smartass remark: try to distinguish between position and displacement. In your wording, and in the equations. In this exercise you need position (you want the positions to be equal).

    Your displacement of the man equation, therefore, is wrong: s = vt is right. But he is 6 m behind at t-0, so his position wrt the door at t=0 is -6 m and develops with time as -6 m + vt. Whereas the door wrt "the door at t=0" has position as you wrote. (As you notice, position and displacement are then the same, because of the choice of origin for the position coordinate).
     
    Last edited: Feb 4, 2015
  4. Feb 4, 2015 #3
    Thank you very much, BvU. Now I understand.
     
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