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Homework Help: Yet Another Motion Word Problem

  1. Feb 12, 2006 #1
    I encountered yet another difficult one but this time, this problems a bit funny because theres 2 different cases that need to be put into consideration and I don't know exactly how to approach this question.

    It goes like this:
    "A Pedestrian is running at her maximum speed of 6.0m/s to catch the bus stopped by a traffic light. When she is 25m from teh bus, the light changes and the bus accelerates uniformly at 1.0m/s^2. Find either (a) how far the pedestrian has to run to catch the bus or (b) her "frustration distance"."

    I understand the first portion of the question, that is to check if she actually does catch the bus within 6 seconds. But, I don't exactly know how to show the work for this portion. What should I do?
  2. jcsd
  3. Feb 12, 2006 #2


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    Well, she will not catch the bus if it exceeds 6 m/s, and is still further from her.

    So the bus, accelerating at 1 m/s2, will achieve a speed of 6 m/s in 6 seconds, by v = at, and will have moved distance of x = 1/2 at2.

    In order to catch the bus, the pedestrian must cover 25 m + x, where x is the distance the bus travels in that time. I presume the "frustration" distance is the distance that the bus does travel, which represents the extra distance the pedestrian travels.
  4. Feb 12, 2006 #3
    the problem doesn't say anything about having only 6 seconds to catch the bus. did you just forget to put in that part when you typed the question? if she really does only have 6 seconds then you can do this:

    the equation for displacement "delta x" is
    delta x=v0t+(1/2)at^2 v0 = initial velocity a=acceleration t=time

    after 6 seconds the bus travels a certain distance and if the person has to catch it then she must travel that same distance the bus moves PLUS 25meters since she's behind it.

    first calculate the distance the bus moves after 6 seconds using
    delta x1=v0(6.0s)+(1/2)a(6.0s)^2

    now calculate the distance the girl moves during that time using
    delta x2=v0(6.0s)+(1/2)a(6.0s)^2

    delta x2 must equal "delta x1 + 25m" if she catches the bus
  5. Feb 12, 2006 #4
    The question is as I have typed. Nothing was left out.

    @Astronuc - So presumably, the bus is closest at 6 seconds then?

    @Steve23063 - So after doing that process, if she does indeed catch the bus, does that mean I wouldn't have to find her frustration distance?

    By trying your method:

    Δd = ViΔt + 1/2(a)(Δt)^2
    = 0(6s) + 1/2(1.0m/s^2)(6s)^2
    = 18m

    Δd1 = ViΔt + 1/2(a)(Δt)^2
    = (6m/s)(6s) + 1/2(0m/s^2)(6s)^2
    = 36m

    But according to what you said, Δd1 = Δd + 25 in order for her to catch the bus. It's already apparent that the distance the woman covers and the bus covers isn't the same. So would this mean that she doesn't catch it?
    Last edited: Feb 12, 2006
  6. Feb 12, 2006 #5


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    She has a maximum speed of 6.0 m/s and the bus is accelerating at 1 m/s2. If she does not catch the bus in 6 seconds it will be going faster than she is and she can't possibly catch it.
  7. Feb 12, 2006 #6
    @HallsofIvy - So that does mean infact that at 6 seconds, she is closest to the bus..?

    Ok so the answer to the problem is 7.0m ..

    From what Astronauc just told me, I'm wondering is the problem really this easy?

    Well we know from what he said was that it the woman is 6 seconds closest to the bus from:

    Vf = Vi + aΔt
    6m/s = 0m/s + (1m/s^2)Δt
    Δt = 6s

    Then by finding the distance of the bus:

    Δd = ViΔt + 1/2(a)(Δt)^2
    = 0(6s) + 1/2(1.0m/s^2)(6s)^2
    = 18m

    Although the distance of the bus is 18m and the woman is 25m away, that means she has to travel 7m? 1 thing I don't get about this logic is that isn't the 18m the distance of the bus AFTER it travels in 6 seconds? So in fact, wouldn't she have to travel 25m + 18m to catch up with the bus assuming she does?
  8. Feb 12, 2006 #7
    yes, good point. sorry it was almost 4am when i typed my response.
    forevergone what you did was fine. you got 18m for the bus and 36m for the person. If the bus traveled 18m she must travel that same distance plus an extra 25m to make up for the fact that she was behind the bus. She should have traveled 18+25 meters but she only traveled 36 so she didn't catch the bus. finally, i'm confused about the meaning of "frustration distance"
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