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Motion Along A Straight Line (Constant Acceleration)

  1. Aug 17, 2014 #1
    1. The problem statement, all variables and given/known data

    Starting from rest, a particle moves along an x axis from point A to point B with constant acceleration a1 = 2.20 m/s2 through displacement d1 = 500 m. Then it slows at constant acceleration a2 through displacement d2 = 300 m from point B to point C where it is again at rest. If the total time is 60.0 s, what is the value of a2?

    2. Relevant equations
    Kinematics Equations
    V = Vo + at
    (X -Xo) = Vt - (1/2)at^2
    V^2 = Vo^2 +2a(X - Xo)
    (X - Xo) = Vot + (1/2)at^2
    (X - Xo) = (1/2)Vo + V)t

    3. The attempt at a solution

    So I layed out a chart with my variables with respect to A1 and A2.

    I tried a a couple things. Solving for the time, then subtracting that time from 60 to use the value for the A2 portion of the kinematics. Kept getting -0.401 as the acceleration whenever I solved for the time. Tried using the final velocity (A1) as the initial velocity (A2) and didn't have much luck.

    Not sure if this is a sig fig thing or if I'm just missing something obvious. Not entirely sure. I'd appreciate the answer + explanation. I've got 3 attempts leftover and would rather not risk losing points on a homework problem. Help please.
     
  2. jcsd
  3. Aug 17, 2014 #2
    What is the final velocity when the 500 m segment is complete? Knowing this, how would you determine the acceleration in the 300 m segment?
     
  4. Aug 17, 2014 #3
    So I got a velocity of 46.904, used that as my initial velocity in the V2 = Vo^2 + 2a(delta X).

    Got an acceleration of 3.66666m/s^2
    Seem right?

    I tried -3.67, but it didn't work. I know those values they gave (500m & 300m) are only one sig fig, but I might be expected to treat them as exact. I'm using Wiley Plus. Any idea how they want the answer? Maybe they just want the value without the negative sign?
     
  5. Aug 17, 2014 #4
    I agree with your value of the final velocity after the first segment. But I obtain another value for acceleration. Show how you get your answer.
     
  6. Aug 17, 2014 #5
    With respect to A2.

    V(final) = 0
    V(initial) = 46.904m/s
    (X - Xo) = 300m

    Using the equation: V^2 = Vo^2 +2a(X - Xo)

    (0)^2 = (46.904)^2 + 2a(300m)
    46.904^2 = 2200

    -2200 = 2a(300m)
    -2200/600 = -3.6666666666 to infinity and beyond.
     
    Last edited: Aug 17, 2014
  7. Aug 17, 2014 #6
    Okay so I ran this by my friend and it looks like the correct answer was -2.02m/s which makes sense to me. Except that when plugging in the values we get a final velocity (with respect to a2) of -31.4m/s or something like that.

    I wouldn't even think twice about it except for the fact that the problem states "from point B to point C where it is again at rest"

    Doesn't this imply that the final velocity (with respect to A2) is 0?
    Even if it didn't the fact that the acceleration was negative and the velocity was negative implies the object sped up, but that makes no sense to me either since the object is supposed to be coming to a halt.

    Looks like I'm missing something conceptual here.
     
  8. Aug 17, 2014 #7

    haruspex

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    The problem seems to be overspecified. It is possible to calculate a2 without using the time information. Indeed, the total time turns out to be about 34 seconds. So the answer you get will depend on which data you use.
     
  9. Aug 17, 2014 #8
    Well given the answer, which was -2.02m/s^2, and the velocity I get when I plug that value into the [V = V0 + at] equation which turns out to be -31.4m/s it looks like the object begins speeding up in the opposite direction. From A forward to B, then backwards towards C.

    Well the final velocity should've been zero no? I don't see why it ended up working out this way?

    With respect to A2.

    t = 38.68s
    a = -2.02m/s^2
    V = -31.4m/s
    Vo = 46.904m/s
    X-Xo = 300m
     
  10. Aug 18, 2014 #9

    haruspex

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    I explained this in post #7. The information provided is inconsistent. The answer you get will depend on which piece of information is considered incorrect.
    That said, I tried discarding each given piece of information in turn, and none gave the book answer. This makes me suspect the book answer uses an inconsistent mix of data.
     
  11. Aug 18, 2014 #10
    Well I'm using Wileyplus for my first physics course (Physics for Science & Engineers), this was one of the problems they assigned.

    So you're saying the information is inconsistent, implying that this problem is flawed right? Goood. I thought I was just stupid for a minute.
     
  12. Aug 18, 2014 #11

    haruspex

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    Have you double checked that you quoted the problem exactly?
     
  13. Aug 18, 2014 #12
    Starting from rest, a particle moves along an x axis from point A to point B with constant acceleration a1 = 2.20 m/s2 through displacement d1 = 500 m. Then it slows at constant acceleration a2 through displacement d2 = 300 m from point B to point C where it is again at rest. If the total time is 60.0 s, what is the value of a2?


    Copy and pasted.
     
  14. Aug 19, 2014 #13

    haruspex

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    fair enough.
     
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