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Motion in one dimension- total trip time

  1. Aug 27, 2013 #1

    QuantumCurt

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    1. The problem statement, all variables and given/known data

    A person takes a trip, driving with a constant speed of 94.0 km/h except for a 22.0-min rest stop. The person's average speed is 71.5 km/h.
    (a) How much time is spent on the trip?

    (b) How far does the person travel?

    This is for introductory general physics. We just finished the first chapter on motion in one dimension. We derived the kinematics equations today, but I can't seem to apply one of the standard kinematics to this. I don't know what I'm missing here.



    3. The attempt at a solution

    I let the total displacement equal

    [tex]\Delta{x}=v_{1}t_{1} \ \ \ \ \ v_{1}=94.0km/hr[/tex]

    [tex]\Delta{t}=t_{1}+t_{2} \ \ \ \ \ t_{2}=22.0min \ \ \ rest \ stop[/tex]

    [tex]\Delta{x}=v_{2}(t_{1}+t_{2}) \ \ \ \ v_{2}=71.5km/hr \ \ average \ speed[/tex]

    Then I set the two [itex]\Delta{x}[/itex] values equal to each other.

    [tex]v_{1}t_{1}=v_{2}(t_{1}+t_{2})[/tex]

    Now I need to solve for [itex]t_{1}[/itex], but I can't find a way of doing so that will work. I feel like I'm making this a lot more complicated than it needs to be. Anyone have any suggestions?
     
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  3. Aug 27, 2013 #2

    TSny

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    Your work looks good. Try using the distributive law of multiplication to write the right hand side of your equation as ##v_{2}t_{1}+v_{2}t_{2}##.
     
  4. Aug 28, 2013 #3

    QuantumCurt

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    I tried that before, and I seemed to hit a dead end.

    [tex]t_{1}=\frac{v_{2}t_{1}+v_{2}t_{2}}{v_{1}}[/tex]

    Now I need to eliminate the [itex]t_{1}[/itex] that's still on the right hand side, correct? I can't see any way of doing that without losing the [itex]t_{1}[/itex] on the left hand side. I could multiply the whole thing by the reciprocal of the [itex]t_{1}[/itex], but that would cancel it on the left hand side, and then I'd still have it in the denominator in the right hand side, which would lead me back to exactly where I'm at now.
     
  5. Aug 28, 2013 #4

    gneill

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    Perhaps a slightly different approach to the equations will help. If you let T be the total time for the trip, and R the time spent resting, then can you write two equations for the total distance traveled, one using the driving speed and one using the average speed?
     
  6. Aug 28, 2013 #5
    I suggest a different approach to the equations like gneill, too. But the only thing to consider is what really is the value of the average speed. I say this because it is very important not losing sight of the goal when you handle the equations.
     
    Last edited: Aug 28, 2013
  7. Aug 28, 2013 #6

    TSny

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    If you want to stick with your original approach, you have ##v_1t_1 = v_2t_1+v_2t_2##

    You already know the value of ##t_2##. So, as you say, you only need to find ##t_1##. You have two terms in the equation with ##t_1##. Can you get those two terms on the same side of the equation?

    It's going to be like solving ##5x = 2x + 6##.

    Of course, you can take a different approach like gneill and LeonhardEu suggest if you want. But I think you're already close to the answer.
     
  8. Aug 28, 2013 #7

    QuantumCurt

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    Can I treat the [itex]v_{2}[/itex] and [itex]v_{1}[/itex] as being coefficients, like this?


    [tex]v_{1}t_{1}=v_{2}t_{1}+v_{2}t_{2}[/tex]

    Then subtract the [itex]v_{2}t_{1}[/itex]

    [tex](v_{1}-v_{2})t_{1}=v_{2}t_{2}[/tex]

    Then divide to get-

    [tex]t_{1}=\frac{v_{2}t_{2}}{v_{1}-v_{2}}[/tex]

    Then solve to get-

    [tex]t_{1}=\frac{(71.5 \ km/hr)(22.0 \ min)(\frac{1 \ hr}{60 \ min})}{(94.0 \ km/hr-71.5 \ km/hr)}[/tex]

    To get 1.17 hours total time spent on the trip?

    edit-No, that wouldn't be total time. I'd need to add the 22.0 min at the rest stop for total time, right? That would give me a total time of 1.54 hours.
     
    Last edited: Aug 28, 2013
  9. Aug 28, 2013 #8

    TSny

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    Yes. That's right. Good work.
     
  10. Aug 28, 2013 #9

    QuantumCurt

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    Awesome, thanks for the help! I didn't think of the fact that I could basically treat the velocities as being coefficients of the time. That seems a lot more clear now.
     
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