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Serway pr 13 pg 49 college phys 8th ed, kinematics vel, dis

  1. Oct 27, 2013 #1

    GreyNoise

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

    I am stumped on a problem (number 13 page 49) from the Serway/Vuille/Faughn text
    College Physics 8th Ed. The problem is

    A person takes a trip driving with a constant speed of 88.5 km/h, except for a 22 min rest stop.
    If the person's average speed is 77.8 km/h, how much time is spent on the trip and how far does
    the person travel?


    2. Relevant equations

    The answer is 2.80 hours and 218 km (from back of book and 218/2.80 = 77.8). I can't even reverse
    engineer the problem from the answer. Presumably all I need are the GT's

    \begin{array}{cll}
    x_2 & = & x_1 + v_1t + \frac{1}{2}at^2 \\
    &&\\
    x_2 & = & \displaystyle x_1 + v_{ave}t \\
    &&\\
    v_{ave} & = & \displaystyle\frac{v_1 + v_2}{2} \\
    &&\\
    v_2 & = & \displaystyle v_1 + at \\
    &&\\
    v_2^2 - v_1^2 & = & \displaystyle 2a(x_2 - x_1) \\
    \end{array}


    3. The attempt at a solution

    For all of these, I keep coming back to the two unknowns (for me anyway) t and x (distance). I tried
    assuming an acceleration from the rest stop to back on the road again,

    [itex]a = \frac{0+88.5}{0.367}[/itex][itex]\frac{km}{h^2}[/itex], where 0.367 h = 22 mins

    I thought that v2 = v1 + at might lead to segmenting the problem into before and after the rest stop,
    but that got me nowhere (I have been reduced to guessing, so I would have been surprised had it worked);
    I tried leveraging the other velocity by

    [itex]v_{ave}[/itex] = 77.8 [itex]\frac{km}{h} = \frac{88.5 - v_{other}}{2}[/itex][itex]\frac{km}{h}[/itex]

    which left me with vother= 66.1 km/h and wondering what the hell that really meant to the problem anyway.
    Can anyone recommend a solution or just a hint (I'll take either) to this problem for me? The book's examples do
    not cover anything quite like this that I have read (and reread).
     
    Last edited: Oct 27, 2013
  2. jcsd
  3. Oct 27, 2013 #2

    SteamKing

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    Homework Helper

    Well, the problem can be simplified a bit by ignoring acceleration. You aren't given a value for the car's acceleration anyway.

    You know the average speed, the duration of the rest stop, and the constant speed at which the car is traveling otherwise. I think by writing the equation for average speed and using d = r*t for the traveling part, you can solve this thing.

    The acceleration bit would only be necessary if the car drove off a cliff somewhere in an ill-advised short-cut to make up for lost time.
     
  4. Oct 28, 2013 #3

    GreyNoise

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    Thnx SteamKing. Yeah the acceleration was an attempt when I couldn't seem to set up the problem for myself. When you wrote

    "I think by writing the equation for average speed and using d = r*t for the traveling part, you can solve this thing",

    did you mean dist = v_ave*t? I am unable to understand how to get distance and the time from this approach; both of them are the unknowns and I keep needing to fall back on one to get the other; it is like I have one eqaution for the two unknowns.
     
  5. Oct 28, 2013 #4

    SteamKing

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    You need to break the problem down into parts. Your trip consists of driving a certain distance at a constant speed. You take a break of known duration during the trip where you aren't moving. What is the equation for average velocity for this trip? Set up the algebra, defining your unknowns.
     
  6. Nov 7, 2013 #5

    GreyNoise

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    Gold Member

    Thnx SteamKing, got it.
     
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