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Average velocity problem

  1. Sep 15, 2006 #1
    Hello Ive been doing my physics homework but sadly I cant solve this problem and dont really know how to approach it. If anyone could, please guide me to how to solve this problem.

    In reaching her destination, a backpacker walks with an average velocity of 1.34m/s due west. This average velocity results becasue she hikes for 6.44km with a average velocity of 2.68 m/s. due west, turns around, and hikes with an average velocity of 0.477 m/s, due east. How far east did she walk?

    If anyone please could, guide me through this problem cause Iam stumped.
     
  2. jcsd
  3. Sep 15, 2006 #2

    Hootenanny

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    Welcome to the Forums,

    One is expect to show ones effort when asking for help. HINT:

    [tex]\vec{v} = \frac{d\vec{s}}{dt}[/tex]

    v is velocity, s is displacement and t is time.
     
    Last edited: Sep 15, 2006
  4. Sep 15, 2006 #3
    Ok this is what Ive done.

    1X = 6440m
    1V = -2.68m/s
    1T = 2402s (I found the time)

    Then for stage 2

    2V = .447m/s
    2X = ?
    2T = ? (These 2 unknowsn are what are throwing me off)

    at first I was going to try to find the final velocity at the end of stage one
    but the it never said acceleration was constant so i couldnt do V = 1/2(Vo + Vf)

    Honeslty Iam just mainly stumped on this one seems like I have 1 to many unknowns
     
  5. Sep 15, 2006 #4

    berkeman

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    The other thing that you know is her overall average velocity. The overall average velocity will apply to her whole hike:

    [tex]V_{ave} = \frac{(V_1 * T_1) + (V_2 * T_2)}{T_1 + T_2}[/tex]
     
  6. Sep 15, 2006 #5

    radou

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    Right. You can get t2 directly from the expression above, and then you can easily retrieve the length of the 'eastern walk' from s2 = v2*t2. Don't forget that the minus sign will stand for a change of direction, and converse either [m/s] to [km/h], or [km] to [m].
     
  7. Sep 15, 2006 #6
    <img src ="https://www.physicsforums.com/latex_images/10/1085467-0.png">

    I am pretty sure I cant use that equation becasue it is no where in the chapter, I iam 100% sure they dont expect me to use something they didnt show. So there has to be another way.

    They also already state the overall average velocity. "a backpacker walks with an average velocity of 1.34m/s due west"
     
    Last edited: Sep 15, 2006
  8. Sep 16, 2006 #7

    Hootenanny

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    Yes, but you have two unknowns (t2 and x2, so you need two equations to solve them simultaneously. You know the average velocity and you must use this in the equation Berkeman stated, I'm pretty sure you would have seen this equation somewhere but perhaps in a different form?

    [tex]\overline{v} = \frac{v_{1}\cdot t_{1} + v_{2}\cdot t_{2}}{t_{1} + t_{2}} = \frac{x_{1} + x_{2}}{t_{1}+t_{2}} = \frac{\Delta x}{\Delta t}[/tex]

    Does this look more familiar now?
     
  9. Sep 16, 2006 #8
    I c. but isnt deltaX and deltaT x2-x1 not x1+x2??
     
  10. Sep 16, 2006 #9

    Hootenanny

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    Not in this case as the hiker started at x = 0, I suppose you could add a term in if you like;

    [tex]\overline{v} = \frac{v_{1}\cdot t_{1} + v_{2}\cdot t_{2} - x_{0}}{t_{1} + t_{2} \right)} = \frac{x_{1} + x_{2} - x_{0}}{t_{1}+t_{2}} = \frac{\Delta x}{\Delta t}[/tex]

    So, in this case the final position of the hiker is given by xf = x1 + x2. Do you follow?
     
  11. Sep 16, 2006 #10
    yes Thanks you guys/girls alot this really helped!!!
     
  12. Jan 26, 2008 #11
    If any one gets a chance... I have the exact same problem except with different numbers and i just cant seem to come up with the correct answer. here is what i have tried... any help appreciated...

    In reaching her destination, a backpacker walks with an average velocity of 1.11 m/s, due west. This average velocity results, because she hikes for 5.37 km with an average velocity of 2.88 m/s due west, turns around, and hikes with an average velocity of 0.342 m/s due east. How far east did she walk (in kilometers)?

    Average V=1.11 due west
    X1= 5370m
    V1= -2.88m/s west
    T1= 1864.58s

    X2= ?
    V2= .342m/s east
    T2= ?

    I have tried it as 1.11= [(-5370)+(.342)(T2)] / (1864.58 + T2)

    I end up w/ T2= 9661.91 and multiplying it by .342 and i get 3.3 kilometers, but it is wrong. Please show me where i messed up, or just tell me if i am WAY off.

    Thanks
     
  13. Jan 27, 2008 #12
    never mind... i took a long break from that problem and came back refreshed and solved it first try... i was way off on my cross multiplication above, man my brain was fried, lol...
    it came out to .777km though incase anyone was wondering.

    Thanks anyways guys. sorry for the long post above, but i was just tired of that problem.
     
  14. Feb 5, 2008 #13
    Hey Jarmen, How did you get that answer? I keep on getting 2.27km, which is no where close to your number.
     
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