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One Dimensional Motion Help

  1. Aug 20, 2009 #1
    1. The problem statement, all variables and given/known data

    A runner hopes to complet the 10,000 m run in less than 30.0 min. After running at constant speed for exactly 27.0 min, there are still 1100 m to go. The runner must tehn acceelerate at .20 s^2 m for how many seconds in order to achieve teh desired time?

    2. Relevant equations

    def of average velocity = t^-1 (X - Xo)

    def of a = t^-1 (V - Vo)

    Def of average velocity (when a is zero) = 2^-1 (V + Vo)

    V = Vo + at

    X = Xo + Vo t + 2^-1 a t^2

    V^2 = Vo^2 + 2a(X - Xo)

    3. The attempt at a solution

    ok for the first part of the race I calculated his velocity has to be

    5.494 s^-1 m

    ok in order to calculate the time to accelerate for the last part of the race

    I saw that

    X = Xo + Vo t + 2^-1 a t^2

    that would do me no good because I do not know the X value for we could stop accelerating and there are two t variables

    V^2 = Vo^2 + 2a(X - Xo)

    doesn't work becasue we don't know the V^2 at the end of the acceleration

    def of a = t^-1 (V - Vo)

    rearanged for t didn't work because we don't know the V at the end of the acceleration...

    So I'm lost...
     
  2. jcsd
  3. Aug 20, 2009 #2

    kuruman

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    You need to figure out the symbols that correspond to the numbers you have.

    Forget the initial 27 minutes of the race.

    What counts here is that he has 3 minutes left to cover 1100 meters, when his initial velocity is 5.494 m/s. So you have a displacement (1100 m), a time (3 min or 180 seconds) and an initial velocity (5.494 m/s). You need to find an acceleration. What equation do you think you should use?

    By the way, when the acceleration is zero, the velocity does not change so the average velocity, the instantaneous velocity and the initial velocity are all the same, v0.
     
  4. Aug 21, 2009 #3
    The question gives us his acceleration for the second part of the race

    The runner must tehn acceelerate at .20 s^2 m for how many seconds in order to achieve teh desired time?

    were asked for time requirered for him to accelerate at the at that speed in order to achieve that desired time
     
  5. Aug 21, 2009 #4
    If he accelerates at .20 s^2 for the rest 180 seconds he would travel

    13130 m total which is to long

    i believe the problem is asking how long must he accelerate at .20 s^2 then stop accelerating and remain at that velocity for the rest of the race...

    Maybe I read it wrong... It's problem 46 in my book so the answer is not in the back of my book and it's a level (III) problem out of 3 levels...
     
  6. Aug 21, 2009 #5
    30.0 min = 1800 s
    27.0 min = 1620 s
     
  7. Aug 21, 2009 #6

    kuruman

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    You didn't read it wrong, I did. Your interpretation is correct. Let's use some symbols.
    Let v0=5.494 m/s

    Then the distance he travels while he is accelerating (call it x1) is

    x1=v0t1+(1/2)at12

    The time that you want is t1 but you need one more equation to find it because you don't know x1. So note that his final speed after he is done accelerating is

    vf=v0+at1

    Then he travels the rest of the distance (1100 m - x1) in time t2. Then
    1100 m - x1 = vf t2 = (v0+at1)t1

    Do you see how to finish the solution now?
     
  8. Aug 21, 2009 #7
    hmmmm let me see

    How do you derive this

    1100 m - x1 = vf t2 = (v0+at1)t1

    ok I agree with this part...
    1100 m - x1 = vf t2
    because there is no acceleration after he stoped acceleration so you can just do
    Vo t to find the X but why do you set it equal to (v0+at1)t1 ?

    vf=v0+at1
    I know were that equation comes from the defintion of acceleration

    I also know how to derive this
    X = Xo + Vo t + 2^-1 a t^2

    I'm just unsure were the equation you gave me came from... Could you please tell me how you got it?

    THANKS!!!
     
    Last edited: Aug 21, 2009
  9. Aug 21, 2009 #8

    kuruman

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    After he is done accelerating, his velocity is constant. Call it vf. The distance that he has left to cover is the initial 1100 m less the distance he already covered under acceleration, 1100 m - x1. Since he travels at constant velocity he covers equal distances in equal times, therefore

    1100 m - xf = vf*t2. OK so far?

    Now use vf = v0 + a t1 to replace vf in the above equation.
     
  10. Aug 21, 2009 #9
    hmmmm ok thanks!
     
  11. Aug 21, 2009 #10
    Wait...
    Is it this
    1100 m - x1 = vf t2 = (v0+at1)t1
    or this
    1100 m - x1 = vf t2 = (v0+at1)t2
     
  12. Aug 21, 2009 #11

    kuruman

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    Sorry, the second one

    1100 m - x1 = vf t2 = (v0+at1)t2

    Too much cutting and pasting ... :redface:
     
  13. Aug 21, 2009 #12
    wait sorry i'm kinda bad

    ok i actually understand that

    1100 m - x1 = vf t2 = (v0+at1)t2

    but how do I solve for t1 the time interval at which the acceleration occuered because I don't know t2 the time that it took to complete the rest of the run after he stopped accelerating I do not know two of the variables

    don't know
    X1 = the distance covered after the acceleration occuered to finish the race
    vf = the velocity at the end of the acceleration
    t2 = the time to complete the rest of the race after accelerating
    t1 = how long the acceleration occuered

    I guess I don't know how to solve for t1 not knowing those variables

    know
    a = the acceleration given to us in the question .20 s^-2 m
    Vo = the velcoity that had to be achieved when he wasn't accelerating in order to cover the first 8900 m at which no acceleration occuered which I calculated to be 5.494 s^-1 m which would be the initial velocity before accelerating
    t1 + t2 = 3.00 min sense that would include the time to accelerate and the time after accelerating sense he started to accelerate after 27.0 min and hopes to get under 30.0 min
     
    Last edited: Aug 21, 2009
  14. Aug 21, 2009 #13

    kuruman

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    But you know that t1 + t2 = 3 min. So ...
     
  15. Aug 21, 2009 #14
    oh.. duh lets see here
    :O
    ok I would believe that I would solve for t2
    right and get
    t2 = 3.00 min - t1

    X = Xo + Vt2 = Xo + (Vo + at1)t2
    X = Xo + V(3.00 min - t1) = Xo + (Vo + at1)(3.00 min-t1)
    :|
     
    Last edited: Aug 21, 2009
  16. Aug 21, 2009 #15
    solve for t1

    X = Xo + Vt2 = Xo + (Vo + a(3.00 min - t2))t2

    not sure what to do
     
  17. Aug 21, 2009 #16

    kuruman

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    Get rid of t2

    t2 = 3 min - t1.

    Now you know that

    1100 m - x1 = (v0+at1)t2
    therefore
    1100 m - x1 = (v0+at1)(3 min - t1)
    and you also know that
    x1=v0t1 + (1/2)at12

    Take this value for x1, put it in the previous equation and solve for t1.
     
  18. Aug 21, 2009 #17
    1100 m - (Xo + Vo t + 2^-1 a t^2) = Xo + (Vo at)(3 min - t1)

    OMG

    I got down to this and didn't know what to do

    Xo + Xo - 1100m = - Vo t - 2^-1 a t^2 - (Vo + at) (3.00 min - t)
     
  19. Aug 21, 2009 #18

    kuruman

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    There is no x0; it is zero. Look at my previous posting and do what I suggested. Do the algebra carefully. You should end up with an equation where t1 is the only unknown quantity.
     
  20. Aug 21, 2009 #19
    ok but is this step right

    1100 m - (Xo + Vo t + 2^-1 a t^2) = Xo + (Vo at)(3 min - t1)

    were I just plugged in X

    then take out the Xo s
    is that correct and then just solve for t???
     
  21. Aug 21, 2009 #20

    kuruman

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    Your equation is sloppy. It mixes t and t1. There is only one kind of time, and that's t1. Also you show (v0 at). It should be (v0+at1). So fix it up and solve for t1.
     
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