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When will Kathy overtake Stan if he starts 1.00s before her with constant a

  1. Sep 25, 2007 #1

    ~christina~

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    1. The problem statement, all variables and given/known data
    Kathy has a sports car that can accelerate at the rate of 4.90m/s^2. She races with Stan. Both cars start from rest but experienced Stan leaves starting line 1.00s before her.
    Stan moves with a constant aceleration of 3.90m/s^2 and Kathy maintains an acceleration of 4.90m/s^2

    a) find time at which Kathy overtakes Stan

    b) The distance she travels before she catches him

    c) speed of both cars the instant she overtakes him

    2. Relevant equations

    Position as a function of time: xf= xi + vxi*t + 1/2*ax*t^2
    velocity as a function of time: vxf= vxi + axt

    3. The attempt at a solution

    Well there was a similar example in the book but it involved a trooper and it started at vi=0 and chased a car with constant acceleration. The car was at a constant v and the trooper started chasing it after 1s after it passed a bilboard where the trooper was stationed. The thing is they gave the velocity and and since you knew that since it was 45.0m/s then after 1 s the initial distance of the car from the billboard was 45.0m and that was when the trooper started to move.

    For this question I'm given acelleration for both. This has gotten me a bit confused as to how to go about finding the displacement of the cars and also since the initial v for both cars is 0.

    Kathy
    vi= 0m/s
    a= 4.90m/s^2

    Stan
    vi= 0m/s
    a= 3.50m/s^2

    I don't know if it is correct but ...

    -don't I need to find the position of Stan when Kathy starts to drive?
    I don't know how to find that though.


    For position as a function of time
    xf= xi + vxi*t + 1/2 ax* t^2

    xf Kathy= 0 + 0 + 1/2 (4.90m/s^2)(t^2)

    xf Kathy= 1/2(4.90m/s^2)*(t^2)

    Would it be the same for Stan? like this:
    however stan starts 1.00 sec ahead of her...HOw do I visualize what I need for the problem?
    I know I need the distance x which equates to both but how I include the 1.00 s is kind of confusing.

    Is the time negative or positive? The book ex of the other one had a diagram like this:
    t= -1.00s ------------t= 0s------------t= ?s

    xf= xi + vxi*t + 1/2 ax* t^2

    xf Stan= 0+ 0t + 1/2( 3.50m/s^2)(t^2)

    Would I equate these 2 equations?? I'm trying to figure out how I incorperate the 1.00s since right now it is like they both started at the same time if I'm not incorrect.


    HELP!:eek:
     
  2. jcsd
  3. Sep 25, 2007 #2

    ~christina~

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    I think I forgot to say that to find the distance of Stan who starts 1.00 s ahead of Kathy wouldn't it be that I can find the velocity of Stan from the acceleration?

    vxf= vxi + axt ?

    Then I could find the v?

    vxf= 0 + 3.50m/s (1.00s)
    then
    vxf= 3.50m/s

    so after 1 sec the
    v=d/t so...
    d= 3.50m/s (1.00s)= 3.50m is Stan's distance away from Kathy??
     
  4. Sep 25, 2007 #3

    learningphysics

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    suppose Kathy starts at t = 0... at any time t.... the time she has been travelling is t - 0 = t. so her equations will be normal...

    Stan starts at t = -1... so at any time t, Stan has been travelling for t - (-1) seconds = t+1 seconds.

    So for stan's equations, wherever you have t, you'd put in t+1 instead...

    there's another way to think about this... we're taking stan's position graph as if he started at t = 0 (same time as Kathy)... but then we shift it to the left by 1s (because he starts 1s earlier)... a left-shift means that you replace t, by t+1.
     
  5. Sep 25, 2007 #4

    ~christina~

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    so If my equations are okay then should ..it be

    for Stan

    xf Stan= 0+ 0(t+1) + 1/2( 3.50m/s^2)(t+1)^2
     
  6. Sep 25, 2007 #5

    learningphysics

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    yeah, but according to the question, it should be 3.90m/s^2.
     
  7. Sep 25, 2007 #6

    ~christina~

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    Oops.. I meant 3.90m/s^2

    xf Stan= 0+ 0(t+1) + 1/2( 3.90m/s^2)(t+1)^2

    Now would I equate this equation and the one for Kathy's position together and solve for t like this:

    0 + 0 + 1/2 (4.90m/s^2)(t^2) = 0+ 0(t+1) + 1/2( 3.90m/s^2)(t+1)^2
    1/2(4.90m/s^2)(t^2) = 1/2 (3.90m/s^2)(t+1)^2
    so..

    2.45t^2 = (1.95)(t+1)^2

    2.45t^2 = 1.95 t^2 + 3.9t +1.95

    0.5t^2- 3.9t - 1.95= 0

    ~I just want to check if this is correct before finding t ..

    Thanks
     
  8. Sep 25, 2007 #7

    learningphysics

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    yup. looks good to me.
     
  9. Sep 25, 2007 #8

    ~christina~

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    Aww...I mixed the #'s b/c it's actually 3.50m/s^2 for Stan's acceleration so it should be..
    ___________________________________________________________________________
    xf Stan= 0+ 0(t+1) + 1/2( 3.50m/s^2)(t+1)^2

    Now would I equate this equation and the one for Kathy's position together and solve for t like this:

    0 + 0 + 1/2 (4.90m/s^2)(t^2) = 0+ 0(t+1) + 1/2( 3.50m/s^2)(t+1)^2
    1/2(4.90m/s^2)(t^2) = 1/2 (3.50m/s^2)(t+1)^2
    so..

    2.45t^2 = (1.75)(t+1)^2

    2.45t^2 = 1.75 t^2 + 3.5t +1.75

    0.7t^2- 3.5t - 1.75= 0

    ___________________________________________________________________________
    Okay so correcting that I get...

    0.7t^2- 3.5t - 1.75= 0

    quadradic formula...

    then I get t= 5.45s

    ~~~~~~~~~~~~~

    for part b.)

    where I have to find the distance she travels before she catches him....

    I would actually plug in the t that I found for the previous right?
    Well assuming it's like that...

    xf Kathy = xi + vxi *t + 1/2 (ax)t^2
    so..

    xf Kathy = 0 + 0*5.45 + 1/2(4.90)(5.45)^2

    xf= 72m

    ~~~~~~~~~~~~~

    part c.)

    speed of both cars the instant she overtakes him....

    hm...
    I'm not sure which equation to plug the values into but I think I need a eqzn to show the velocity as a function of time..am I correct??

    eqzn= vxf= vxi + ax*t


    ~Thank you~
     
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