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Physics kinematics question

  1. Sep 13, 2008 #1
    1. The problem statement, all variables and given/known data


    To demonstrate the tremendous acceleration of a top fuel drag racer, you attempt to run your car into the back of a dragster that is "burning out" at the red light before the start of a race. (Burning out means spinning the tires at high speed to heat the tread and make the rubber sticky.)

    You drive at a constant speed of v0 toward the stopped dragster, not slowing down in the face of the imminent collision. The dragster driver sees you coming but waits until the last instant to put down the hammer, accelerating from the starting line at constant acceleration, a . Let the time at which the dragster starts to accelerate be t=0

    What is tmax, the longest time after the dragster begins to accelerate that you can possibly run into the back of the dragster if you continue at your initial velocity?


    please help me, just get me started . I've look this question from all possible angles but couldn't figure out how to start

    2. Relevant equations

    vf = vi +a(delta)t

    sf = si +vi(delta)t + 1/2a(delta)t^2

    vf^2= vi^2 +2a(delta)s

    ps. "s" is position
    "V" velocity
    "f" final
    "i" initial
    I use the word delta instead of the little triangle, I dont know how to put it

    3. The attempt at a solution

    My first attempt was trying to figure out the constant velocity of the car, but the thing is that there's not enough given values in order to solve (or are there enough???)


    Am I using the wrong formulas?

    Any kind of help is truly appreciated
     
  2. jcsd
  3. Sep 14, 2008 #2

    Redbelly98

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    The condition to use is that sf is the same for the car and the dragster. So I'd work with that equation.
     
  4. Sep 14, 2008 #3
    oo true man, thanx I'll give it a shot
     
  5. Sep 16, 2008 #4
    so I gave the question a try and her is what I came up with

    Equation for the accelerating dragster

    (sf)d = s0 +v0t + 1/2at2

    Equation for the car going at a constant velocity

    (sf)c = s0 + v0t


    then....

    (sf)d= (sf)c

    1/2at2 = v0t

    at2 = 2v0t

    at = 2v0

    tmax = 2v0/a
     
  6. Sep 16, 2008 #5
    please let me know if it is right or wrong or give me some pointers
     
  7. Sep 16, 2008 #6

    Redbelly98

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    You're on the right track, but there is a detail or two we need to think about more carefully.

    Looks like you're saying v0 is zero for the dragster, and v0 is not zero for the car. That's correct. (It is potentially confusing since you used the same term, v0, for the car and for the dragster.)

    It also looks like you're saying s0 is the same for the car and dragster. That would be a problem, since it means they have already collided at t=0.
     
  8. Sep 16, 2008 #7
    so would this be better then

    Equation for the accelerating dragster

    (sf)d = s0)d +(v0)dt + 1/2at2

    Equation for the car going at a constant velocity

    (sf)c = (s0)c + (v0)ct


    then....

    (sf)d= (sf)c

    1/2at2 = (s0)c + (v0)t

    at2 = 2(v0t + (s0)c)


    at2 = 2v0t + 2(s0)c

    does it makes more sense now?

    or I'm completely off
     
  9. Sep 17, 2008 #8

    Redbelly98

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    Yes, that's correct. Proceed.
     
    Last edited: Sep 17, 2008
  10. Sep 17, 2008 #9
    so if what I did was correct then now I have to Isolate for t,

    at2 = 2v0t + 2(s0)c

    t2 = (2v0t + 2(s0)c)/a

    t2 = (2(v0t + (s0)c))/a


    t = [(2(v0t + (s0)c))/a]1/2


    but my question is, what can I do with the 't' that is inside of the equation

    does it seem right?
     
  11. Sep 17, 2008 #10

    Redbelly98

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    By the way, this is not a very "straightforward" problem ... as you might suspect by now.

    Note we have a quadratic equation in "t":
    a t2 = 2 v0 t + 2 (s0)c

    And there is a formula for solving quadratics (think back to high school algebra) ...
     
  12. Sep 17, 2008 #11
    ooo!!

    0 = at2 - 2v0t - 2(s0)c

    t = (2v0t +- [(-2v0)2 - 4(a)(-2(s0)c)]1/2/2a

    since time is only positive then.... I only take the positive side of the obtained values..

    tmax = (2v0t + [4(v0)2 - 8(a)(s0)c)]1/2/2a
     
  13. Sep 18, 2008 #12

    Redbelly98

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    You're getting there.

    Get rid of the "t" on the right hand side, it's the variable we're solving for.

    Uh, not so fast.
    The minus-sign I've highlighted should be a +. And the "negative" part of the +- doesn't necessarily give a negative time, so we should keep it as +- for now.
     
  14. Sep 19, 2008 #13

    Redbelly98

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    I've (hopefully) thought of a better way to explain what's going on.

    A graph of the car's position is a straight line, with slope v0 and unknown y-intercept s0.

    A graph of the dragster's position is a parabola, x = (1/2) a t^2

    The two graphs can intercept at two values of t, given by the quadratic formula solution for t. A key thing to realize is that the earlier-time (lower t, "-√") solution is when the crash occurs. So take the "-√" solution.

    After that, find the value of s0 which gives the largest value of t.
     
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