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A car is slowing down at a constant pace

  1. May 23, 2010 #1
    A car is slowing down at a constant rate...

    1. The problem statement, all variables and given/known data
    WARNING: This is a translation. If you find something you can't understand, i'll try to explain.
    A car is slowing down at a constant rate. The car travels half of its stopping distance in t1=4s. How long does the car take to stop?

    2. Relevant equations
    t1=4s
    S1=S2

    S1 - first part of the path
    S2 - second part of the path
    v0 - starting speed
    v1 - speed after the first half of the path
    t2 - whole time spent traveling
    x0 - starting position (0?)

    3. The attempt at a solution
    Err:
    v=v0t?+at?2 (i'm guessing that v=0?)
    v0=at?
    v1=a(t2-t1) (?)

    So what i am guessing is that i should make
    S1=x0+v0t?+at?2/2
    S2=x0+v1(t2-t1)+a(t2-t1)2/2

    and then put in the v0 and v1 i got earlier. And then since S1=S2 i put what i got from the above.. I think the a's should get removed if i divide by "a" and i should get a quadratic equation. Here's the problem, i've tried doing it (filled around 3 pages of my notes) and i get the wrong answer. Which, i think, is mainly due to the fact i don't know what times to put in where. The answer should be:
    t22-4t2t1+t12=0

    And i get always something around that ^. But never it. Sometimes i'm missing just 1 number, sometimes a letter. What to do? :(
     
    Last edited: May 23, 2010
  2. jcsd
  3. May 23, 2010 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi Avathacis! :smile:
    ("A car is slowing down at a constant rate. The car travels half of its stopping distance in t1=4s. How long does the car take to stop?" :wink:)

    Hint: find v1 by using a formula which doesn't involve t. :smile:
     
  4. May 23, 2010 #3
    [tex]
    v^2 = v_1^2 + 2 a \Delta x?
    [/tex]
    I don't think i'm supposed to know this formula :>. Or at least i haven't used it. Or i'm not getting something?
    Also thanks for the fixes.

    Wouldn't that make (assuming i am correct about v being 0)
    [tex]
    v_1 = root - 2 a S
    [/tex] Also i assume that delta x is S in my problem? :>

    Also could anyone tell me if my guesses are correct and why it would be great :>.
     
    Last edited: May 23, 2010
  5. May 23, 2010 #4

    Doc Al

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    Staff: Mentor

    You don't need that formula. Just use your equations for distance for each half of the distance.

    Hint: Express both V1 and V0 in terms of a, T1, T2. (Careful with signs.)
     
  6. May 23, 2010 #5
    v0=at1 (?)
    v1=a(t2-t1)
     
  7. May 23, 2010 #6

    Doc Al

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    This one is not quite right. V0 goes to 0 in time T2.
    Good.

    Note that here you are using 'a' to represent the magnitude of the acceleration. I suggest that you revise your earlier equations to use the same convention.
     
  8. May 23, 2010 #7
    1) I'm supposed to use the time the speed gets down to 0? It'll be helpful!

    Though i'm not sure i get what you mean by magnitude :<.
     
  9. May 23, 2010 #8

    Doc Al

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    Sure. That time is what you call T2--the total time.
    Magnitude means the quantity without regard for its sign. For example: The magnitude of the number -11 is just 11.

    So if you call the magnitude of the acceleration 'a', then the actual acceleration will be -a, since it's slowing down. For example:

    Vf = V0 - at

    Applying that to the final speed, we get:
    0 = V0 - at2
    or
    V0 = at2
     
  10. May 23, 2010 #9
    Kinda like modules? (|-11|=11?)
    I guess you're kinda turning to the point that i'm making a + or - mistake somewhere right?
     
  11. May 23, 2010 #10

    Doc Al

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    Staff: Mentor

    Exactly like that.
    It's hard to tell. Why don't you write down your four equations, using 'a' as the modulus of the acceleration. (Thus, you may have to put '-a' in some of your formulas.) You'll have your two distance equations and the two velocity equations.
     
  12. May 23, 2010 #11
    [tex]
    S_1 = x_0 + v_0t_2 + at_2^2/2
    [/tex]

    [tex]
    S_2 = x_0 + v_1(t_2 - t_1) + a(t_2 - t_1)^2/2
    [/tex]

    [tex]
    v_0 = at_2
    [/tex]

    [tex]
    v_1 = a(t_2 - t_1)
    [/tex]

    Right, i know i have to put in the v_0 and V_1 into S_1 and S_2.
     
  13. May 24, 2010 #12

    Doc Al

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    Since S1 is the first part of the distance, shouldn't the time be t1, not t2?

    Since the car is slowing down, you need to replace 'a' with '-a' in both of those equations.

    OK.

    Yes, then set S1 = S2.
     
  14. May 24, 2010 #13
    I think i can solve it now. Well explained, thanks. I will post again if i need more help :>.
     
  15. May 24, 2010 #14
    I have a quick question:
    After solving the equation a little i get:
    t2=2t2t1+t2t1sqrt2
    Can i divide by t2 to remove it?
     
  16. May 24, 2010 #15

    Doc Al

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    Algebraically, sure you can. But that equation cannot be correct. (You cannot have a physically meaningful equation where terms have different units. You can't have a time term equal to a time-squared term.)

    So go over it again carefully.
     
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