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Schoolbook question.

  1. Oct 18, 2007 #1
    Hello there!
    I am a swedish student (16 years old) and I would need help with a part of a question in my book. I have a test tomorrow, and it would be nice if I could get some help with this. The thing I need to know is, is it possible to calculate time by only knowing acceleration and lenght?

    The question was about a car breaking. I know the acceleration is -7,856 m/s^2, and the length is 22m.

    I've been trying a lot but havn't found a way to calculate time with this information. The question wanted to know the velocity before the car started to break, but i figured I need time before I can calculate the velocity.

    Ask if it was something you didn't understand, and thanks in advance for all the help! :)

    Karl
     
  2. jcsd
  3. Oct 18, 2007 #2

    Doc Al

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    Staff: Mentor

    You could find the time first, but you don't have to. It depends on what kinematic equations you have at your disposal. (You can find a list here: Basic Equations of 1-D Kinematics )

    What's the most basic relationship between speed, acceleration, and time?

    Hint: Try to solve it symbolically before plugging in any values.
     
  4. Oct 18, 2007 #3
    Forgot to mention my english ain't the best, so have understanding if I would misunderstand anything :)

    The most basic relationship must be;

    a = deltav / deltat

    but, i don't know the starting velocity, only the ending velocity, which means I can't calculate time through this formula with this info, or can I?
     
  5. Oct 18, 2007 #4

    Doc Al

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    Staff: Mentor

    That's a good start. No, you can't calculate the time directly, but you can express the time in terms of other variables. Since the final speed is zero, we can write: a = v/t or t = v/a. (Where v is the initial velocity, which is what we really want to find.)

    How can you combine that with the distance that you're given? What's the most basic relationship between distance, time, and velocity?
     
  6. Oct 18, 2007 #5
    v=s/t

    maybe we could replace v with s/t in the quation

    t = v/a
    t = (s/t)/a
    ta = s/t
    t^2*a = s
    t^2 = s / a

    t^2 = 22 / 7,865 ( the acceleration is negative, but since you can't "root" any negative numbers i put it as a positive number, is this right or wrong? )
    t^2 = 2,78
    t = 1,67s
     
  7. Oct 18, 2007 #6

    Doc Al

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    Staff: Mentor

    Almost. But the speed is not constant, so you need the average velocity = s/t. How can you write the average velocity in terms of the initial and final velocities?
     
  8. Oct 18, 2007 #7
    That would be ( Vzero + V ) / 2 :

    Vzero = initial velocity
    V = the final velocity

    t = v/a
    t = ( ( Vzero + V ) / 2 ) / a
    ta = ( Vzero + V ) / 2
    2ta = Vzero + V
    Vzero = 2ta - V

    BUT, this is 2 unknown variables ( Vzero and t ), which makes the equation uncalculable (as far as I know). Maybe its possible to replace t with something else that we do know, but the two equations that has been into discussion both include v.
     
  9. Oct 18, 2007 #8

    Doc Al

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    Staff: Mentor

    Good. But I thought the final speed was zero and the initial speed is V. In any case, you can now write:
    v/2 = s/t

    Combine this with:

    Eliminate t, since you don't really need it.
     
  10. Oct 18, 2007 #9
    maybe we could replace the Vzero with something instead.

    V = Vzero + at
    -Vzero = -V + at
    Vzero = V - at

    Vzero = 2ta - V
    V - at = 2ta - V
    V = 3at - V
    2V = 3at

    Since V = 0 the equation would be

    0 = 3at
    0 = 3 * 7,856 * t
    0 = 23,56t

    Didn't work. :/

    EDIT : Was written before your post. Will check into it now.
     
  11. Oct 18, 2007 #10
    You must mean Vzero + V = s / t ?
    if we replace the t with v/a we get a deltav, and the delta v should be the negative of Vzero.

    Because out of that I got:

    ( Vzero + V ) / 2 = s / ( -Vzero / a )
    Vzero + V = 2s / (v / a )
    -Vzero/a ( Vzero + V ) = 2s
    -Vzero (Vzero+V) = 2as
    -Vzero^2 - Vzero * V = 2as

    Since V equals zero, we can delete that term (is it called term in english ?).

    -Vzero^2 = 2as
    -Vzero^2 = 345,7
    -Vzero = 18,6 m/s
    Vzero = -18,6 m/s

    The answer is right, only thing is its negative. Maybe I put a negative number somewhere in the equation when I shouldn't have.
     
  12. Oct 18, 2007 #11

    malty

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    Gold Member

    How is this number +ve when car is decelerating??:smile:
     
  13. Oct 18, 2007 #12
    Thanks :) Now the equation works, awesome! :D

    Thanks a lot Doc Al for helping me through this, it means a lot :)
     
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