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Homework Help: Embarassing question- Displacement equation

  1. Apr 18, 2010 #1
    Embarassing question-- Displacement equation

    1. The problem statement, all variables and given/known data

    Im just trying to understand why the equation for displacement along a line is [tex]x= 1/2(v_{0}+v)t[/tex]. If the other equation for displacement is [tex]v*t[/tex], where does the 1/2 come from?



    2. Relevant equations
    [tex]v*t[/tex]
    [tex]x= 1/2(v_{0}+v)t[/tex]


    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 18, 2010 #2
    Re: Embarassing question-- Displacement equation

    Derive the formula you are asking about by using these other two kinematic equations

    [tex] \Delta x = v_0t + \frac{1}{2}at^2 [/tex]

    [tex] v = v_0 + at [/tex]
     
  4. Apr 18, 2010 #3
    Re: Embarassing question-- Displacement equation

    Where does the 1/2 come from in the first equation you've given??
     
  5. Apr 18, 2010 #4

    Borek

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    Staff: Mentor

    Re: Embarassing question-- Displacement equation

    From integration needed to calculate displacement in uniformly accelerated linear motion.
     
  6. Apr 18, 2010 #5
    Re: Embarassing question-- Displacement equation

    you must take in consideration difference between Average velocity and Instantaneous velocity

    x = vt v is average velocity

    x= 1/2* (V0 + V )t now v is velocity at time t
     
    Last edited: Apr 18, 2010
  7. Apr 18, 2010 #6

    Matterwave

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    Science Advisor
    Gold Member

    Re: Embarassing question-- Displacement equation

    Perhaps it's easy to see that the two equations are valid for different situations. Let's just consider the situation of a constant velocity v.

    You will notice that if I have a constant velocity v, then the v0 the second equation: .5(v0+v)t is simply v (since velocity is constant, the initial velocity is equal to the velocity always). Therefore the second equation just reduces to: .5(v+v)t=vt the first equation!
     
  8. Apr 18, 2010 #7
    Re: Embarassing question-- Displacement equation

    Yeah, the equation x = vt is not accurate for an object with an acceleration. Perhaps that is why you are confused?
     
  9. Apr 18, 2010 #8
    Re: Embarassing question-- Displacement equation

    [tex] a = \frac{dv}{dt} [/tex]

    [tex]a dt = dv [/tex]

    [tex] \int a dt = \int dv [/tex]

    [tex] at = v - v_0 [/tex]

    [tex] at + v_0 = \frac{dx}{dt}[/tex]

    [tex] atdt + v_0dt = dx[/tex]

    [tex] \int atdt + \int v_0dt = \int dx[/tex]

    [tex]
    \Delta x = v_0t + \frac{1}{2}at^2
    [/tex]
     
  10. Jun 11, 2010 #9
    Re: Embarassing question-- Displacement equation

    To simplify, the average of 2 speeds is 1/2 their sum!
     
  11. Jun 11, 2010 #10
    Re: Embarassing question-- Displacement equation

    This is of course true only when dealing with a constant acceleration, or that at least isnt a function of time.
     
  12. Jun 11, 2010 #11
    Re: Embarassing question-- Displacement equation

    Yes, but so is the original question.
     
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