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Derivative of Distance

  1. Sep 5, 2012 #1
    Hello. Apparently (according to one of my math textbooks) if you want to calculate the time it takes for an object to fall a distance, given constant acceleration, the equation is thus:
    [tex]t=\sqrt{\frac{D}{4.9}}[/tex]
    Where: t=time D=distance
    I recongnized "4.9" has half of Earth's Gravitational Field Strength, "9.8". Giving us the formula:
    [tex]t=\sqrt{\frac{D}{(\frac{g}{2})}}[/tex]
    Since "g" can be treated as acceleration, we get:
    [tex]t=\sqrt{\frac{D}{(\frac{a}{2})}}[/tex]

    I asked a teacher of mine who has a general interest in science, he said this was true, but after that, he didnt know.

    My question is whether or not velocity can be thought of as the deriviative of distance (in a one dimensional context). The reason why I'm asking this is because the above equation rearranges into this: [itex]D=\frac{g{t^2}}{2}[/itex]

    This looks awefully like a second antiderviative of acceleration in respect to time. I thought about it and considerd the fact that the first antiderivative of acceleration is velocity. That got me thinking, why would you integrate velcoty? Well a basic algebriac equation says that the product of velocity and time gives you distance traveled, assuming velocity is constant. Given constant velocity, the antiderivative of velocity is velocity by time. Since velocity is not constant, we must find the velocity by anti-differentiating the acceleration, then anti-differentiate that, giving us the equation above.

    I thought about my idea and figured that if velocity is positive, distance would increase, if velocity was negative, the distance would decrease. If the velocity was negative you would be moving back towards the place you started. If it was positive, you would be moving away, thus increasing distance. If the distance was negative, you would be behind the place you started.

    I came up with this (which looks like the more well known D=vt) [itex]D(t)=vt+{D_i}[/itex]
    (Di=initial distance)
    I also tried integrating velcoity, which gave me a logical answer.

    If I'm right:
    [tex]\frac{dD}{dt}=v[/tex]

    Is this correct? please correct any mistakes I may have made, thank you.
     
  2. jcsd
  3. Sep 5, 2012 #2
    Yes, this is the standard way of talking about the velocity of objects. Given [itex]D = D(t)[/itex], [itex]dD/dt[/itex] is the velocity, and [itex]d^2D/dt^2 = a[/itex], the acceleration.

    It sounds like you have a good familiarity with calculus and are just discovering how this is related to classical physics. Consider this:

    [tex]F = \frac{dp}{dt}[/tex]

    Force is the time derivative of momentum. This is a modern expression of Newton's laws---or a general statement of conservation of momentum in the absence of forces.

    I know nowadays some college intro physics courses are trying to teach mechanics using just that expression of Newton's laws, rather than the traditional approach of teaching [itex]d = d_0 + vt + \frac{1}{2} at^2[/itex] out of nothing and such.
     
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