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Homework Help: Speed and acceleration

  1. Mar 8, 2016 #1
    1. The problem statement, all variables and given/known data
    Hello this is my first post here.

    I have studied quite maths and physics when I was young and I was quite good about it but now I am stucked by a very very basic question so much that I feel like a fool and I am even ashamed to be asking it here but I would very much appreciate if someone help me

    My problem is with speed and acceleration. I am reading now a paper on this and I am trying to understand what it says

    2. Relevant equations

    ω(t)=∫ω.τ= ω.t

    3. The attempt at a solution
    Now I know (or at least remember) that Δω/Δt=ω. right? Acceleration is the change of speed. So far so good. If my acceleration is 0 then the change of speed is 0 too, (meaning the speed is constant) ,. Makes sense.
    If I apply integration to Δω and Δt then I suppose I get something like the equation above...

    but that blew me up... in the equation above, if ω. is 0 (accel is 0) then ω(t)=0 right??? that doesnt make sense!! because speed is not 0!

    What in my line of reasoning is a mistake???

    I know it must be the most foolish question but Please help
  2. jcsd
  3. Mar 8, 2016 #2


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    Staff Emeritus
    Science Advisor
    Homework Helper

    I don't know where you found the equation in section 2, but it's meaningless. It's not in the proper form for an integral anyway.

    In most physics work, angular motion and rectilinear motion are analyzed separately, even though the basic equations for each type of motion are similar in structure.

    For rectilinear motion, which is motion in a straight line, we are dealing with the position of a body, call this s(t); its velocity, v(t); and its acceleration, a(t). All of these properties of the body, namely position, velocity, and acceleration, are functions of time, which is why the '(t)' is used.

    Now, the velocity of a body is the change of its position with respect to time, just like the acceleration is the change in velocity with respect to time.

    We indicate these changes by using derivatives like so:

    ##v(t)=\frac{d\,s(t)}{dt}## and ##a(t) = \frac{d\,v(t)}{dt}##

    If we know the position of a body as a function of time, we can calculate its velocity and acceleration using the equations above. Further, if the velocity is constant, this implies that the acceleration is zero, since a constant velocity does not change with respect to time. Ditto for the position of a body. If position does not change, then velocity is zero since there is no movement.

    But what if we know only the acceleration of the body and we want to calculate its velocity and find its position? Then, we work backwards, like so:

    ##v(t) = \int a(t) \,dt + C_1## and

    ##s(t) = \int v(t) \,dt + C_1 ⋅ t + C_2##

    The constants of integration, C1 and C2, can be worked out if we know some extra details about the initial motion of the body, namely its initial velocity and initial position.
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