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Homework Help: Calculus in Physics

  1. Aug 20, 2005 #1
    Can anyone tell me how to solve the s= ut+1/2 at^2 equation in calculus method ? It's urgent!!! :confused: The one with most detailed answer will get a free Gmail account!!

    Email me with the answer at riast.ullah@gmail.com

  2. jcsd
  3. Aug 20, 2005 #2
    woohoo, a gmail account. they are already free. at this point, almost everyone who wants one has one.
  4. Aug 20, 2005 #3


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    can you post an intelligible question, please?
  5. Aug 20, 2005 #4
    I don't wonna such an account. So I'll post your an answer directly in this forum.

    First you have to rewrite these all in integral and differential form. i.e. for
    s=1/2at^2 you can do this by rewriting s=v*t by s=integral[v,dt]. If you go on with this (a=dv/dt and solve this for v you get v=integral[a,dt]. Thus plugging this in the above equation yields s=integral[integral[a,dt],dt]. do this with the ut part and you get what you want I think
  6. Aug 20, 2005 #5


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    I can comprehend neither this nor the original post! We are asked in the original post to "solve the s= ut+1/2 at^2 equation in calculus method ?". I have no idea what that means! It is a quadratic equation in t so you could solve for t by using the quadratic formula. It is a linear equation in u and a so you could solve for either of those by using elementary methods. None of those have anything to do with "calculus method" so apparently that isn't what is intended.

    I think that Kruger is interpreting this as "derive s= ut+ (1/2)at^2, s is the distance an object moved in time t, a is the constant acceleration and u is the speed at t=0".
    He may well be write but that is a heckuvalot more than was said! Assuming that was the intended question, I'm not sure I would phrase exactly as Kruger did.

    Acceleration is defined as "the derivative of the speed function" so speed is the anti-derivative of the acceleration. If acceleration is a constant, a, then its antiderivative is simply u(t)= at+ C (C is the arbitrary "constant of integration"). Since (presumably) u(0)= u, u(0)= a(0)+ C= u so C=u and u(t)= at+ u.
    But speed is the derivative of the distance function so distance is the anti-derivative of speed, u(t). An anti-derivative of t is (1/2)t^2 and an anti-derivative of the constant u is, again, ut. The anti-derivative of at+ u is (1/2)at^2+ ut+ C where C is again the "constant of integration". Assuming that s is measure from the position at t=0, we have s(0)= (1/2)(0)^2+ u(0)+ C= 0 so C= 0 and s(t)= (1/2)t^2+ ut.

    Rush007: Please take the time to state the entire problem as clearly as possible! Showing us what you have already tried or what you think you do understand about the problem will also help us understand it.
    Last edited by a moderator: Aug 20, 2005
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