I would like to obtain a relation between initial velocity and distance covered till velocity is zero, with time varying acceleration. For a constant acceleration we can solve: \begin{align*} \int_{0}^{x_0} m \, a(t) dx &= \frac{1}{2} m v_0^2\\ \end{align*}

What can we do if the acceleration is not constant?

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    $\begingroup$ The equation you have there doesn't make sense. Those units will not give distance, that is energy. That can be a step in solving it, but that equation is not distance. $\endgroup$ – JMac Apr 7 '17 at 16:34
  • $\begingroup$ @JMac those equation will definitely give a relation between distance and velocity . At RHS is velocity and LHS is distance what are you talking about. I mean of course both sides resemble energy but due to it you get relation between required two $\endgroup$ – Utkarsh futous Apr 7 '17 at 16:50
  • $\begingroup$ Although if a is time variable it's not possible to integrate but that's a different thing! $\endgroup$ – Utkarsh futous Apr 7 '17 at 16:52
  • $\begingroup$ @Utkarshfutous I don't even understand where he got this integration. He wants to solve for distance but has 0 and $x_0$ as the integral limits, yet those appear nowhere in the equation beside it, but were still unknown :S It's weird to talk about distance formulas then show a different formula and skip a bunch of steps in deriving an equivalent expression. $\endgroup$ – JMac Apr 7 '17 at 16:54
  • $\begingroup$ @JMac I think you have misunderstood it, the given equation is work-energy theorem. For sake of understanding just integrate LHS for constant...you will get max now simplify the equation you will notice that you got nothing but third equation of motion $\endgroup$ – Utkarsh futous Apr 7 '17 at 17:15
  1. Write the given equation of acceleration $a = f(t)$
  2. Now use $ dv/dt = a $ in place of a
  3. Take dt to RHS and integrate
  4. Integrate again to get relation between x and t
  5. Assume t to be parametric variable get the relation between v and x
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  • $\begingroup$ this is what I suspected but was hoping for a more direct way $\endgroup$ – fadiak Apr 7 '17 at 16:36
  • $\begingroup$ I don't think there is other way unless you could relate a and distance or distance and time . $\endgroup$ – Utkarsh futous Apr 7 '17 at 16:46

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