-1
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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
0
$\begingroup$
  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
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.