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Finding $v$ in $v = L[1-(v^2/c^2)]/t$

Closest attempt: $[1-(c^2/v^2)]v = L/t$

I've been working on this since yesterday. I think I need some help.

EA304GT
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user114027
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1 Answers1

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\begin{equation} v = L(1-v^2/c^2)/t \hspace{10mm}\Rightarrow \hspace{10mm} (t/L)v = 1 - (1/c^2)v^2 \hspace{10mm}\Rightarrow \hspace{10mm} (1/c^2)v^2 + (t/L)v - 1 = 0. \end{equation}

Now use the abc formula.

Marc
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  • Thanks! I have one question. Is there a way that one could isolate a single v on one side of the equation? My professor used v = L/sqrt[t^2+(L^2)/c^2] but didn't explain how he derived it. – user114027 Dec 07 '14 at 15:02
  • The last equation is a quadratic in $v$. You can use the quadratic formula to get your professor's answer. – Ross Millikan Dec 07 '14 at 15:17
  • After Marc's answer and your question in the comments, the solutions of the quadratic Marc wrote are given by $$v_{\pm}=-\frac{c^2t\pm\sqrt{c^4 t^2+4 c^2 L^2}}{2 L}$$ I suppose that we need to discard $v_{-}$. So, keep $v_+$ and multiply numerator and denominator by the conjugate of the current numerator. Expand and simplify. – Claude Leibovici Dec 07 '14 at 15:19
  • @user114027 I tried substituting your professor's value of $v$ on the right-hand side of your first equation and simplified. The result did not evaluate to the same value of $v$. There is a stray factor of $\sqrt{1 + L^2/(c^2 t^2)}$. Maybe I made a mistake, but you might want to try this yourself. – David K Dec 07 '14 at 16:31