So I know the Initial Velocity ($V_i$), Time ($t$), and Distance ($d$).

I know that $$d = V_it + \frac{1}{2} at^2$$

If I rearrange this, would acceleration $a = \dfrac{2(d - V_it)}{t^2}$ ?

Then assume Final Velocity ($V_f$) will be $V_i + at$

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    $\begingroup$ Yes, you are correct. $\endgroup$ – N.S.JOHN Apr 15 '16 at 7:40

Under the assumption that acceleration is constant, yes, you are correct. Notice that $a$ can be negative, in which case the object in question is decelerating. Still, the equation $Vi+at=V_f$ will hold.

  • $\begingroup$ Thank you. I forgot to mention it is constant acceleration. Great. $\endgroup$ – That Guy Apr 15 '16 at 10:18

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