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How do we calculate time that we need for specific distance if we have non-constant acceleration. I have some formulas for other variable but don't have for time:

Speed at time: $$ v(t) = v_0+at+\frac{j(t^2)}{3} $$

Distance at time: $$ x(t)=x_0+v_0t+\frac{a(t^2)}{2}+\frac{j(t^3)}{6} $$

Acceleration at time: $$ a(t)=a_0+jt $$

user2704821
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    You need to solve $x(t) = L$. As it is a polynomial of degree 3, you'll need Cardano's method (or an equivalent method) to get an explicit solution. – Tryss Dec 16 '15 at 12:47
  • If $j$ is constant speed of acceleration change, $x(t)=x_0+v_0t+{a_0t^2 \over 2} + {jt^3 \over 6}$, where $x_0, v_0, a_0$ are coordinate, speed and acceleration at $t=0$. Do you have exactly this situation (i.e. for $t>0$ you move with constant speed of acceleration increase)? – Abstraction Dec 16 '15 at 13:05
  • I don't know if I understand you question. But yes j (jerk) is constant in my case and $v_0$ is not 0 at t=0 also $v_0$ is not 0 at t=0; I found very helpful Cardano's method mentioned in comment above and answer belowe for cubie equation – user2704821 Dec 16 '15 at 15:12

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What you have there is a cubic equation, so if you want to solve for a specific distance, you would take your formula for $x(t)$, subtract the specific distance you want to get a cubic equaling 0, and then you would use the cubic formula, one version of which is posted here: http://www.math.vanderbilt.edu/~schectex/courses/cubic/

Warning, its not pretty.

Alan
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