-3
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$$x = A + Bt + Ct^2 + Dt^3 $$

where $$C = 0.14~\rm m/s^{2} $$ and $$D = 0.01~\rm m/s^{3} $$

After how much time after the start of movement does the acceleration become 1 m/s^2 and what is the medium acceleration during this time interval?

I am not asking you to solve this, I need to know where shall I start from?

Is this formula enough?

$$x = x_0 + v_0t + a (t^2) / 2$$

or should I start from $$a = \frac{dv}{dt}$$

I am really confused, thanks!

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closed as off-topic by Kyle Kanos, ZeroTheHero, sammy gerbil, Chris, stafusa Feb 25 '18 at 11:31

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  • $\begingroup$ Hint: the acceleration is the second time derivative of the position. $\endgroup$ – ndrearu Feb 24 '18 at 20:38
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    $\begingroup$ You don't need the first formula, the reformulation of the second should suffice. $\endgroup$ – Anjan Feb 24 '18 at 20:41
  • $\begingroup$ Your equation that starts $x=x_0 + \ldots$ is valid only when the acceleration is constant, which is not the case here. $\endgroup$ – garyp Feb 24 '18 at 21:11
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You should be able to tell that $x = x_0 + v_0t + a \frac{t^2}{2}$ won't do because you will be missing the $t^3$ term.

Start with $$a = \frac{{\rm d}^2x}{{\rm d}t^2} $$ and plug and chug the time values.

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  • $\begingroup$ so because v = dx / dt and a = dv / dt, it results that a = d2x/ dt2? You really saved me here! Is the average acceleration supposed to be calculated like some integral from time 1 to time 2? Thank you again! $\endgroup$ – NuSuntStudent Feb 24 '18 at 21:42

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