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I started studying instantaneous velocity derivatives using only now. It may seem stupid but really I'm not sure whether that's right:
I have an equation:
$$x (t) = 1.5t - 9,75t³$$ To set the time when this object stops would velocity = zero , right? If velocity is v (t) = -29,25t² (v = dx / dt = - 9,75*3t²), then the object will be stopped with t = 0s? It's the second equation in which it occurs so I came resort here.

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1 Answer 1

up vote 2 down vote accepted

First of all, derivative of $1.5t-9.75t^3$ is $1.5-29.25t^2$, you missed first part.

Secondly, consider ideal pendulum. It's speed is zero at extreme points, however it never stops for ever. it depends on the problem and additional conditions (like, object hitting ground, so speed=0 is good enough condition for finding time of impact)

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Really didn't realize the mistake, thanks! The question doesn't cite any specific situation, only the equation and want to know when that stops. I answer as √0.05s or 0s? –  Daniela Morais Jun 14 at 1:23

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