Displacement $(x)$ of a particle is related to time $(t)$ as $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = at + bt^{2} - ct^{3}$ Where $a,\ b\ and\ c$ are constants of the motion. The velocity of the particle when its acceleration is zero is given by

Question

Dr. Abubacker · Accepted Answer

(c) Displacement $x = at + bt^{2} - ct^{3}$
Velocity, $v = \frac{dx}{dt} = a + 2bt - 3ct^{2}\ \ \ \ \ \ldots(i)$
And acceleration, $a = \frac{a^{2}x}{dt^{2}} = 2b - 6ct\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
When acceleration $a = 0,\ t = \frac{b}{3c}$
Substituting the value of $t$ in Eq. (i), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = a + \frac{b^{2}}{3c}$

Q. Displacement $(x)$ of a particle is related to time $(t)$ as $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = at + bt^{2} - ct^{3}$ Where $a,\ b\ and\ c$ are constants of the motion. The velocity of the particle when its acceleration is zero is given by

Explanation

Reviewed by Dr. Abubacker

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