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

a + \frac{b^{2}}{c}

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a - \frac{b^{2}}{3 c}

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a + \frac{b^{2}}{3 c}

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a - \frac{b^{2}}{4 c}

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Solution

The correct option is C $a - \frac{b^{2}}{3 c}$
$x = a t + b t^{2} + c t^{3} v = d x / d t = a + 2 b t + 3 c t^{2} d v / d t = 2 b + 6 c t$ at $t = - b / 3 c, a = 0$ $V = a + 2 b t + 3 c t^{2}, a t t = - b / 3 c v = a + 2 b (- b / 3 c) + 3 c (- b / 3 c)^{2} v = a - 2 b^{2} / 3 c + b^{2} / 3 c = a - b^{2} / 3 c$

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Displacement (x) of a particle is related to time (t) as x=at+bt2+ct3 where a,b and c are constants of motion. The velocity of the particle when its acceleration is zero, is given by:

The correct option is C a−b23cx=at+bt2+ct3v=dx/dt=a+2bt+3ct2dv/dt=2b+6ctat t=−b/3c,a=0 V=a+2bt+3ct2,att=−b/3cv=a+2b(−b/3c)+3c(−b/3c)2v=a−2b2/3c+b2/3c=a−b2/3c

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