The position of a particle as a function of time t, is given by x(t)=at + bt2 – ct3 x(t)=at + bt2 – ct3 where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be:

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The position of a particle as a function of time t, is given by x(t)=at + bt² – ct³ $x (t) = at + {bt}^{2} - {ct}^{3}$ where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be:

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

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

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

$a + \frac{b^{2}}{4 c}$

answer is A.

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Detailed Solution

Given x = at + bt² − ct³

and v = dx/dt = a + 2bt − 3ct²

a = dv/dt = 2b−6ct = 0

⇒ t = b/3c

v (at t = b/3c)

= a+2b(b/3c)−3c(b/3c)

=a+b²/3c.

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