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Power, Force and Velocity

The position ...

Question

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

A

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

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B

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

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C

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

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D

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

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Solution

The correct option is D $a + \frac{b^{2}}{3 c}$
$x = a t + b t^{2} - c t^{3}$
$v = \frac{d x}{d t} = a + 2 b t - 3 c t^{2}$
$a = \frac{d v}{d t} = 2 b - 6 c t = 0 \Rightarrow t = \frac{b}{3 c}$
$v_{⎛ ⎝ a t t = \frac{b}{3 c} ⎞ ⎠} = a + 2 b (\frac{b}{3 c}) - 3 c (\frac{b}{3 c})$
$= a + \frac{b^{2}}{3 c}$ .

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