Q73 / 90 · JEE Main 2019 (09 Apr Shift 2) · Physics

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 particles zero acceleration, then its velocity will be:

Correct answer
A
$a + \frac{b^{2}}{3 c}$
B
$a + \frac{b^{2}}{2 c}$
C
$a + \frac{b^{2}}{c}$
D
$a + \frac{b^{2}}{4 c}$

Solution

x (t) = a t + b t^{2} - c t^{3}

velocity

v (t) = \frac{d}{d t} (x) = \frac{d}{d t} (a t + b t^{2} - c t^{3})

= a + 2 b t - 3 c t^{2}

Acceleration

= \frac{d}{d t} v (t) = 2 b - 6 t c

acceleration = 0 \Rightarrow 2 b - 6 t c = 0

t = \frac{b}{3 c}

∴

velocity when

t = \frac{b}{3 c},

v_{(t = \frac{b}{3 c})} = a + 2 b (\frac{b}{3 c}) - 3 {(\frac{b}{3 c})}^{2}

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

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

Hence, the correct answer is option: 'A'.

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