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:
Correct Option: , 2
Solution:
(2) $x=a t+b t^{2}-c t^{3}$
Velocity, $v=\frac{d x}{d t}=\frac{d}{d t}\left(a t+b t^{2}+c t^{3}\right)$
$=a+2 b t-3 c t^{2}$
Acceleration, $\frac{d v}{d t}=\frac{d}{d t}\left(a+2 b t-3 c t^{2}\right)$
or $0=2 b-3 c \times 2 t$
$\therefore t=\left(\frac{b}{3 c}\right)$
and $v=a+2 b\left(\frac{b}{3 c}\right)-3 c\left(\frac{b}{3 c}\right)^{2}$
$=a+\frac{b^{2}}{3 c}$
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