The position of a particle as a function of t | vedclass

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Question

$\begin{array}{l}
X = at + b{t^2} - c{t^3}\
V = \frac{{dx}}{{dt}} = a + 2bt - 3c{t^2}\
a = \frac{{dv}}{{dt}} = 2b - 6ct\
Put\,acceleration\, =

Vedclass · Accepted Answer

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

Vedclass · Answer

$\begin{array}{l}
X = at + b{t^2} - c{t^3}\
V = \frac{{dx}}{{dt}} = a + 2bt - 3c{t^2}\
a = \frac{{dv}}{{dt}} = 2b - 6ct\
Put\,acceleration\, = 0\
 \Rightarrow \,\,t = \frac{b}{{3c}}\
Find\,V\,at\,t = \frac{b}{{3c}}\
V = \,a + \frac{{{b^2}}}{{3c}}
\end{array}$

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

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