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2.Motion in Straight Line
hard
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
A$a + \frac{{{b^2}}}{{4c}}$
B$a + \frac{{{b^2}}}{{c}}$
C$a + \frac{{{b^2}}}{{2c}}$
D$a + \frac{{{b^2}}}{{3c}}$
(JEE MAIN-2019)
Solution
$\begin{array}{l}
X = at + b{t^2} – c{t^3}\\
V = \frac{{dx}}{{dt}} = a + 2bt – 3c{t^2}\\
a = \frac{{dv}}{{dt}} = 2b – 6ct\\
Put\,acceleration\, = 0\\
\Rightarrow \,\,t = \frac{b}{{3c}}\\
Find\,V\,at\,t = \frac{b}{{3c}}\\
V = \,a + \frac{{{b^2}}}{{3c}}
\end{array}$
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}$
Standard 11
Physics
