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A particle is moving in a straight line. At time $t$, the distance between the particle from its starting point is given by $x=t-6 t+z$. Its acceleration will be zero at
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The correct answer is:
$t=2$ unit time
Hints : $x=t-6 t^2+t^3$
$$
\begin{aligned}
& \frac{d x}{d t}=1-12 t+3 t^2 \\
& \frac{d^2 x}{d t^2}=-12+6 t
\end{aligned}
$$
Acceleration $=\frac{d^2 x}{d t^2}$
$\therefore$ Acceleration $=0 \Rightarrow 6 t-12=0 \Rightarrow t=2$
$$
\begin{aligned}
& \frac{d x}{d t}=1-12 t+3 t^2 \\
& \frac{d^2 x}{d t^2}=-12+6 t
\end{aligned}
$$
Acceleration $=\frac{d^2 x}{d t^2}$
$\therefore$ Acceleration $=0 \Rightarrow 6 t-12=0 \Rightarrow t=2$
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