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The position ' $x$ ' of a particle varies with a time as $x=a t^2-b t^3$ where ' $a$ ' and ' $b$ ' are constants. The acceleration of the particle will be zero at
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Verified Answer
The correct answer is:
$\frac{a}{3b}$
$x=a t^2-b t^3$
Differentiating the displacement, we get velocity
$\mathrm{V}=2 \mathrm{at}-3 \mathrm{bt} \mathrm{t}^2$
Differentiating, we get acceleration
$A=2 a-6 b t$
Substituting $\mathrm{A}=0$
$\begin{aligned}
0 & =2 \mathrm{a}-6 \mathrm{bt} & \therefore \quad 6 \mathrm{bt}=2 \mathrm{a} \\
\therefore \quad \mathrm{t} & =\frac{2 \mathrm{a}}{6 \mathrm{~b}}=\frac{\mathrm{a}}{3 \mathrm{~b}} &
\end{aligned}$
Differentiating the displacement, we get velocity
$\mathrm{V}=2 \mathrm{at}-3 \mathrm{bt} \mathrm{t}^2$
Differentiating, we get acceleration
$A=2 a-6 b t$
Substituting $\mathrm{A}=0$
$\begin{aligned}
0 & =2 \mathrm{a}-6 \mathrm{bt} & \therefore \quad 6 \mathrm{bt}=2 \mathrm{a} \\
\therefore \quad \mathrm{t} & =\frac{2 \mathrm{a}}{6 \mathrm{~b}}=\frac{\mathrm{a}}{3 \mathrm{~b}} &
\end{aligned}$
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