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  4. If a particle moves according to the law s = t3/3 -3t2 + 8t, then the distance covered before it first comes to rest is

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Q. If a particle moves according to the law $s = t^3/3 -3t^2 + 8t$, then the distance covered before it first comes to rest is

Application of Derivatives

Solution:

$\upsilon = \frac{ds}{dt} = \frac{3t^{2}}{3}-6t+8$
$=t^{2}-6t+8=\left(t-4\right)\left(t-2\right)$
$a=\frac{d^{2}s}{dt^{2}}=2t-6$
$\upsilon = 0$
$\Rightarrow \frac{ds}{dt}=0$
$\Rightarrow \left(t -^{ }4\right) \left(t- 2\right) = 0$
$\Rightarrow t = 2$ or $4$
Required distance $= s$ when $t = 2$
$= \frac{8}{3}-3\left(4\right)+8\left(2\right)$
$= \frac{8}{3}+4=\frac{20}{3}$