A particle moves along a straight line such that its displacement at any time t is given by s = {t^3

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2.Motion in Straight Line

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A particle moves along a straight line such that its displacement at any time $t$ is given by $s = {t^3} - 6{t^2} + 3t + 4$ metres. The velocity when the acceleration is zero is........$m{s^{ - 1}}$

A

$3$

B

$ - 12$

C

$42$

D

$ - 9$

(AIPMT-1994)

Solution

(d) $v = \frac{{ds}}{{dt}} = 3{t^2} – 12t + 3$ and $a = \frac{{dv}}{{dt}} = 6t – 12$

For $a = 0$, we have $t = 2$ and at $t = 2,\;v = – 9\;m{s^{ – 1}}$

Standard 11

Physics

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