Relation 3t = √ {3x} + 6 describes displacement of a particle in one direction where x is in metre

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2.Motion in Straight Line

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The relation $3t = \sqrt {3x} + 6$ describes the displacement of a particle in one direction where $x$ is in metres and $t$ in sec. The displacement, when velocity is zero, is.........$metres$

A

$24$

B

$12$

C

$5$

D

$0$

Solution

(d) $3t = \sqrt {3x} + 6 \Rightarrow 3x = {(3t – 6)^2}$

$ \Rightarrow x = 3{t^2} – 12t + 12$

$v = \frac{{dx}}{{dt}} = 6t – 12$, for $v = 0,\;t = 2\sec $

$x = 3{(2)^2} – 12 \times 2 + 12 = 0$

Standard 11

Physics

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