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

24 m

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12 m

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5 m

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z e r o

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Solution

The correct option is D $z e r o$

$3 t = \sqrt{3 x} + 6 \Rightarrow 3 t - 6 = \sqrt{3 x}$

$x = \frac{1}{3} (3 t - 6)^{2}$

$v = \frac{d x}{d t} = \frac{1}{3} {2 (3 t - 6) \times 3} \Rightarrow v = 6 t - 12$

Now $v = 0 = 6 t - 12 \Rightarrow t = 2 s$

so at $t = 2 s e c \Rightarrow x = \frac{1}{3} (3 \times 2 - 6)^{2} = 0$

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The relation 3t=√3x+6 describes the displacement of a particle in one direction where x is in meters and t in seconds. The displacement when velocity is zero is:

The correct option is D zero 3t=√3x+6⇒3t−6=√3x x=13(3t−6)2 v=dxdt=13{2(3t−6)×3}⇒v=6t−12 Now v=0=6t−12⇒t=2s so at t=2sec⇒x=13(3×2−6)2=0

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