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Standard XIII

Physics

Uniform and Non Uniform

The relation3...

Question

The relation $3 t = \sqrt{3 x} + 6$ 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

24 m e t r e s

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12 m e t r e s

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5 m e t r e s

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Z e r o

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Solution

The correct option is D $Z e r o$
$3 t + \sqrt{3 x} + 6 \Rightarrow 3 x = (3 t - 6)^{2}$

$\Rightarrow x = 3 t^{2} - 12 t + 12$

$v = \frac{d x}{d t} = 6 t - 12, f o r v = 0, t = 2 s e c$

$x = 3 (2)^{2} - 12 \times 2 + 12 = 0$

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The relation3t=√3x+6describes the displacement of a particle in one direction where x is in metres and t in sec. The displacement, when velocity is zero, is

The correct option is D Zero3t+√3x+6 ⇒ 3x=(3t−6)2 ⇒ x=3t2−12t+12 v=dxdt=6t−12, for v=0,t=2 sec x=3(2)2−12×2+12=0

The relation $3 t = \sqrt{3 x} + 6$ 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

The correct option is D $Z e r o$
$3 t + \sqrt{3 x} + 6 \Rightarrow 3 x = (3 t - 6)^{2}$

$\Rightarrow x = 3 t^{2} - 12 t + 12$

$v = \frac{d x}{d t} = 6 t - 12, f o r v = 0, t = 2 s e c$

$x = 3 (2)^{2} - 12 \times 2 + 12 = 0$