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Standard XII

Physics

The 'e' Equation

The relation ...

Question

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

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12 metres

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5 metres

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Zero

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Solution

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

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

v= $\frac{d x}{d t}$ =6t-12, for v=0, t=2sec

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

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The relation 3t=√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

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

The relation 3t= $\sqrt{3 x}$ +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 Zero
3t= $\sqrt{3 x} + 6 \Rightarrow 3 x = (3 t - 6)^{2}$

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

v= $\frac{d x}{d t}$ =6t-12, for v=0, t=2sec

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