The displacement x of a particle at the instant when its velocity is v is given by v-3 x-16- Its acceleration and initial velocity areA- 1-5 unit- 4 unitB- 3 unit- 4 unitC- 16 unit- 1-6 unitD- 16 unit- 3 unit

The correct option is A 1.5 unit, 4 unit v = √(3x +16) Squaring both sides ⇒ v^2 = 3x + 16 ⇒ v^2 16 = 3x Compare with amp; v^2 u^2 = 2aS 2a=3 amp; u ...

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Standard XII

Physics

Displacement

The displacem...

Question

The displacement x of a particle at the instant when its velocity is v is given by $v = \sqrt{3 x + 16}$ . Its acceleration and initial velocity are

1.5 unit, 4 unit

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3 unit, 4 unit

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16 unit, 1.6 unit

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16 unit, 3 unit

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Solution

The correct option is A
1.5 unit, 4 unit

$v = \sqrt{3 x + 16}$
Squaring both sides
$\Rightarrow v^{2} = 3 x + 16 \Rightarrow v^{2} - 16 = 3 x$
$\begin{matrix} C o m p a r e w i t h & v^{2} - u^{2} = 2 a S 2 a = 3 & u^{2} = 16 ⇓ & ⇓ a = 1.5 m / s & u = 4 m / s \end{matrix}$

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The displacement x of a particle at the instant when its velocity is v is given by v=√3x+16. Its acceleration and initial velocity are

The correct option is A 1.5 unit, 4 unit v=√3x+16 Squaring both sides ⇒v2=3x+16⇒v2−16=3x Compare withv2−u2=2aS2a=3u2=16⇓⇓a=1.5m/su=4m/s

The displacement x of a particle at the instant when its velocity is v is given by $v = \sqrt{3 x + 16}$ . Its acceleration and initial velocity are

The correct option is A
1.5 unit, 4 unit

$v = \sqrt{3 x + 16}$
Squaring both sides
$\Rightarrow v^{2} = 3 x + 16 \Rightarrow v^{2} - 16 = 3 x$
$\begin{matrix} C o m p a r e w i t h & v^{2} - u^{2} = 2 a S 2 a = 3 & u^{2} = 16 ⇓ & ⇓ a = 1.5 m / s & u = 4 m / s \end{matrix}$