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- 3 units- 4 unitsB- 16 units- 1-6 unitsC- 1-5 units- 4 unitsD- 16 units- 3 units

The correct option is C 1.5 units, 4 unitsv=√(3x+16) or v^2=3x+16 or v^2 16=3x Comparing with v^2 u^2=2aS, we get, u = 4 units, 2a = 3 or a = 1.5 units ...

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Standard XII

Mathematics

Monotonically Increasing Functions

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 units, 4 units

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3 units, 4 units

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16 units, 1.6 units

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16 units, 3 units

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Solution

The correct option is A 1.5 units, 4 units
$v = \sqrt{3 x + 16}$ or $v^{2} = 3 x + 16$ or $v^{2} - 16 = 3 x$

Comparing with $v^{2} - u^{2} = 2 a S$ , we get, u = 4 units,

2a = 3 or a = 1.5 units

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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 units, 4 unitsv=√3x+16 or v2=3x+16 or v2−16=3x Comparing with v2−u2=2aS, we get, u = 4 units, 2a = 3 or a = 1.5 units

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 units, 4 units
$v = \sqrt{3 x + 16}$ or $v^{2} = 3 x + 16$ or $v^{2} - 16 = 3 x$

Comparing with $v^{2} - u^{2} = 2 a S$ , we get, u = 4 units,

2a = 3 or a = 1.5 units