The velocity of a body is dependent on its position as v - x - where V and x are in SI units- Then its accelerationA- increases linearlyB- decreases linearlyC- remains constantD- first increases and then decreases

The correct option is C remains constantThe acceleration a is given by a = vdvdx=√(x).d(√(x))dx a=√(x)×1/2√(x)=1/2 Hence a remains constant. ...

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Standard XII

Physics

Uniform and Non Uniform

The velocity ...

Question

The velocity of a body is dependent on its position as $v = \sqrt{x}$ , where $v$ and $x$ are in SI units. Then its acceleration

increases linearly

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decreases linearly

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remains constant

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first increases and then decreases

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Solution

The correct option is C remains constant
The acceleration $a$ is given by
$a = v \frac{d v}{d x} = \sqrt{x} . \frac{d (\sqrt{x})}{d x}$
$⟹ a = \sqrt{x} \times \frac{1}{2 \sqrt{x}} = \frac{1}{2}$
Hence $a$ remains constant.

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The velocity of a body is dependent on its position as v=√x, where v and x are in SI units. Then its acceleration

The correct option is C remains constantThe acceleration a is given by a=vdvdx=√x.d(√x)dx ⟹a=√x×12√x=12 Hence a remains constant.

The velocity of a body is dependent on its position as $v = \sqrt{x}$ , where $v$ and $x$ are in SI units. Then its acceleration

The correct option is C remains constant
The acceleration $a$ is given by
$a = v \frac{d v}{d x} = \sqrt{x} . \frac{d (\sqrt{x})}{d x}$
$⟹ a = \sqrt{x} \times \frac{1}{2 \sqrt{x}} = \frac{1}{2}$
Hence $a$ remains constant.