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Byju's Answer

Standard XII

Physics

Uniform and Non Uniform

A particle st...

Question

A particle starts from the origin at $t = 0$ and moves in the x-y plane which constant acceleration 'a' in the y direction. An equation of motion is $y = b x^{2}$ . The x-components of its velocity is?

Variable

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\sqrt{\frac{2 a}{b}}

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\frac{a}{2 b}

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\sqrt{\frac{a}{2 b}}

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Solution

The correct option is D $\sqrt{\frac{a}{2 b}}$
Equation of vertical displacement is given by y $= \frac{1}{2} a t^{2} and given y = b x^{2}$
$2 b x^{2} = a t^{2} x^{2} = \frac{a t^{2}}{2 b} \frac{x}{t} = \sqrt{\frac{a}{2 b}} v = \sqrt{\frac{a}{2 b}}$

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A particle starts from the origin at t=0 and moves in the x-y plane which constant acceleration 'a' in the y direction. An equation of motion is y=bx2. The x-components of its velocity is?

The correct option is D √a2bEquation of vertical displacement is given by y=12at2and giveny=bx22bx2=at2x2=at22bxt=√a2bv=√a2b

A particle starts from the origin at $t = 0$ and moves in the x-y plane which constant acceleration 'a' in the y direction. An equation of motion is $y = b x^{2}$ . The x-components of its velocity is?

The correct option is D $\sqrt{\frac{a}{2 b}}$
Equation of vertical displacement is given by y $= \frac{1}{2} a t^{2} and given y = b x^{2}$
$2 b x^{2} = a t^{2} x^{2} = \frac{a t^{2}}{2 b} \frac{x}{t} = \sqrt{\frac{a}{2 b}} v = \sqrt{\frac{a}{2 b}}$