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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 \( y= \) \( b x^{2} . \) The \( x \) -components of its velocity is?

1. Variable
2. \( \sqrt{\frac{2 a}{b}} \)
3. \( \frac{a}{2 b} \)
4. \( \sqrt{\frac{a}{2 b}} \)

Correct answer: \( \sqrt{\frac{2 a}{b}} \)

Solution

With constant acceleration only in the y-direction, the x-velocity stays constant. Using y = (1/2)at^2 and x = v_x t, substitute into y = b x^2 to relate the constants and solve for v_x.

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