AP EAMCET · PHYSICS · Motion In Two Dimensions
A particle crossing the origin at time \(t=0\), moves in the \(x y\)-plane with a constant acceleration ' \(a\) ' in \(y\)-direction. If the equation of motion of the particle is \(y=b x^2\) (where b is a constant), then its velocity component in the x -direction is
- A \(\sqrt{\frac{2 \mathrm{~b}}{\mathrm{a}}}\)
- B \(\sqrt{\frac{a}{2b}}\)
- C \(\sqrt{\frac{a}{b}}\)
- D \(\sqrt{\frac{b}{a}}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{\frac{a}{2b}}\)
Step-by-step Solution
Detailed explanation
\(y = b x^2\) \(v_y = \frac{dy}{dt} = 2bx \frac{dx}{dt} = 2bx v_x\) \(a_y = \frac{dv_y}{dt} = 2b \left( \frac{dx}{dt} v_x + x \frac{dv_x}{dt} \right) \) Given \(a_y = a\) and \(a_x = \frac{dv_x}{dt} = 0\). \(a = 2b (v_x^2 + x \cdot 0)\) \(a = 2b v_x^2\) \(v_x^2 = \frac{a}{2b}\)…
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