Physics Two Dimensional Motion Question 139
Question: A particle crossing the origin of co-ordinates at time t = 0, moves in the xy-plane with a constant acceleration a in the y-direction. If its equation of motion is $ \text{y = b}{{\text{x}}^{\text{2}}} $ (b is a constant), its velocity component in the x-direction is
Options:
A) $ \sqrt{\frac{2\text{b}}{\text{a}}} $ B)$ \sqrt{\frac{\text{a}}{2\text{b}}} $ C) $ \sqrt{\frac{\text{a}}{\text{b}}} $ D)$ \sqrt{\frac{\text{b}}{\text{a}}} $
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Answer:
Correct Answer: B
Solution:
[b] $ \text{y = b}{{\text{x}}^{\text{2}}} $ . Differentiating w.r.t to t an both $ \frac{\text{dy}}{\text{dx}}=\text{b2x}\frac{\text{dx}}{\text{dt}}\Rightarrow {{v _{y}}=2\text{bx}{{v _{x}} $ Again differentiating w.r.t to t on both sides we get $ \frac{\text{d}{{v _{y}}}{\text{dt}}=2\text{b}{{\text{v}} _{\text{x}}}\frac{\text{dx}}{\text{dt}}+2\text{bx}\frac{\text{d}{{v _{x}}}{\text{dt}}=2\text{bv} _{x^{2}+0 $ [$ \frac{\text{d}{{v _{x}}}{\text{dt}}=0, $ because the particle had constant acceleration along y-direction] $ \text{Now, }\frac{\text{d}{{v _{y}}}{\text{dt}}=\text{a=2bv} _{x^{2}; $
$ \text{v} _{x^{2}=\frac{\text{a}}{2\text{b}}\Rightarrow {{v _{x}}=\sqrt{\frac{\text{a}}{\text{2b}}} $
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