Kinematics Question 591
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 it’s equation of motion is $ \text{y = b}{{\text{x}}^{\text{2}}} $ (b is a constant), it’s 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{dv_y}{dt}=2{b}{v_x}\frac{dx}{dt+2bx}\frac{dv_x}{dt}=2{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}}; $
$ {v} _{x^{2}}=\frac{a}{2{b}}$
$\Rightarrow {{v _{x}}}=\sqrt{\frac{\text{a}}{\text{2b}}} $
