A particle moves in the xy plane with a const | vedclass

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(D) Given that the particle has acceleration only in the negative $y$-direction,the acceleration in the $x$-direction is $a_x = 0$. Since $a_x = 0$,th

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All of the above

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(D) Given that the particle has acceleration only in the negative $y$-direction,the acceleration in the $x$-direction is $a_x = 0$. Since $a_x = 0$,the $x$-component of velocity $v_x$ is constant.From the equation $y = ax - bx^2$,differentiate with respect to time $t$: $\frac{dy}{dt} = a\frac{dx}{dt} - 2bx\frac{dx}{dt}$,which gives $v_y = v_x(a - 2bx)$.Differentiating again with respect to time: $\frac{d^2y}{dt^2} = a\frac{d^2x}{dt^2} - 2b\left[\left(\frac{dx}{dt}ight)^2 + x\frac{d^2x}{dt^2}ight]$.Since $a_x = 0$ and $a_y = -g$,we get $-g = -2bv_x^2$,which implies $v_x = \sqrt{\frac{g}{2b}}$.At the origin $(x = 0)$,$v_y = a v_x = a\sqrt{\frac{g}{2b}}$.The angle $	heta$ with the $x$-axis is given by $	an 	heta = \frac{v_y}{v_x} = \frac{a v_x}{v_x} = a$,so $	heta = 	an^{-1}(a)$.Thus,all statements are correct.

$A$ particle moves in the $xy$ plane with a constant acceleration $g$ in the negative $y$-direction. Its equation of motion is $y = ax - bx^2$,where $a$ and $b$ are constants. Which of the following are correct?

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