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

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Question

As in this it is given that the particle will only have the acceleration in the negative $y-$ direction so the acceleration in the $x-$direction is ze

Vedclass · Accepted Answer

All of the above

Vedclass · Answer

As in this it is given that the particle will only have the acceleration in the negative $y-$ direction so the acceleration in the $x-$direction is zero so $\quad a_{x}=0$

so the velocity in the $x$ -direction will be constant as acceleration in the $x$ -direction is

zero now $y=a x-b x^{2}$

so now differentiate the equation so we get

$\frac{\mathrm{d} y}{\mathrm{d} t}=a \frac{\mathrm{d} x}{\mathrm{d} t}-b 2 x \frac{\mathrm{d} x}{\mathrm{d} t}$

now differentiate this again $\frac{\mathrm{d}^{2} y}{\mathrm{d} t^{2}}=a \frac{\mathrm{d}^{2} x}{\mathrm{d} t^{2}}-2 b\left(\left(\frac{\mathrm{d} x}{\mathrm{d} t}\right)^{2}+x \frac{\mathrm{d}^{2} x}{\mathrm{d} t^{2}}\right)$

that is $a_{y}=a a_{x}-2 b\left(v_{x}^{2}+2 x a_{x}\right)$

as $a_{x}=0$ and $a_{y}=-g$

$\mathrm{so}-g=-2 b v_{x}^{2}$

so $v_{x}=\sqrt{g / 2 b}$

so as we know $v_{y}=a v_{x}-2 b x v_{x}$

as at origin means $x=0$ $v_{y}=a v_{x}$

so $v_{y}=a \sqrt{g / 2 b}$

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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