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Question

# A particle moves in the xy plane with a constant acceleration g in the negative y-direction. Its equation of motion is y=ax−bx2, where a and b are constants. Which of the following are correct?

A
The x-component of its velocity is constant
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B
At the origin, the y-component of its velocity is ag2b
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C
The particle moves exactly like a projectile
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D
None of these
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Solution

## The correct options are A The x-component of its velocity is constant B At the origin, the y-component of its velocity is a√g2b C The particle moves exactly like a projectileAs 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 zeroso ax=0so the velocity in the x-direction will be constant as acceleration in the x - direction is zeronow y=ax−bx2so now differentiate the equation so we get $\frac{\mathrm{d} y}{\mathrm{d} t}= a \frac{\mathrm{d} x}{\mathrm{d} t} - b2x\frac{\mathrm{d} x}{\mathrm{d} t}$now differentiate this again$\frac{\mathrm{d}^{2} y}{\mathrm{d} t^{2}}= a \frac{\mathrm{d}^{2} x}{\mathrm{d} t^{2}} - 2b((\frac{\mathrm{d} x}{\mathrm{d} t})^{2} + x\frac{\mathrm{d} ^{2}x}{\mathrm{d} t^{2}})$that is $a_{y} = aa_{x} - 2b(v_{x}^{2} + 2xa_{x})$as ax=0 and ay=−gso $-g = - 2bv_{x}^{2}$so $v_{x} = \sqrt{g/2b}$so as we know $v_{y} = av_{x}- 2bxv_{x}$as at origin means x =0$v_{y} = av_{x}$so $v_{y} = a\sqrt{g/2b}$and the graph or trajectory of is like parabola that like projectile as shown aboveso option A , B , C are correct

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