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Differentiation from 1st Principles

A particle mo...

Question

A particle moves along the parabolic path $y = 2 x - x^{2} + 2$ , in such a way that the x-component of velocity vector remains constant ( $5 m / s$ ). Find the magnitude of acceleration of the particle.

$- 10 m / s^{2}$

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$- 20 m / s^{2}$

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$- 50 m / s^{2}$

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$- 100 m / s^{2}$

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Solution

The correct option is C
$- 50 m / s^{2}$

$y = 2 x - x^{2} + 2 \frac{d y}{d t} = 2 \frac{d x}{d t} - 2 x \frac{d x}{d t} \Rightarrow v_{y} = 2 v_{x} - 2 x v_{x} \Rightarrow v_{y} = 10 - 10 x \Rightarrow \frac{d}{d t} v_{y} = - 10 \frac{d x}{d t} \Rightarrow a_{y} = - 50 m / s^{2}$

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A particle moves along the parabolic path y=2x−x2+2, in such a way that the x-component of velocity vector remains constant (5 m/s). Find the magnitude of acceleration of the particle.

The correct option is C −50 m/s2 y=2x−x2+2dydt=2dxdt−2xdxdt⇒vy=2vx−2xvx⇒vy=10−10x⇒ddtvy=−10dxdt⇒ay=−50 m/s2

A particle moves along the parabolic path $y = 2 x - x^{2} + 2$ , in such a way that the x-component of velocity vector remains constant ( $5 m / s$ ). Find the magnitude of acceleration of the particle.

The correct option is C
$- 50 m / s^{2}$

$y = 2 x - x^{2} + 2 \frac{d y}{d t} = 2 \frac{d x}{d t} - 2 x \frac{d x}{d t} \Rightarrow v_{y} = 2 v_{x} - 2 x v_{x} \Rightarrow v_{y} = 10 - 10 x \Rightarrow \frac{d}{d t} v_{y} = - 10 \frac{d x}{d t} \Rightarrow a_{y} = - 50 m / s^{2}$