If the position vector of a point P relative to the origin varies with time as- r -2 - t - t 2- where - and - are constants- then the equation for trajectory of the particle is A- y- x2-2 -2B- y - x 2-2 -2C- y - x 2-4 - v 2D- y- x2-4 -2

The correct option is D y = β x^24 α^2Given, r = 2 α t î β t^2 ĵ = x î + y ĵ then, x = 2 α t ⇒ t = x2 α and y = β t^2 ⇒ y = β x^24 α^2 ...

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Byju's Answer

Standard XII

Physics

2D Motion as Two 1D Motions

If the positi...

Question

If the position vector of a point $P$ relative to the origin varies with time as, $\to r = 2 α t^i - β t^{2}^j$ , where $α$ and $β$ are constants, then the equation for trajectory of the particle is

y = \frac{- β x^{2}}{4 α^{2}}

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y = \frac{- β x^{2}}{2 α^{2}}

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y = \frac{- β x^{2}}{2 α}

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y = \frac{- β x^{2}}{4 α}

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Solution

The correct option is A $y = \frac{- β x^{2}}{4 α^{2}}$
Given,
$\to r = 2 α t^i - β t^{2}^j = x^i + y^j$
then,
$x = 2 α t \Rightarrow t = \frac{x}{2 α}$
and $y = - β t^{2} \Rightarrow y = \frac{- β x^{2}}{4 α^{2}}$

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If the position vector of a point P relative to the origin varies with time as, →r=2αt ^i−β t2^j , where α and β are constants, then the equation for trajectory of the particle is

The correct option is A y=−βx24α2Given, →r=2αt^i−βt2^j=x^i+y^j then, x=2αt⇒t=x2α and y=−βt2⇒y=−βx24α2

If the position vector of a point $P$ relative to the origin varies with time as, $\to r = 2 α t^i - β t^{2}^j$ , where $α$ and $β$ are constants, then the equation for trajectory of the particle is

The correct option is A $y = \frac{- β x^{2}}{4 α^{2}}$
Given,
$\to r = 2 α t^i - β t^{2}^j = x^i + y^j$
then,
$x = 2 α t \Rightarrow t = \frac{x}{2 α}$
and $y = - β t^{2} \Rightarrow y = \frac{- β x^{2}}{4 α^{2}}$