The radius vector of a point A relative to the origin varies as r-a t -b t2- where a and b are positive constants- Find the equation of trajectory- A- y-b-a2 x2B- y 2- b - a 2 xC- y-a2-b x2D- none of these

The correct option is A y=ba^2x^2Given r= at î+bt^2ĵ=x nbsp;î+yĵ nbsp; x=at, y=bt^2 ∴ y=b ( xa )^2 ...

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Standard XII

Physics

2D Motion as Two 1D Motions

The radius ve...

Question

The radius vector of a point A relative to the origin varies as $\to r = a t^i + b t^{2}^j$ where a and b are positive constants. Find the equation of trajectory.

y = \frac{b}{a^{2}} x^{2}

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y^{2} = \frac{b}{a^{2}} x

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y = \frac{a^{2}}{b} x^{2}

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none of these

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Solution

The correct option is A $y = \frac{b}{a^{2}} x^{2}$
Given $\to r = a t^i + b t^{2}^j = x^i + y^j$
$x = a t, y = b t^{2}$

$∴ y = b {(\frac{x}{a})}^{2}$

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The radius vector of a point A relative to the origin varies as →r=at^i+bt2^j where a and b are positive constants. Find the equation of trajectory.

The correct option is A y=ba2x2Given →r=at^i+bt2^j=x^i+y^j x=at,y=bt2 ∴y=b(xa)2

The radius vector of a point A relative to the origin varies as $\to r = a t^i + b t^{2}^j$ where a and b are positive constants. Find the equation of trajectory.

The correct option is A $y = \frac{b}{a^{2}} x^{2}$
Given $\to r = a t^i + b t^{2}^j = x^i + y^j$
$x = a t, y = b t^{2}$

$∴ y = b {(\frac{x}{a})}^{2}$