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Standard XII

Physics

Radius of Curvature

A particle mo...

Question

A particle moves in the x-y plane such that $X = (t^{2}, 2 t$ ) where t is time and X is the coordinate of the particle . The trajectory of the path is

circle

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Straight line

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parabola

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None of these

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Solution

The correct option is C parabola
Given that,

Time $= t$

$X = t^{2}, 2 t$

Now, given co-ordinate of the particle as $(t^{2}, 2 t)$

Means at any time t $x = t^{2}, y = 2 t$

Now, $t = \frac{y}{2}$

So,

$x = {(\frac{y}{2})}^{2}$

$4 x = y^{2}$

Hence, the trajectory of the particle is a parabola with equation $y^{2} = 4 x$

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A particle moves in the x-y plane such that X=(t2,2t) where t is time and X is the coordinate of the particle . The trajectory of the path is

The correct option is C parabolaGiven that, Time =t X=t2,2t Now, given co-ordinate of the particle as (t2,2t) Means at any time t x=t2,y=2t Now, t=y2 So, x=(y2)2 4x=y2 Hence, the trajectory of the particle is a parabola with equation y2=4x

A particle moves in the x-y plane such that $X = (t^{2}, 2 t$ ) where t is time and X is the coordinate of the particle . The trajectory of the path is