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The velocity of a particle moving in th...

The velocity of a particle moving in the `x-y` plane is given by
`(dx)/(dt) = 8 pi sin 2 pi t and (dy)/(dt) = 5 pi sin 2 pi t`
where, `t = 0, x = 8 and y = 0`, the path of the particle is.

A

a straight line

B

an ellipse

C

a circle

D

a parabola

Text Solution

Verified by Experts

The correct Answer is:
B
`y-x` graph gives the shape of path of particle
`(dx)/(dt) = 8 pi sin 2 pi t`
`int_8^x dx = int_0^t 8 pi sin 2 pi t dt`
`x - 8 = -(8 pi)/(2 pi)[cos 2 pi]_0^t`
`x - 8 = 4(1 - cos 2 pi t) rArr cos 2 pi t = (x - 12)/(4)`
`(dy)/(dt) = - 5 pi cos 2 pi t`
`int_0^y dy = 5 pi t int_0^t cos 2 pi t rArr y = (5)/(2) sin 2 pi t`
`((x - 12)/(4))^2 + (y^2)/((5/2)^(2)) = 1 rArr ((x - 12)^2)/((4)^2) + (y^2)/(5/((2)^(2))) = 1`.
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