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A particle moves in the plane according to the law $x=kt$, $y=kt(1- \alpha t)$ where $k$ and $\alpha$ are positive constants, t is time. The particle trajectory $ y(x) $ is

a) circle b) parabola c) straight line d) hyperbola

1 Answer

$x=kt,y=kt(1-\alpha t)$
$y=kt-k\alpha t^2$
$y=x-k\; \alpha \;t^2$
$y=x-k \;\alpha\;\large\frac{x^2}{k^2}$
ie parabola
Hence b is the correct answer.


answered Jul 2, 2013 by meena.p
edited Jan 26, 2014 by meena.p

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