$x$ and $y -$ coordinates of a particle moving in $x - y$ plane are, $x = 1 - 2 t + t^{2}$ and $y = 4 - 4 t + t^{2}$ For the given situation match the following two columns.

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Solution

$x = 1 - 2 t + t^{2}$
$y = 4 - 4 t + t^{2}$
It crosses $y -$ axis when $x = 0$
therefore, $1 - 2 t + t^{2} = 0$
this gives $t = 1 s$
at $t = 1, v_{y} = \frac{d y}{d t} = - 4 + 2 t = - 2$
$A \to 2$
when $y = 0$
$4 - 4 t + t^{2} = 0$
therefore $, t = 2$
At $t = 2$
$v_{x} = \frac{d x}{d t} = - 2 + 2 t = 2$
$B \to 1$
at $t = 0$
$v_{x} = - 2$
$v_{y} = - 4$
therefore $, v = \sqrt{(v_{x}^{2} + v_{y}^{2})}$
$v = \sqrt{(20)}$
$C \to 4$
$a_{x} = \frac{d^{2} x}{d t^{2}} = 1$
$a_{y} = \frac{d^{2} y}{d t^{2}} = 1$
$a = \sqrt{a_{x}^{2} + a_{y}^{2}}$
$a = \sqrt{2}$
$D \to 4$

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