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Physics-

General

Easy

Question

A particle moves in the $x minus y$ plane with velocity $v subscript x end subscript equals 8 t minus 2$ and $v subscript y end subscript equals 2$ . If it passes through the pointy $x equals 14$ and $y equals 4$ at $t equals 2 s$ , find the equation ( $x minus y$ relation) of the path

$x = y^{2} - y + 2$
$x = 2 y^{2} + 2 y - 3$
$x = 3 y^{2} + 5$
Can not be found from above data

The correct answer is: $x equals y to the power of 2 end exponent minus y plus 2$

$v subscript x end subscript equals 8 t minus 2$
or $fraction numerator d x over denominator d t end fraction equals 8 t minus 2$
or $not stretchy integral from 14 to x of d equals not stretchy integral from 2 to t of open parentheses 8 t minus 2 close parentheses d t$
or $x minus 14 equals open square brackets 4 t to the power of 2 end exponent minus 2 t close square brackets subscript 2 end subscript superscript t end superscript equals 4 t to the power of 2 end exponent minus 2 t minus 12$
or $x equals 4 t to the power of 2 end exponent minus 2 t plus 2$ (i)
Further, $v subscript y end subscript equals 2$
or $fraction numerator d y over denominator d t end fraction equals 2$
$therefore blank not stretchy integral from 4 to y of d y equals not stretchy integral from 2 to t of 2 d t$
or $y minus 4 equals open square brackets 2 t close square brackets subscript 2 end subscript superscript t end superscript equals 2 t minus 4$
or $y equals 2 t$
or $t equals fraction numerator y over denominator 2 end fraction$ (ii)
Substituting value of $t$ from Eq. (ii) in Eq. (i), we have
$x equals y to the power of 2 end exponent minus y plus 2$

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