Question
If $x \propto {t^{5/2}}$ , then
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Answer
Given $x \propto t^2 \quad \Rightarrow x= Kt ^2 \quad$ where $K = constant$
$\therefore$ Velocity of the particle $v =\frac{ dx }{ dt }=2 Kt$
$\therefore$ Acceleration of the particle $a =\frac{ dv }{ dt }=2 K = constant$
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