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Question Number 30875 by Tinkutara last updated on 27/Feb/18

Commented by ajfour last updated on 28/Feb/18

$s o r r y, i t s j u s t (3) t h a t i s i n c o r r e c t .$
	Answered by ajfour last updated on 28/Feb/18

	$\tan θ = \frac{y}{x} = 3 t \Rightarrow y^{2} = 9 t^{2} x^{2}$ $r^{2} = x^{2} + y^{2} = 4 t^{2} (1 + 9 t^{2})$ $\Rightarrow x^{2} + y^{2} = \frac{4 y^{2}}{9 x^{2}} (\frac{x^{2} + y^{2}}{x^{2}})$ $\Rightarrow y^{2} = \frac{9}{4} x^{4} o r y = \frac{3}{2} x^{2} . . . . (i)$ $(c o n s i d e r i n g t h e u p w a r d p a r a b o l a)$ $s i n c e y = 3 t x$ $x = 2 t, y = 6 t^{2}$ $v_{x} = 2, a_{x} = 0$ $v_{y} = \frac{d y}{d t} = 12 t \Rightarrow a_{y} = 12$ $i n i t i a l s p e e d = \sqrt{v_{x 0}^{2} + v_{y 0}^{2}} = 2 m / s$ $i n i t i a l r a t e o f s p e e d i n g i s$ $i n i t i a l t a n g e n t i a l a c c e l e r a t i o n$ $N o w a t t = 0, \bar{v} = 2 \hat{i}$ $w h i l e \bar{a} = 12 \hat{j}$ $s o n o i n i t i a l t a n g e n t i a l a c c e l e r a t i o n .$ $a s a_{t} = \frac{\bar{a} . \bar{v}}{∣ \bar{v} ∣} = 0 a t t = 0 .$ $\frac{v^{2}}{ρ} = a_{n o r m a l} (ρ i s r a d i u s o f c u r v a t u r e)$ $s o ρ = \frac{v^{2}}{a_{n o r m a l}}$ $a t t = 0 w e h a v e m i n . r a d i u s o f$ $c u r v a t u r e s i n c e v i s m i n i m u m$ $a n d a_{n o r m a l} t h e m a x i m u m$ $ρ_{m i n i m u m} = \frac{4}{12} = \frac{1}{3}$ $b u t ρ_{m a x i m u m} \to \infty s i n c e a s$ $t i m e p r o c e e d s v k e e p s i n c r e a s i n g$ $w h i l e a_{n o r m a l} k e e p s d e c r e a s i n g .$
	Commented by Tinkutara last updated on 01/Mar/18
	Thank you very much Sir! I got the answer.
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