A particle P moves on a straightline from a fixed point O and the distance x from O after t seconds is given as x = (1/(4 )) t^4 − (3/2) t^2 + 2t. Find: a) the velocity of P when t = 2, b) the acceleration of P when t = 2, c) the time at which the speed P is Minimum.


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Question Number 38365 by Rio Mike last updated on 24/Jun/18

$A p a r t i c l e P m o v e s o n a s t r a i g h t l i n e$ $f r o m a f i x e d p o i n t O a n d t h e d i s t a n c e$ $x f r o m O a f t e r t s e c o n d s i s g i v e n a s$ $x = \frac{1}{4} t^{4} - \frac{3}{2} t^{2} + 2 t .$ $F i n d :$ $a) t h e v e l o c i t y o f P w h e n t = 2,$ $b) t h e a c c e l e r a t i o n o f P w h e n t = 2,$ $c) t h e t i m e a t w h i c h t h e s p e e d P$ $i s M i n i m u m .$
	Answered by MJS last updated on 25/Jun/18

	$s (t) = \frac{1}{4} t^{4} - \frac{3}{2} t^{2} + 2 t$ $v (t) = \frac{d s}{d t} = t^{3} - 3 t + 2 \Rightarrow v (2) = 4$ $a (t) = \frac{d v}{d t} = 3 t^{2} - 3 \Rightarrow a (2) = 9$ $local \min (v (t)) / \max (v (t)) = v (x) \Rightarrow v^{'} (x) = 0 \Rightarrow a (x) = 0$ $v (x) = \min (v (t)) \Rightarrow v^{″} (x) > 0$ $v (x) = \max (v (t)) \Rightarrow v^{″} (x) < 0$ $3 x^{2} - 3 = 0$ $x^{2} = 1$ $x_{1} = - 1$ $x_{2} = 1$ $v^{″} (t) = a^{'} (t) = \frac{d a}{d t} = 3 t$ $a^{'} (- 1) = - 3 < 0 \Rightarrow local \max (v (t)) = v (- 1) = 4$ $a^{'} (1) = 3 > 0 \Rightarrow local \min (v (t)) = v (1) = 0$
	Commented by Rio Mike last updated on 25/Jun/18

	$p e r f e c t!$
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