A particle starts from rest with acceleration(30+6t) ms^(−2) at time t. Where will the particle come to rest again?


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Question Number 43354 by pieroo last updated on 10/Sep/18

$A particle starts from rest with acceleration (30 + 6 t)$ ${ms}^{- 2} at time t . Where will the particle come to rest$ $again ?$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18

	$\frac{d v}{d t} = 30 + 6 t$ $h e r e a c c e l a r a t i o n i s + v e s o n o q u e s t i o n$ $a r i s e t h a t p a r t c l e w i l l c o m e t o r e s t a g a i n$ $d v = (30 + 6 t) d t$ $v = 30 t + 3 t^{2} + c$ $t = 0 v = 0 s o c = 0$ $v = 30 t + 3 t^{2} i f p a r t i c l e a t r e s t v = 0$ $30 t + 3 t^{2} = 0$ $3 t (10 + t) = 0$ $t = 0 b u t t c a n n o t b e n e g e t i v e (t = - 10)$ $s o p a r t i c l e c a n n e v e r c o m e t o r e s t a g a i n . . .$
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