A particle moves in a linear scare such that acceleration after t seconds is a ms^(−2) where a= 2t^2 + t.If its initial velocity was 3ms^(−1) find an expression for S,the distance in meters traveled from start t seconds.


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Question Number 47967 by Rio Michael last updated on 17/Nov/18

$A p a r t i c l e m o v e s i n a l i n e a r s c a r e s u c h t h a t a c c e l e r a t i o n$ $a f t e r t s e c o n d s i s a {m s}^{- 2} w h e r e a = 2 t^{2} + t . I f i t s i n i t i a l$ $v e l o c i t y w a s 3 {m s}^{- 1} f i n d a n e x p r e s s i o n f o r S, t h e d i s t a n c e i n m e t e r s$ $t r a v e l e d f r o m s t a r t t s e c o n d s .$
	Answered by ajfour last updated on 17/Nov/18

	$a = \frac{d v}{d t} = 2 t^{2} + t$ $\Rightarrow \int_{3}^{v} d v = \int_{0}^{t} (2 t^{2} + t) d t$ $\Rightarrow v = \frac{d x}{d t} = \frac{2}{3} t^{3} + \frac{t^{2}}{2} + 3$ $\Rightarrow \int_{x_{9}}^{x} d x = \int_{0}^{t} (\frac{2}{3} t^{3} + \frac{t^{2}}{2} + 3) d t$ $s = x - x_{0} = t (\frac{t^{3}}{6} + \frac{t^{2}}{6} + 3)$ $\Rightarrow s = \frac{t}{6} (t^{3} + t^{2} + 18) .$
	Commented by Rio Michael last updated on 17/Nov/18

	$p l e a s e s i r i {d o n}^{'} t u n d e r s t a n d t h e p o i n t w h e r e$ $\int_{x_{9}}^{x} d x = \int_{0}^{t} (\frac{2}{3} t^{3} + \frac{t^{2}}{2} + 3) d t$
	Commented by ajfour last updated on 17/Nov/18

	$x i s t h e p o s i t i o n c o o r d i n a t e$ $a t t i m e t, x_{0} t h e i n i t i a l p o s i t i o n$ $c o o r d i n a t e; s t h e d i s p l a c e m e n t$ $i s a l w a y s = △ x = x - x_{0} .$
	Commented by Rio Michael last updated on 18/Nov/18

	$t h a n k s s i r$
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