A particle Q is moving with constant velocity (−5i+3j)m/s. At time t=5sec , Q is at the same point with position vector (−i−5j)m. Find the distance of Q from the origin at time t=2sec


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Question Number 191052 by otchereabdullai last updated on 17/Apr/23

$A p a r t i c l e Q i s m o v i n g w i t h c o n s t a n t$ $v e l o c i t y (- 5 i + 3 j) m / s . A t t i m e$ $t = 5 s e c, Q i s a t t h e s a m e p o i n t w i t h$ $p o s i t i o n v e c t o r (- i - 5 j) m . F i n d t h e$ $d i s t a n c e o f Q f r o m t h e o r i g i n a t$ $t i m e t = 2 s e c$
	Answered by mr W last updated on 17/Apr/23

	$p o s i t i o n a t t = 2 :$ $(- 1, - 5) - 3 \times (- 5, 3) = (14, - 14)$ $d i s t a n c e t o o r i g i n :$ $\sqrt{{(14)}^{2} + {(- 14)}^{2}} = 14 \sqrt{2} \approx 19.8 m$
	Commented by otchereabdullai last updated on 17/Apr/23

	$A m a l w a y s g r a t e f u l t o y o u f o r t h e$ $h e l p b u t P r o f W p l e a s e i w a n t t o$ $u n d e r s t a n d i t v e r y w e l l e x p e c i a l l y$ $t h e 3 m u l t i p l y i n g (- 5, 3)$
	Commented by mr W last updated on 17/Apr/23

	$p o s i t i o n (t = 5) = p o s i t i o n (t = 2) + (5 - 2) \times v$ $\Rightarrow p o s i t i o n (t = 2) = p o s i t i o n (t = 5) - 3 \times v$
	Commented by otchereabdullai last updated on 17/Apr/23

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