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-

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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