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Question Number 171614 by Tawa11 last updated on 18/Jun/22

A particle is moving along a straight line such that it's position from a fixed point is S = ( 12 - 15t² + 5t³ )m where t is in seconds. Determine: A. Total distance travelled by the particle from t = 1sec to t = 3sec B. The average speed of the particle during this time.
Commented by Tawa11 last updated on 18/Jun/22

$Is total distance 14 m ?$
Commented by Beginner last updated on 18/Jun/22

$t o t a l d i s t a n c e = s 1 + s 2 . s u b s t i t u t e t h e p a r a m e t e r s d i r e c t l y$
Commented by mr W last updated on 18/Jun/22

$30 m$
Commented by mr W last updated on 19/Jun/22

$a t 1 o^{'} c l o c k y o u a r e 2 k m e a s t e r l y$ $a p a r t f r o m y o u r h o u s e,$ $a t 2 o^{'} c l o c k y o u a r e 8 k m w e s t e r l y$ $a p a r t f r o m y o u r h o u s e,$ $t h a t m e a n s f r o m 1 o^{'} c l o c k t o 2 o^{'} c l o c k$ $y o u w a l k e d 10 k m i n d i r e c t i o n$ $f r o m e a s t t o w e s t .$ $a t 3 o^{'} c l o c k y o u a r e a g a i n 12 k m e a s t e r l y$ $a p a r t f r o m y o u r h o u s e,$ $t h a t m e a n s f r o m 2 o^{'} c l o c k t o 3 o^{'} c l o c k$ $y o u w a l k e d 20 k m i n d i r e c t i o n$ $f r o m w e s t t o e a s t .$ $t h a t m e a n s f r o m 1 o^{'} c l o c k t o 3 o^{'} c l o c k$ $y o u w a l k e d t o t a l l y 30 k m .$
Commented by mr W last updated on 19/Jun/22

$y o u s h o u l d k n o w t h a t t h e p a r t i c l e$ ${d o e s n}^{'} t m o v e i n o n e d i r e c t i o n o n l y .$ $j u s t s t u d y s = 12 - 15 t^{2} + 5 t^{3} i n d e t a i l!$
Commented by mr W last updated on 19/Jun/22

Commented by Tawa11 last updated on 19/Jun/22

$God bless you sirs$
Commented by mr W last updated on 19/Jun/22

$Distance traveled is the total length$ $of the path traveled between two$ $position s .$ $I n c u r r e n t e x a m p l e t h e p a r t i c l e t r a v e l e d$ $a p a t h o f 30 m l e n g t h i n t h e s e c o n d$ $a n d t h i r d s e c o n d . s o t h e d i s t a n c e$ $t r a v e l e d i s 30 m .$
Commented by mr W last updated on 19/Jun/22

Commented by mr W last updated on 19/Jun/22

$a n o b j e c t m o v e d f r o m A t o B$ $f o l l o w i n g t h e r e d p a t h w i t h l e n g t h s .$ $d i s p l a c e m e n t = \vec{A B}$ $d i s t a n c e =∣ \vec{A B} ∣= d$ $d i s t a n c e t r a v e l e d = l e n g t h o f p a t h = s$
Commented by Tawa11 last updated on 21/Jun/22

$Wow, I just finished studying your concept sir .$ $Here displacement still end up in 10 m$ $That is (- 8 - 2) + [12 - (- 8)] = 10 m .$ $And$ $Distance = 8 + 2 + 20 = 30 m$
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