motion

Derive the general formula for the distance covered by a body in its nth second of motion

u+1/2a(2n+1) v+1/2a(2n-1) u+1/2a(2n-1) u+1/2a(4n-1)

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1 solution

L e t u b e t h e i n i t i a l v e l o c i t y a n d l e t a b e t h e c o n s t a n t a c c e l e r a t i o n o f t h e b o d y . D i s t a n c e c o v e r e d i n n t h s e c o n d i s g i v e n b y D i s t a n c e c o v e r e d i n n s e c o n d s D i s t a n c e c o v e r e d i n ( n 1 ) s e c o n d s : s n s n 1 u n + a 2 ( n 2 ) [ u ( n 1 ) + a 2 ( n 1 ) 2 ] u n u n + u + a n 2 2 a n 2 2 a 2 + a n u + a n a 2 u + a 2 [ ( 2 n 1 ) Let\quad 'u'\quad be\quad the\quad initial\quad velocity\quad and\quad let\quad 'a'\quad be\quad the\quad constant\quad \\ acceleration\quad of\quad the\quad body.Distance\quad covered\quad in\quad { n }^{ th }\quad second\\ is\quad given\quad by\\ Distance\quad covered\quad in\quad n\quad seconds-Distance\quad covered\quad in\quad (n-1)\\ seconds:\\ \\ { s }_{ n }-{ s }_{ n-1 }\\ un+\frac { a }{ 2 } ({ n }^{ 2 })-[u(n-1)+\frac { a }{ 2 } { (n-1) }^{ 2 }]\\ un-un+u+\frac { a{ n }^{ 2 } }{ 2 } -\frac { a{ n }^{ 2 } }{ 2 } -\frac { a }{ 2 } +an\\ u+an-\frac { a }{ 2 } \\ u+\frac { a }{ 2 } [(2n-1)

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