Question
Wed April 13, 2011 By: Ananya Vyas

# A body starting from rest moves with constant acceleration, then the ratio of the distance covered in 5th second to 5 seconds is?

Wed April 13, 2011

Let us assume that constant acceleraion to be A.
u = o since the body starts from rest.
and we know that Dispacement in nth sec = snth = supto n - supto(n-1)
[un+(1/2)an2]-[u(n-1)+(1/2)a(n-1)2
Snth=u+(1/2) (2n-1)a
Now,
Distance covered in 5th second is
S = 0+1/2 (2 x5-1)a =(9/2)a

Distance covered in 5 seconds is
S=ut+(1/2)at2
S=0+(1/2) (ax5x5) = (25/2) xa
Therefore the ratio of the distance covered in 5th second to 5 second is =
Related Questions
Mon April 17, 2017

# Pandit Dev Raj Shastri Ji is world famous best astrologer situated In Surat, Gujarat, India. He has lots of powers and blesses Baba Ramdev Runiche, he do work with 100% guaranteed result for your problems within some days or hours. He visited all over India for solve the life related problems for clients. He also solve life related problems in all over worlds like:- love marriage, lost love, vashikaran, black magic, love spells, inter caste marriage, children problems, business problems, jobs and career problems, husband wife disputes, divorces issues, You can contract with guru ji 24 x 7 Contact Details +91-9723634914 Watsup no. +91-8866090899 Email:- devrajastrologer8866090899@gmail.com

Wed March 15, 2017