reduce the relationship for the distance travelled in 'n' th the second
Asked by sidchopra | 29th May, 2010, 07:21: AM
v=velocity, vo=initial velocity (or u)
s=vot + 1/2at^2
To calculate the distance travelled during the nth second, you calculate the distance covered in n seconds and subtract the distance covered in n-1 seconds and get
s = von-vo(n-1) + 1/2an^2-1/2a(n-1)^2
simplifying gives us
vo term = vo(n-(n-1)) = vo*1 = vo t The value u (velocity) is multiplied by 1 second giving u feet (or meters).
(n-1)^2 = n^2-2n+1
a term = 1/2a(n^2-(n-1)^2) = 1/2a(n^2-n^2+2n-1) the n^2's cancel and give us 1/2a(2n-1) The n^2's cancel.
s = vo + 1/2a(2n-1)
it's just that vo was multiplied by 1 second, and the n^2 terms in the second part of the equation cancel each other.
Answered by | 29th May, 2010, 09:07: AM
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