if the initial velocity of a body be u and acceleration be a then the expression of distance travel by body in nth second is....
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Let displacement of the object in nth of its motion
S,nth=S,n - S,n-1
where S,n and S,n-1 are displacements of the object in n and n-1 seconds.
we know, S = ut+a/2.t^2
So, therefore equation,
S,nth = [un+a/2.n^2] - [u(n-1) + a/2(n-1)^2]
=[un+a/2.n^2] - [u(n-1) + 1/2a(n^2+1-2n)]
= u(n-n+1) + a/2(n^2-n^2-1+2n)
so we found,
S,nth= u+a/2(2n-1) ........
I hope it will helpful in study..
S,nth=S,n - S,n-1
where S,n and S,n-1 are displacements of the object in n and n-1 seconds.
we know, S = ut+a/2.t^2
So, therefore equation,
S,nth = [un+a/2.n^2] - [u(n-1) + a/2(n-1)^2]
=[un+a/2.n^2] - [u(n-1) + 1/2a(n^2+1-2n)]
= u(n-n+1) + a/2(n^2-n^2-1+2n)
so we found,
S,nth= u+a/2(2n-1) ........
I hope it will helpful in study..
kamleshgupta:
ooo
Please derive the equation and show necessary calculation. Thanks! ^_^
okk
We know, S=it+1/2.at^2 let displacement of object in nth, S,nth=S,n-S,n
Now from equation, S,nth=[un+1/2.an^2]-[u(n-1) +1/2.a(n-1) ^2]
Now this equation solve, you get. S,nth= u+1/2.a(2n-1)
Ravi399 please complete your answer. Make editions there in your answer ^_^
OK
S,nth= S,n-S,n-1. We know, S=ut+a/2.t^2. From equation, S,nth=[un+1/2.an^2]-[u(n-1) +1/2a(n-1)] = [un+1/2.an^2]-[u(n-1) +1/2.a(n^2+1-2n)] = u(n-n+1) +1/2.a(n^2-n^2-1+2n) so, we found, S,nth = u+ a/2(2n-1).
OK...
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