prove Snth = u + a/2 (2n-1)
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Here, S=u+(1/2)*a*(2n-1) this equation is used to find distance travelled by body in nth second.
LHS
So, S(nth) = distance travelled in nth second
i.e S(nth) = [L]/[T] = [LT^-1] dimensionally
RHS
Now, U+(1/2)*a*(2n-1) = where,
U= initial velocity = [LT^-1]
a= acceleration = [LT^-2]
2n-1= n=time = [T]
1/2=dimensionless = [M°L°T°]
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