Question

prove that formula of finding displacement in 'n'th second is dimensionally correct....... the formula is:-Sn=u+a/2(2n-1)

Answer 1

S = ut + 1/2at²

let t = n second particle covered d distance ,

d = un +1/2an² --------(1)

now, distance covered in t = (n -1) sec is D

now,
D = u( n -1) + 1/2a(n -1)² -------(2)

here both equation follow , rules and also dimensionally correct now,

for t = nth second

Sn = d - D = u( n - n +1) + 1/2a( n² -n² +2n -1) = u + 1/2a (2n -1)

Sn = u +1/2a (2n -1)

here you see clearly why , distance covered in nth second is dimensionally correct . becoz , in nth second is a time of special second magnitude =1

actually here if you see deeply then ,
here Snth = u.1 + 1/2a(2.1.n -1)
here 1 is what ????
this is the special time , eg nth time
now, if you take dimension,
then,

LHS = dimension of Snth = [ L ]

RHS = dimension of u.1 = [ LT-¹][T] = L =
dimension of 1/2a(2.1.n -1) = [ LT-²][ T²] =[L]

hence, LHS = RHS
so, this is dimentionally correct