# The formula S=u+a(n-1/2) is dimensionally not correct.But still we use it in variety of applications. Why do we still get the correct answer?Please reply!!!

509 Points
11 years ago

Sn=un+an^2/2  ...........1               (displacement in n seconds)

Sn-1=u(n-1) +a(n-1)^2 /2.........2   (displacement in n-1 seconds)

Snth=Sn-Sn-1

Snth  =u + a(n-1/2) ..............3

in eq 1 , dimension of n is time and in eq 2 ,n-1 second that means 1 in n-1 second is having dimension of time......

1sec is a constant but it is having dimension ....so in same manner in eq 3 ,n-1/2 has dimension of time .....

so this relation is correct

approve my ans if u like

DIVYANK DUVEDI
33 Points
11 years ago

he you have taken this q frm dc pandey na

aakash
33 Points
7 years ago
a particle covers 10m in first 5s and 10m in next 3s. assuming constant acceleration find intial speed, acceleration , distance covered in next 2s.
PRAJVAL98
18 Points
7 years ago
it is dimentionally correct.
S in nth second in the sense u r talking about the diplacement in that second which will basically have the dimentional formula [M^0*L^1*T^-1]
the rhs also has the same dimentional formula.

prajval98
Karan sahu
24 Points
4 years ago
St=(ut+1/2at2)-(u (t-1)+1/2a (t-1)2)=(u)(1)+1/2 (a)(2t)-1/2(a)(1)2the first term is (u)(1), which we are writing only u. Dimentions of u=[LT-1]and dimentions of 1 which we call 1 sec is [T]So, [(u)(1)]=[LT-1][T]=[L]=Sttherefore dimention of (u)(1) are same as the dimentions of St. Same logic can be applied with other terms too.👍👍
Simra
11 Points
4 years ago
Sn-Sn-1=u(n-n+1)+1/2a(n²-n²+2n-1) =u+1/2a(2n-1)Sn=u+1/2(2n-1)Snth=u.1+1/2(2.1.n-1)D.f of LHS=(LT-¹)(T) (L) D. F of RHS=(LT-¹)(T) (L) D. F of 1/2a(2.1.n-1)=(LT-²)(T²) (L) This could be done like this, thus S=ut+1/2at² is dimensional ly correct