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```        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?
7 years ago

510 Points
```										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
```
7 years ago
DIVYANK DUVEDI
33 Points
```										he you have taken this q frm dc pandey na
```
7 years ago
aakash
33 Points
```										a particle covers 10m in first 5s and 10m in next 3s. assuming constant acceleration find intial speed, acceleration , distance covered in next 2s.
```
3 years ago
PRAJVAL98
18 Points
```										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
```
3 years ago
Karan sahu
24 Points
```										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.👍👍
```
3 months ago
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