General Physics> S=u+a/2(2t-1) Is it dimensionally correct...

S=u+a/2(2t-1)
Is it dimensionally correct? If yes, how?

Poulomi Mukherjee , 10 Years ago

Grade 11

3 Answers

Neeti

This formula is used to determine distance travelled in n^thsecond of journey so yes it is correct

Last Activity: 10 Years ago

Neeti

I’m sorry i didn’t see the “if yes, how”.

In the first part, u is basically u x 1second therefore m/s x s = m . and in the 2t-1 part there is t² but it get’s cancelled so it basically is m/s² x s²which = m so both sides have the same unit which is metres hence the formula is dimensionally correct.

Last Activity: 10 Years ago

malayala.lakshmi Narayana

Yes, it is dimensionally correct.

To begin, if we look at the formula “S=u+a/2(2t-1)”, at the LHS, we have “S”, which is the distance travelled in n^thsecond,

so dimensionally it is the velocity as it is the distance travelled in the given time(i.e, v=d/t), so its dimensional formula is [M⁰LT^-1].

And on RHS, we have “u+a/2(2t-1)”, and according to rules od dimensionall analysis we have two terms on RHS seperated by plus operation(+),

1- “u”

2- “a/2(2t-1)”

and the dimensional formula of the first term is [M⁰LT^-1] as it is the velocity(u).

Whereas, the dimensional formula of the second term ( a/2(2t-1) ),

having terms a for acceleration and t for time, and by the rules of dimensional analysis in the expression – a/2, 2 is also having the same dimensions as that of a(i.e, of acceleration) and similarly the other expression -2t-1, having dimensions of time. So, acceleration(a)=velocity(v)/time(t) ,(a=v/t) and by subtituting in the expression( a/2(2t-1) ) we have (v/t)*t=v so the dimensional formula of the second term is [M⁰LT^-1].

Hence the equation [M⁰LT^-1]=[M⁰LT^-1]+[M⁰LT^-1], is dimensionally correct.

Last Activity: 6 Years ago