Guest

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

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

Grade:11

3 Answers

Neeti
571 Points
8 years ago
This formula is used to determine distance travelled in nth second of journey so yes it is correct
Neeti
571 Points
8 years ago
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 swhich = m so both sides have the same unit which is metres hence the formula is dimensionally correct.
malayala.lakshmi Narayana
22 Points
4 years ago
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 nth second,
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.

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free