Question

Check the equation v =u+ at by the method of dimensions Wherev=final velocity ,
u= initial velocity, a= acceleration ,
t=time

Answer 1

$\huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

We have to check wheather the equation is correct or not by using Dimensional method.

Let's begin !

Take the given equation :

$\Large{\boxed{\sf{v \: = \: u \: + \: at}}}$

As we know that dimension formula of Velocity is ${\sf{(v \: = \: LT^{-1})}}$

And for acceleration is ${\sf{(a \: = \: LT^{-2})}}$

⇒ Also put initial velocity (u) = 0

By putting these values we get,

$\rightarrow {\sf{\big[ LT^{-1} \big] \: = \: \big[ LT^{\cancel{-2}} \big] \: \big[ \cancel{T} \big]}}$

$\rightarrow {\sf{\big[ LT^{-1} \big] \: = \: \big[ LT^{-1} \big]}}$

Hence,

$\Large{\mathbb{L.H.S \: = \: R.H.S}}$

So, the given equation is dimensionaly correct

Answer 2

Answer:

Given:

An Equation has been provided as follows :

v = u + at

To Check:

Validity of the equation by dimensional analysis.

Calculation:

For dimensional analysis, we have to check whether the LHS and RHS match

LHS:

$v = ( {L}^{1} \: {T}^{ - 1} )$

RHS:

$1. \: u = ( {L}^{1} \: {T}^{ - 1} )$

$2. \: a \times t \: = ( {L}^{1} {T}^{ - 2} ) \times (T)$

$= > a \times t \: = ( {L}^{1} \: {T}^{ - 1} )$

As per addition rule in dimensions, only quantities having similar dimensions can be added :

$u + at \: = ( {L}^{1} \: {T}^{ - 1} )$

So, LHS = RHS

Hence Equation is dimensionally correct.