Question

Q.By the method of dimensional analysis, derive the

Answer 1

Answer:

LHS=RHS=[L]

Explanation:

Given mathematical expression:

$s=u.t+\frac{1}{2} a.t^2$

where:                                             dimension:

s = displacement length                $[L]$

u = initial velocity                           $[L.T^{-1}]$

t = time                                                 $[T]$

a = acceleration                             $[L.T^{-2}]$

now using dimensional analysis:

$[L]=[L.T^{-1}]\times [T]+[L.T^{-2}] [T]^2$

we know that the ratio and constants have no dimension.

$[L]=[L]+[L]$

as we know that only similar dimensions can be added or subtracted therefore we get a correct conclusion.

However we can deduce the operators between the equations and can neither check for the validity of the constants. We can only check for the dimension of the terms involved.