Question

please answer this question​

Answer 1

Question : check the correctness of:

• v² - u² = 2as

Dimensions:-

$\\\bullet\quad\sf Dimensions\:of\:velocity = [M^{0} L^{1} T^{-1} ]$

$\\\bullet\quad\sf Dimensions\: of\: acceleration = [M^{0} L^{1} T^{-2} ]$

$\\\bullet\quad\sf Dimensions\:of\:displacement= [L^{1} ]$

Principle of homogeneity of dimensions : Dimensions of each term of the given dimensional equation must be equal on both sides.

Applying this principle to check the correctness of 3rd kinematical equation of motion:

$\\\quad\longrightarrow\quad\sf v^{2} - u^{2} = 2as$

$\\\quad\longrightarrow\quad\sf [M^{0}L^{1}T^{-1} ]^{2} -[M^{0} L^{1}T^{-1} ]^{2} = [M^{0}L^{1}T^{-2} ][L]$

$\\\quad\longrightarrow\quad\sf [M^{0}L^{2}T^{-2} ] -[M^{0} L^{2}T^{-2} ] = [M^{0} L^{2} T^{-2} ]$

$\\\quad\longrightarrow\quad\sf [M^{0}L^{2}T^{-2} ] = [M^{0} L^{2} T^{-2} ]$

∵ LHS = RHS

Hence the equation is dimensionaly correct!!