Question

V^2=u^2+2as
Show that this equation is dimensionaly correct

Answer 1

To show that v² = u² + 2as is dimensionally correct

$\sf{Here}\begin{cases}\sf{u,v \longrightarrow Velocity \longrightarrow LT^{-1}}\\ \sf{a \longrightarrow Acceleration \longrightarrow LT^{-2}}\\\sf{s \longrightarrow Distance \longrightarrow L}\end{cases}$

By Dimensional Analysis

LHS

$\sf v^2 = [LT^{-1}]^2 \\ \\ \longmapsto \sf v^2 = L^2 T^{-2}-----------(1)$

RHS

$\sf u^2 + 2as = [LT^{-1}]^2 + [LT^{-2}][L] \\ \\ \longmapsto \sf u^2 + 2as = L^2T^{-2} + L^2T^{-2} --------------(2)$

By Principle Of Homogeneity,I conclude equations (1) and (2) are dimensionally similar

Thus,

v² = u² + 2as is dimensionally correct