Question

Test dimensionally if the equation $v^2=u^2+2aS$ may be correct.

Answer 1

Since, $v^2-u^2=2as$

$\Rightarrow v^2=u^2+2as$

$\therefore \left[v\right]=\left[u\right]=$ velocities

$=\left[M^0{L}^1{T}^{-1}\right]$

$a=$ acceleration ,$\left[a\right]=\left[M^0{L}^1{T}^{-2}\right]$

$s=$ distance, $\left[s\right]=\left[M^0{L}^1{T}^0\right]$

hence $LHS:-[v{]}^2=[M^0{L}^1{T}^{-1}{]}^2=\left[M^0{L}^2{T}^{-2}\right].....\left(1\right)$

$u^2=[M^0{L}^1{T}^{-1}{]}^2=[M^0{L}^2{T}^{-2}].......(2)$

and

$\left[2as\right]=\left[M^0{L}^1{T}^{-2}\right]\left[M^0{L}^1{T}^0\right]$

$=\left[M^0{L}^2{T}^{-2}\right].........\left(3\right)$

Two quantities can be added / Subtracted only when having same dimension.

From equation $\left(A\right),\left(1\right),\left(2\right),\left(3\right)$

$LHS=RHS$