Question

Check whether the relation $S=ut+\frac{1}{2}a{t}^2$ is dimensionally correct or not, where symbols have their usual meaning.

Answer 1

We have $S=ut+\frac{1}{2}a{t}^2$
Checking the dimensions on both sides, LHS = $\left[M^0{L}^1{T}^0\right]$
RHS = $\left[L{T}^{-1}\right]\left[T\right]+\left[L{T}^2\right]\left[T^2\right]$
= $\left[M^0{L}^1{T}^0\right]+\left[M^0{L}^1{T}^0\right]=\left[M^0{L}^1{T}^0\right]$
Comparing the LHS and RHS, we get
LHS = RHS
Hence the formula is dimensionally correct.