Solution:
Here, the given expression is
v = u + 5 at
The dimension of LHS
= \(\left[LT^1\right]\)
The dimension of RHS
= \(\left[LT^1\right]+5\left[LT^{-2}\right]\left[T\right]\)
= \(\left[LT^1\right]+5\left[LT^{-1}\right]\)
= \(6\left[LT^{-1}\right]\)
Here, 6 is a dimensionless quantity. So, dimension of LHS = dimension of RHS. Thus, the given equation is dimensionally correct.