Answer to Question #285892 in Physics for Precy

Question

Show on the dimensional Analysis that v^2 - u^2 = 2as

Answer 1

Dimension of a speed

$$[v]=[u]=\rm m/s=LT^{-1}$$

Dimension of an acceleration

$$[a]=\rm m/s^2=LT^{-2}$$

Dimension of a distance

$$[s]=\rm m=L$$

Let's check an equation

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

We get

$$[v^2]-[u^2]=[a][s]$$

$$\rm (LT^{-1})^2-(LT^{-1})^2=(LT^{-2})L$$

$$\rm L^2T^{-2}=L^2T^{-2}$$

We get an identity, so equation is dimensionally correct.