As dimensions do not depend on the magnitude of the quantity, therefore both v_i and v_f have same dimensions = [ms^-1] = [LT^-1]                                                     ….                                           (1)

Dimensions of acceleration, a = ms^-2 = [LT^-2]        …                             (2)

Dimensions of distance, S = [L]                                                                  (3)

Dimensions of time, t = [T]                                                                          (4)

Now put the values of the dimensions of velocity, acceleration and time in equation (a).

[LT^-1] =  [LT^-1] + [LT^-2][T] = [LT^-1] + [LT^-2T] = [LT^-1] + [LT^-1]

⇒[LT^-1] = 2[LT^-1].

Since 2 is dimensionless number, therefore, [LT^-1] = [LT^-1]. Therefore, the equation is dimensionally correct.

• Put the respective dimensions from the above four equations, we have [L] = [LT^-1][T] + ½ [LT^-2][T²] ⇒ [L] = [LT^-1T] + ½ [LT^-2T²] = [LT^-1+1] + ½ [LT^-2+2] = [LT⁰] + ½ [LT⁰] = 3/2 [L]. And since 3/2 is dimensionless number, therefore, the equation becomes [L] = [L]. Thus the equation is dimensionally correct.