Source: Extra Practice

According to the principle of dimensional homogeneity, for the equation x=vt+12at2x = vt + \frac{1}{2}at^2 to be dimensionally correct, which of the following must be true?

Options

Option A

Only the dimensions of xx and vtvt must be the same.

Option B

Only the dimensions of vtvt and 12at2\frac{1}{2}at^2 must be the same.

Option C is correct

The dimensions of xx, vtvt, and 12at2\frac{1}{2}at^2 must all be the same.

Option D

The dimensions of xx and aa must be the same.

Explanation

The principle of dimensional homogeneity states that an equation is dimensionally correct if the dimensions of all the terms on both sides of the equation are the same. In the given equation:

  1. The dimension of xx (displacement) is [L][L].
  2. The dimension of vtvt (velocity [LT1][LT^{-1}] multiplied by time [T][T]) is [LT1][T]=[L][LT^{-1}][T] = [L].
  3. The dimension of 12at2\frac{1}{2}at^2 (a dimensionless constant 12\frac{1}{2} multiplied by acceleration [LT2][LT^{-2}] and time squared [T2][T^2]) is [LT2][T2]=[L][LT^{-2}][T^2] = [L]. For the equation to be dimensionally correct, all terms must have the same dimensions, which in this case is [L][L].