To show that the equation [tex]\( x = x_0 + v_0 t + \frac{1}{2} \alpha t^2 \)[/tex] is dimensionally consistent, we need to ensure that each term on the right-hand side has the same dimension as [tex]\( x \)[/tex], which is a distance.
1. Identify the dimensions:
- Distance [tex]\( x \)[/tex]: The dimension of [tex]\( x \)[/tex] is length, denoted as [tex]\([L]\)[/tex].
- Initial distance [tex]\( x_0 \)[/tex]: The dimension of [tex]\( x_0 \)[/tex] is also length, [tex]\([L]\)[/tex].
- Velocity [tex]\( v_0 \)[/tex]: Velocity is distance per unit time, so its dimension is [tex]\([L][T]^{-1}\)[/tex].
- Time [tex]\( t \)[/tex]: The dimension of time is [tex]\([T]\)[/tex].
- Acceleration [tex]\( \alpha \)[/tex]: Acceleration is velocity change per unit time, so its dimension is [tex]\([L][T]^{-2}\)[/tex].
2. Check each term in the equation:
- Dimension of [tex]\( x_0 \)[/tex]:
[tex]\[
x_0 \text{ has dimension } [L]
\][/tex]
- Dimension of [tex]\( v_0 t \)[/tex]:
[tex]\[
v_0 \text{ has dimension } [L][T]^{-1} \text{ and } t \text{ has dimension } [T]
\][/tex]
[tex]\[
\Rightarrow \text{Dimension of } v_0 t = [L][T]^{-1} \cdot [T] = [L]
\][/tex]
- Dimension of [tex]\(\frac{1}{2} \alpha t^2 \)[/tex]:
[tex]\[
\alpha \text{ has dimension } [L][T]^{-2} \text{ and } t^2 \text{ has dimension } [T]^2
\][/tex]
[tex]\[
\Rightarrow \text{Dimension of } \alpha t^2 = [L][T]^{-2} \cdot [T]^2 = [L]
\][/tex]
3. Summing these dimensions:
Each term on the right-hand side of the equation [tex]\( x_0 + v_0 t + \frac{1}{2} \alpha t^2 \)[/tex] has dimension [tex]\([L]\)[/tex].
Therefore, the equation is dimensionally consistent because all terms add up dimensionally to [tex]\( [L] \)[/tex], which matches the dimension of [tex]\( x \)[/tex].
Thus, the equation [tex]\( x = x_0 + v_0 t + \frac{1}{2} \alpha t^2 \)[/tex] is dimensionally consistent.
