To check if the given equation is dimensionally correct, we will analyze and confirm the dimensions for each term in the equation. The equation is:
\[ x = x_i + v_{xi} t + \frac{1}{2} a t^2 \]
where:
- \(x\) and \(x_i\) are positions (in meters, \( \text{L} \)),
- \(v_{xi}\) is velocity (in meters per second, \( \text{LT}^{-1} \)),
- \(t\) is time (in seconds, \( \text{T} \)),
- \(a\) is acceleration (in meters per second squared, \( \text{LT}^{-2} \)).
Analyzing Each Term:
-
First Term: \(x\)
- Dimension: \( \text{L} \)
-
Second Term: \(x_i\)
- Dimension: \( \text{L} \)
-
Third Term: \(v_{xi} t\)
- Dimension of \(v_{xi}\): \( \text{LT}^{-1} \)
- Dimension of \(t\): \( \text{T} \)
- Combined Dimension: \( \text{LT}^{-1} \cdot \text{T} = \text{L} \)
-
Fourth Term: \(\frac{1}{2} a t^2\)
- Dimension of \(a\): \( \text{LT}^{-2} \)
- Dimension of \(t^2\): \( \text{T}^2 \)
- Combined Dimension: \( \text{LT}^{-2} \cdot \text{T}^2 = \text{L} \)
Assembling the Equation
Now we can reassemble the dimensions of the entire equation:
- Left side: \(x\) has dimensions of \( \text{L} \).
- Right side: \(x_i + v_{xi} t + \frac{1}{2} a t^2\) consists of:
- \(x_i\) (dimension \( \text{L} \))
- \(v_{xi} t\) (dimension \( \text{L} \))
- \(\frac{1}{2} a t^2\) (dimension \( \text{L} \))
Adding terms of the same dimension results in a final dimension of \( \text{L} \).
Conclusion
Since both sides of the equation have dimensions of \( \text{L} \), the equation:
\[ x = x_i + v_{xi} t + \frac{1}{2} a t^2 \]
is dimensionally correct.