Question

Which of the equations below are dimensionally correct? (where v is velocity, x is
position, a is acceleration, and t is time)𝑥s
=
𝑥
𝑖
+
𝑣
𝑥
𝑖
𝑡
+
1
2
𝑎
𝑡
2
x=x
i
​
+v
xi
​
t+
2
1
​
at
2

Answer 1

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:

1. **First Term: \(x\)**
- Dimension: \( \text{L} \)

2. **Second Term: \(x_i\)**
- Dimension: \( \text{L} \)

3. **Third Term: \(v_{xi} t\)**
- Dimension of \(v_{xi}\): \( \text{LT}^{-1} \)
- Dimension of \(t\): \( \text{T} \)
- Combined Dimension: \( \text{LT}^{-1} \cdot \text{T} = \text{L} \)

4. **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.