Which expression is dimensionally consistent with an expression that would yield a value for time $$ { }^{-1} $$ ? ( $$ v $$ is velocity, $$ x $$ is distance, and $$ t $$ is time) 1) $$ v / x $$ 2) $$ v^{2} / x $$ 3) $$ x / t $$ 4) $$ v^{2} t $$ 5) $$ x / v $$

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To determine which expression is dimensionally consistent with yielding a value for time $$ { }^{-1} $$ (i.e., units of $$ \text{T}^{-1} $$), we'll analyze the dimensions of each given option.

**Given:**
- **Velocity ($$ v $$)** has dimensions: $$[L][T]^{-1}$$
- **Distance ($$ x $$)** has dimensions: $$[L]$$
- **Time ($$ t $$)** has dimensions: $$[T]$$

**Analyzing Each Option:**

1. **$$ \frac{v}{x} $$**
   $$
   \frac{[L][T]^{-1}}{[L]} = [T]^{-1}
   $$
   **Dimensions:** $$ \text{T}^{-1} $$

2. **$$ \frac{v^{2}}{x} $$**
   $$
   \frac{([L][T]^{-1})^{2}}{[L]} = \frac{[L]^2 [T]^{-2}}{[L]} = [L][T]^{-2}
   $$
   **Dimensions:** $$ \text{L}\text{T}^{-2} $$

3. **$$ \frac{x}{t} $$**
   $$
   \frac{[L]}{[T]} = [L][T]^{-1}
   $$
   **Dimensions:** $$ \text{L}\text{T}^{-1} $$

4. **$$ v^{2} t $$**
   $$
   ([L][T]^{-1})^{2} [T] = [L]^2 [T]^{-2} [T] = [L]^2 [T]^{-1}
   $$
   **Dimensions:** $$ \text{L}^2\text{T}^{-1} $$

5. **$$ \frac{x}{v} $$**
   $$
   \frac{[L]}{[L][T]^{-1}} = [T]
   $$
   **Dimensions:** $$ \text{T} $$

**Conclusion:**

Only **option 1** ($$ \frac{v}{x} $$) has the dimensions $$ \text{T}^{-1} $$, making it the correct choice.

**Answer:**  
1) $$ v / x $$
$$ \frac{v}{x} $$ is dimensionally consistent with time $$ { }^{-1} $$.

Which expression is dimensionally consistent with an expression that would yield a value for time \( { }^{-1} \) ? ( \( v \) is velocity, \( x \) is distance, and \( t \) is time) 1) \( v / x \) 2) \( v^{2} / x \) 3) \( x / t \) 4) \( v^{2} t \) 5) \( x / v \)

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