Question

a student has derived the following dimensionally inhomogeneous equation: ( a = \frac{x}{t^2} - vt + \frac{f}{m} ) where ( v ) is a velocity’s magnitude, ( a ) is an acceleration’s magnitude, ( t ) is a time, ( m ) is a mass, ( f ) is a force’s magnitude, and ( x ) is a distance (or length). which terms are dimensionally homogeneous? check all that apply. view available hint(s) options: ( \frac{x}{t^2} ), ( a ), ( \frac{f}{m} ), ( vt )

Explanation:

To determine which terms are dimensionally homogeneous with acceleration \( a \) (dimensions of \( \text{length}/\text{time}^2 \) or \( \text{L}/\text{T}^2 \)), we analyze each term:

Step 1: Analyze \( \boldsymbol{\frac{x}{t^2}} \)

• \( x \) (distance) has dimensions \( \text{L} \).
• \( t \) (time) has dimensions \( \text{T} \), so \( t^2 \) has \( \text{T}^2 \).
• Dimensions of \( \frac{x}{t^2} \): \( \frac{\text{L}}{\text{T}^2} \), which matches \( a \) (acceleration).

Step 2: Analyze \( \boldsymbol{a} \)

• \( a \) is acceleration, with dimensions \( \text{L}/\text{T}^2 \). It is trivially homogeneous with itself.

Step 3: Analyze \( \boldsymbol{\frac{F}{m}} \)

• From Newton’s second law (\( F = ma \)), \( \frac{F}{m} = a \).
• Thus, \( \frac{F}{m} \) has dimensions \( \text{L}/\text{T}^2 \), matching \( a \).

Step 4: Analyze \( \boldsymbol{vt} \)

• \( v \) (velocity) has dimensions \( \text{L}/\text{T} \).
• \( t \) (time) has dimensions \( \text{T} \).
• Dimensions of \( vt \): \( \frac{\text{L}}{\text{T}} \times \text{T} = \text{L} \) (length), which does not match \( \text{L}/\text{T}^2 \).

Answer:

The dimensionally homogeneous terms are \( \boldsymbol{\frac{x}{t^2}} \), \( \boldsymbol{a} \), and \( \boldsymbol{\frac{F}{m}} \). So check the boxes for \( \frac{x}{t^2} \), \( a \), and \( \frac{F}{m} \).