SOLUTION: Dimensional analysis, dimensional analysis physics ... - Free Printable

Let’s go step by step to check the dimensional homogeneity of the two equations shown in your notes.

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First Equation:

vₜ = vᵢ + at

We need to check if all terms have the same dimensions.

- vₜ (final velocity) → dimension is [LT⁻¹]
- vᵢ (initial velocity) → also [LT⁻¹]
- a (acceleration) → [LT⁻²]
- t (time) → [T]

So, at has dimension:
[LT⁻²] × [T] = [LT⁻¹]

Now compare all terms:
- Left-hand side (LHS): vₜ → [LT⁻¹]
- Right-hand side (RHS): vᵢ + at → [LT⁻¹] + [LT⁻¹] = [LT⁻¹] (since you can only add quantities with same dimensions)

✔ So LHS = RHS → Dimensionally correct

*(Note: In your notes, it says “2[LT¹]” — that’s not quite right. You don’t multiply by 2; you just say both terms are [LT⁻¹], so they match. The “2” might be a typo or miswriting.)*

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Second Equation:

S = vᵢt + ½at²

Check each term:

- S (displacement) → [L]
- vt → [LT⁻¹] × [T] = [L]
- at² → [LT⁻²] × [T²] = [L]
→ So ½at² is still [L] (the ½ is a number, no dimension)

So:
- LHS: S → [L]
- RHS: vt + ½at² → [L] + [L] = [L]

✔ So LHS = RHS → Dimensionally correct

*(In your notes, you wrote: [L] = [LT⁻¹][T] + ½[LT⁻²][T]² — which simplifies correctly to [L] = [L] + [L]. Good!)*

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Final Check:

Both equations are dimensionally consistent. That means they *could* be physically correct (though dimensional analysis doesn’t prove they’re definitely correct — just that they’re not obviously wrong).

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Final Answer:
Both equations are dimensionally correct.