Check the dimensional correctness of v = u + at²
A wrong first equation of motion. Looks like the first equation of motion, but the last term has the wrong dimensions.
Equation checker
Matching dimensions are necessary, not sufficient. A check cannot see pure numbers such as 1/2 or 2π, so a consistent equation can still be wrong.
v = u + a t2v[L T-1]=u[L T-1]+at2[L]Step by step
1. What each symbol stands for
- v velocity [L T-1]
- u velocity [L T-1]
- a acceleration [L T-2]
- t time [T]
2. The dimensions of each term
- v [L T-1] [L T-1]
- u [L T-1] [L T-1]
- +at2 [L T-2] × [T]2 [L]
3. Compare
"u" and "at2" are added or subtracted, so they need the same dimensions, but "u" is [L T-1] and "at2" is [L].
The term at2 does not match
A likely corrected form:
v = u + at
The final speed after time t with a steady acceleration a.
Check v = u + atWhat it proves and what it does not
It can show
- Every term that is added or subtracted has the same dimensions, so the sum makes sense.
- Both sides of the equals sign have the same dimensions.
- Angles inside sin, cos and similar functions are pure numbers.
It cannot show
- A pure number such as 1/2 or 2π has no dimensions, so a check cannot see a missing or wrong one.
- Two quantities with the same dimensions, such as work and torque, cannot be told apart.
- A sign error or a wrong power that keeps the dimensions equal goes unnoticed.
When the dimensions of the terms differ the equation cannot be correct, whatever numbers are in it, as long as each symbol means what the table says.
Look up each quantity
v = u + at²: quick answers
Is v = u + at² dimensionally correct?
No. "u" and "at²" are added or subtracted, so they need the same dimensions, but "u" is [L T⁻¹] and "at²" is [L].
Does a dimension check prove v = u + at² is right?
No. A check can show that an equation is wrong, but it cannot show that it is right. Pure numbers such as 1/2 and 2π have no dimensions, so the check cannot see them.