LIVETUTORIALS LEARNING LIBRARY
Check a Physics Equation with Dimensions
Test the dimensional consistency of motion formulas, convert speed units, and understand why matching dimensions do not prove a formula.
Understand the concept
A dimensional check compares the kinds of physical quantities represented by an equation. Use M for mass, L for length and T for time. Square brackets can denote dimensions: [v] = LT^-1 means that velocity has the dimension length divided by time. The notation describes a relationship, not a particular choice of metres, kilometres or seconds.
For straight-line motion, let s be displacement, u initial velocity, a acceleration and t elapsed time. Their dimensions are [s] = L, [u] = LT^-1, [a] = LT^-2 and [t] = T. Therefore [ut] = L, [at] = LT^-1 and [at^2] = L. Only terms with the same dimensions can be added as contributions to s.
Consider a proposed expression s = ut + at. The first term is a length, while the second has the dimension of velocity. The expression fails the check. Replacing the final term with a quantity proportional to at^2 makes the dimensions compatible, but dimensional reasoning alone does not fix the proportionality constant.
Under constant acceleration, the familiar relation is s = ut + (1/2)at^2. Its dimensions match, but an expression with a different dimensionless multiplier would also pass a dimensional check. Establishing the actual factor requires the motion model or a derivation; dimensions cannot supply it by themselves.
For numerical work, convert to compatible units first and retain them through multiplication and division. A speed of 72 kilometres per hour is 72 × 1,000/3,600 = 20 metres per second. The unit factors cancel kilometres and hours while leaving metres per second.
Finally, distinguish a consistent formula from a proven physical statement. Dimensional analysis can reject certain mistakes, but it cannot choose between different quantities with the same dimensions, recover missing dimensionless constants, or replace the assumptions of a model.
Worked example
A body has initial velocity 3 m/s and constant acceleration 2 m/s^2 for 4 s. Check the formula s = ut + (1/2)at^2 and calculate its displacement.
First term: (3 m/s)(4 s) = 12 m. Second term: (1/2)(2 m/s^2)(4 s)^2 = (1/2)(2)(16) m = 16 m. Both terms are lengths, so adding them is dimensionally meaningful. The displacement is 12 m + 16 m = 28 m.
The alternative expression ut + at would attempt to add 12 m to 8 m/s. Those units are incompatible. The expression ut + at^2 would have compatible units but would produce the wrong constant-acceleration result because it lacks the factor one half. This separates what a dimensional check can detect from what it cannot.
Practice and answer
1. Convert 54 km/h to m/s.
Answer: 54 × 1,000/3,600 = 15 m/s.
2. Force has dimension MLT^-2 and area has dimension L^2. Find the dimensions of pressure = force/area.
Answer: MLT^-2/L^2 = ML^-1T^-2. In SI units this is kg m^-1 s^-2, equivalent to N/m^2.
3. Does v = u + at^2 pass a dimensional check if a is acceleration?
Answer: no. v and u have dimension LT^-1, while at^2 has dimension L.
4. Can dimensional analysis decide whether the coefficient of at^2 in a displacement formula is 1 or 1/2?
Answer: no. Both coefficients are dimensionless; the physical model must determine the correct coefficient.
What you will learn
- Express velocity, acceleration and force in base dimensions.
- Test each term of an equation before inserting numbers.
- Convert speed units and state the limits of dimensional analysis.
Before you start
- Basic multiplication, division and powers.