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Question. 1. Dimensional analysis can be used to spot incorrect equation.

Example: Given the equation , is this equation correct if V has the dimension [L]/[T], x has dimension [L] and a has dimension [L]/[T2]?

Answer: the equation is incorrect because ax2 has the dimension [L3]/[T2] which is not the same as the dimension of V2 which is [L2]/[T2].

Question: Which of the following equations is INCORRECT? The dimension of x, v, a and t are [L], [L]/[T], [L]/[T2] and [T].

A. X= 1/2at2

B. V= x-xi /t

C. V=vi + at

D. X=xi +vt2

The idea of dimensional analysis also works with units.

Example: Given the equation v = at where v has the unit of m/s and t has the unit of s, what is the unit of a?

Answer: a must have a unit of m/s2 so that the product at has the unit m/s just like the unit of v. Think of it as an equation for the units

Solve for unit of a by dividing both sides by the unit s.

Question: Given the equation: T=2π√I/g . If T has the unit of second (s) and l has the unit of meter (m), what is the unit of g?

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