Nadia S.
asked 01/19/22Which of the following equations are dimensionally consistent? (a) ๐ = ๐ ๐ ๐๐ ๐ , (b) t=v/x, (c) t= (๐๐/๐) ๐/๐
Which of the following equations are dimensionally consistent? (a) ๐ = ๐ ๐ ๐๐ ๐ , (b) t=v/x, (c) t= (๐๐/๐) ๐/๐
1 Expert Answer
James C. answered 01/20/22
Experienced (30+ years) high school physics teacher, conceptual to AP
Each of the equations you listed above consist of various combinations of the dimensions length (L) and time (T). To determine whether an equation is dimensionally consistent, begin by writing out the dimensions of each component on each side of the equation. For instance, in the first equation, x = 1/2 at2, we have the following:
x = 1/2 a t2
L = L/T2 T2 (Note that no dimensions are associated with the 1/2)
Next, simplify the combinations of dimensions on each side of the equation. Here the left is already simplified. On the right, multiplying (L/T2) by T2 results in L.
Finally, compare the left and right sides of the equation. Here, with L on each side, we have agreement, so the equation is dimensionally consistent.
For the second equation, we have T on the left and (L/T) / L on the right. Simplifying the right hand side gives us 1/T, which is not the same as the T on the left, so this equation is not consistent. (The correct expression here would instead be t = x/v.)
For the third equation, we have T on the left. On the right, we find the square root of L/ (L/T2). This expression reduces to the square root of T2, which is consistent with the T on the left.
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Stanton D.
Hi Nadia S., It would really help if you took the trouble to write the equations as they were presented to you. "x = 1 2 at 2" suggests that at some (other) variable having a value of 2, x is a matrix of value [1 2] . Which is bizarre. Now, only the classification under Physics saves you -- equation (a) is for distance travelled vs. time at a constant acceleration (and that's a "(1/2)" not a "1 2"!) and that's OK for dimensions, and so on. So rewrite the equations (correctly!) including their typical units, and solve as identities for just the units alone. That's what "dimensionally consistent" means. So for (a), a has units m s^-2, t has units s, so t^2 has units s^2. Therefore, (1/2) at^2 has units m, which is appropriate for a distance, isn't it? You should be using units throughout all your calculations, to check that you've (we hope) chosen the proper equations and done the math correctly. If your units DON'T come out right, then you've done something wrong, big time. -- Cheers, --Mr. d.01/19/22