How do i find 'p' and 'q' to make v = (a^p)(t^q) dimensionally consistent? ?

Question

Velocity is related to acceleration and time by the following expression, v = (a^p)(t^q). Find the powers p and q that make this equation dimensionally consistent.

No idea how to go about this. Any help would be great.

Answer 1

Well, you know that velocity is (usually) expressed as 'm/s'.

Acceleration is m/s².....and time is usually in seconds (s).

Together they give "at" = (m/s²)(s) = m/s........right?

If you raise "a" to anything greater than one you will end up with the mass "m" also raised to that power. There's nothing that can cancel that in the units of "t", and having "m" to anything other than to the first power will make the velocity dimensionally incorrect.

So this means that "p" has to be one.

If "p" is one that means its units will be "m/s²". Therefore you only need one unit of time to cancel one unit of "s" in the "s²" in the denominator. So this means "q" has to also be one.

So....p = q = 1 to give velocity its correct dimensions.

Had the problem been given as m/s^p and t^q...then you'd have a different situation, in which case "q" would have to be greater than "p" by one....q = p + 1 to give the correct dimensionality of velocity.

Does this help and make sense???

Answer 2

v = (a^p)(t^q)

d/t = (d/t²)^p * t^q

if p = 1 and q = 1, this translates to

d/t = (d/t²) * t = d/t

.