## SI Units question in physics conceptual question

1. The problem statement, all variables and given/known data
If speed, v, of an object depends on time, t, according to the equation v = A + Bt + Ct^4, what are the SI units of A, B, C?

2. Relevant equations
v = A + Bt + Ct^4

3. The attempt at a solution
For this, I've not known where to start, so I've be plugging in the variables to try to lighten the question:

m/s = A + B(s) + C(s)^4

m/s = s + m/s(s) + m/s^2(s)^4

m/s^5 = s + m/s(2) + m/s^2

m/s^4 = s + m/s + m/s^2

m/s^4 = s + m/s + m

m/s^2 = s + m/s

m/s = s + m

m/s^2 = m

Obviously this is incorrect, can someone help me understand this question? I've gotten past the concepts of this question and just started mindlessly trial and error to plug in values because it's so frustrating but I cannot complete this question. Can anyone point me in the right direct? Any help is appreciated!

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 You can't do math exactly like that with units of measure So try this assume B and C are zero what are the units of measure of A. One down. Now repeat setting A and C to zero, what are uom for B?...

Recognitions:
Homework Help
 Quote by xCrusade 2. Relevant equations v = A + Bt + Ct^4

Think of the right side like this, each combined term must have the same SI units as the 'v' on the left. This is so since you can't add meters to seconds and so on. (like apples and oranges)

so A, Bt and Ct4 should have the same units as 'v'.

What the units of 'v' and hence what are the units of 'A'?

## SI Units question in physics conceptual question

If I substitute B and C with zero, does this cancel out t and t^4 making A = m/s?

Mentor
 Quote by xCrusade If I substitute B and C with zero, does this cancel out t and t^4 making A = m/s?
I'd suggest using Rock.Freak's suggestion. That's the best way to start, IMO.

 I still don't understand, A, B, and, C have to be a combination of m/s, m, s, or m/s^2? I am having trouble grasping this concept.

Mentor
 Quote by xCrusade I still don't understand, A, B, and, C have to be a combination of m/s, m, s, or m/s^2? I am having trouble grasping this concept.
No. The quantity on the lefthand side (LHS) of the equation has units of velocity, or [m/s]. So that means all that each of the 3 terms on the RHS have to also have units of [m/s]. If they had different units and you tried to just add them up and set them equal to the LHS, that would be an error.

You have a term on the RHS that is Bt. the time t has units of [s], so what units must B have to give you the overall units of [m/s]? Remember that you can cancel the same units when they appear on both the top and bottom of a division.

So for example, if I divide a distance by a time, I get the units of velocity:

D[m] / t[s] = V[m/s]

Or if I multply velocity by time, I get distance:

V[m/s] * t[s] = D[m]

See how the seconds units in the numerator and denominator cancel out?

Does that help? What do you need for units for the RHS constants to make each term have units of velocity [m/s] ?

 Based on your description, would this make A = m/s, B = m/s^2, C = m/s^5 since B will cancel out the s from time and C will cancel out the s^4 from time?

Mentor
 Quote by xCrusade Based on your description, would this make A = m/s, B = m/s^2, C = m/s^5 since B will cancel out the s from time and C will cancel out the s^4 from time?
Perfect!

Carrying units along in equations like that can really help you find typos and mistakes as you work. Especially in large algebraic manipulations, checking units along the way can help you find a mistake early and fix it before you waste time on pages of calcs.