# Determine units of variables in formula

1. Aug 10, 2011

### 1MileCrash

1. The problem statement, all variables and given/known data

During a short interval of time, the velocity v in m/s of an automobile is given by v=at^2 + bt^3 where the time is t in seconds. The units of a and b are...

2. Relevant equations

3. The attempt at a solution

Alright, here's the thing. A friend gave me some old physics tests, I'm taking it in the fall and familiarizing. I don't know how to do it mathematically. I tried just "plugging in" the units to see if it works out (it's multiple choice here) but I want a way to do it mathematically.

Someone, just give me a nudge in the right direction so I can attempt here if you'd be so kind. I don't want the answer, or even too big of a hint.

2. Aug 10, 2011

### SteamKing

Staff Emeritus
Hint: The units on both sides of the equation must match after all mathematical operations are completed.

3. Aug 10, 2011

### HallsofIvy

Staff Emeritus
v(m/s)= a(?) t^2(m^2/s^2)+ b(?)t^3(m^3/t^3)
The "(?)" are the unknown units. As SteamKing said the units on both sides of the equation must be the same.

4. Aug 10, 2011

### 1MileCrash

Thanks.

How does t^2 and t^3 affect the units?

For example, (1 m/s)t^2 does what to m/s? Adds exponends to s, making m/s^3? Or add exponents to both, making m^3/s^3, or neither?

5. Aug 10, 2011

### 1MileCrash

Could someone give me more info? I worked out all of the choices and I never get the same units. HallsofIvy, what are you doing when you say v(m/s)= a(?) t^2(m^2/s^2)+ b(?)t^3(m^3/t^3) ??? Why are those units after t^2 and t^3? Where did they come from?

v = at^2 + bt^3

Option one, m/s^2 and m/s^3 respectively.

v = (m/s^2)(s^2) + (m/s^3)(s^3)

v = (ms^2/s^2 ) + (ms^3/s^3)

v = m + m

No.

Option Two, m/s^4 and m/s^5 respectively.

v = (m/s^4)(s^2) + (m/s^5)(s^3)
v = (ms^2/s^4) + (ms^3/s^5)
v = m/s^2 + m/s^2

No.

Option Three, s^3/m and s^4/m, respectively.

v = (s^3/m)(s^2) + (s^4/m)(s^3)
v = s^5/m + s^7/m

Hell no.

Option Four, m*s^2 and m*s^4, respectively.

v = (m*s^2)(s^2) + (m*s^4)(s^4)
v = ms^2s^4 + ms^3s^7

Bigger Hell no.

I am completely and utterly lost.

Last edited: Aug 10, 2011
6. Aug 10, 2011

### Redbelly98

Staff Emeritus
But there are infinitely many choices! Random guessing is no way to solve the problem.

v units are m/s. So units of a*t^2 must be m/s as well. And we know t has units of s:

a*s^2 = m/s​

Solve for a.

7. Aug 10, 2011

### 1MileCrash

It's a multiple choice problem..

at^2 and I want m/s, my logic says that:

m/s = a(s^2)

(1/s^2)(m/s) = a
m/s^3 = a

Yet this is not an option as the a unit.

8. Aug 10, 2011

### Redbelly98

Staff Emeritus
Ah, okay.
Then the choices are all wrong, since m/s3 are the correct units for a.

9. Aug 10, 2011

### Staff: Mentor

Let's assume for a moment that m/s3 is an option. You have m/s3 * s2 = m/s, the correct units for velocity. Why do you think your answer is not an option?

10. Aug 10, 2011

### 1MileCrash

Again, because it is a multiple choice problem. I'm not saying that the unit is physically impossible, meaningless, or anything else like that. It's just not an option on the test paper out of a, b, c, and d.

That's why I tested the four "options" above.

11. Aug 10, 2011

### PeterO

As a sceptical reader, I would ask that you try to provide a copy of the original question and optional answers.
Not a transcription, but a scan.

12. Aug 10, 2011

### PeterO

First, let's just look at a

v = at2

dimensionally [in units] that is

[m/s] = [a] x [s2]

"divide" both sides my s2 we get [a] = [m/s3]

doing a similar operation for b means that dimensionally

= [m/s4]