# Works only when using seconds,but not minutes -really need help

1. Sep 13, 2005

### beginner16

hello

I just started learning physics and already am stuck

Train starts moving with constant acceleration a . In five minutes it travels 5 kilometers . Find a !

S = a * t^2 / 2

a = 2 * S / t^2

My question is why does this formula work only if we express time in seconds ?
If I don't convert 5 minutes into seconds prior to putting variable t into formula , the result is wrong .
Why not correct when t is expressed in minutes , and wrong when expressed in seconds ?
When deriving that formula , I found nothing nothing in it that was specific only for seconds

thank you

2. Sep 13, 2005

### Curious3141

You can express distance and time in whatever units you want, as long as you keep the units consistent.

If you put distance in km and time in minutes, you get a in km/min^2.

You can interconvert like so :

$$1 km/{min}^2 = 1 km/(60s)^2 = \frac{1}{3600} km/s^2 =( \frac{1}{3600})(1000m)/s^2 = \frac{1000}{3600} m/s^2 = \frac{5}{18}m/s^2$$

3. Sep 13, 2005

### FredGarvin

When you are rearranging the variables in an equation, you are not changing any of the relationships between them, just restating them. When you deal with units, you have to have a base from which to build on. It will depend on the system you are using, but for the SI system, the base units are kg for mass, m for length and sec for time. All other units you use will depend on combinations of these basic units. For example, the unit for force is a Newton (N). That can be broken down into (kg*m)/(sec^2). That is the definition of a Newton. Can you now see why you have to use seconds as your time base instead of minutes? The definition of the derived unit demands that you use seconds.

In your example, if you were to calculate the acceleration in the units of km/min^2, you would be incorrect ONLY if you are asked to calculate it in the units of m/sec^2. Calculate the acceleration any way you want, but depending on what you are going to do with that number, you may need to convert it to other units. So, if your only end was to calculate the acceleration of the train, you could simply calculate it in km/min^2 and be done. However, if you needed that acceleration value to then calculate a force, you would HAVE to convert it to m/sec^2 for the reasons I stated above.

I hope this helps. If not, keep asking questions.

4. Sep 13, 2005

### beginner16

I'm sorry but it still confuses me .

I know that looking at it from pure mathematical stand point we could consider minutes and seconds as variables

x=min and y=sec and as such

1x = 60y -> ( x^2 ) = ( 60^2 * y^2 )

But for some reason I can't figure out why it all works out just as in real world when we apply those conversion to physic formulas ?

I'm probably not making much of a sense huh?!

5. Sep 13, 2005

### lightgrav

So, a = 2*S/t^2 = 2*(5[km])/(5[min])^2 = 0.4 [km/min^2] .
Another way of writing this is a = 0.4 [(km/min)/min] .

We can, if we want, replace either [ /min] with [ /60s],
so that a = .00667 [(km/min)/s] = .00667 [(km/s)/min]
= 6.67 [(m/min)/s] = 6.67 [(m/s)/min] ,
after repacing the [km ] with [1000m ] .

If you want, you can replace the other [ /min] also, to get
a = 0.111 [(m/s)/s]

It's your choice. SI prefers seconds, but ...
you probably DO want to convert to the units that the book uses,
to check with the "answer-in-the-book".

6. Sep 13, 2005

### HallsofIvy

Staff Emeritus
No, we should consider minutes and seconds as units! It's the numbers associated with those units that are variables!

Do you mean x= 1 min and y= 1 second? Then, yes, 1 min equals 60 seconds and 1 "square minute" equals 3600 "square seconds"- but I refuse to even speculate on what "square minutes" and "square seconds" are- those are yours, not mine!

Well, not a whole lot! But the point is that units in our formulas are expressed in such a way as to make those ideas work- you can treat units as if they were algebraic expressions. speed, distance divided by time, is written by "normal people" as "miles per hour", or "meters per second" but physicists write them as mi/hr or m/s for the express purpose of making the formulas work!

As for your first question, "My question is why does this formula work only if we express time in seconds ?", That's simply not true.

The formulas you gave
S = a * t^2 / 2

a = 2 * S / t^2

work just as well taking t to be minutes (which is how time is given to you).

The exercise was "A train starts moving with constant acceleration a . In five minutes it travels 5 kilometers . Find a !"

Okay, a= 2*(5 km)/(5 minutes)2= 10 km/25 min2= (2/5) km/min2. Of course, because I used minutes, my answer gives acceleration in "km/min2" or "kilometers per minute per minute".

You may have just finished working with "gravity" problems where the acceleration is automatic: 9.81 m/s2 or 32.2 ft/s2. In those problems, because the physical constant, g, is given in terms of meters and seconds, or ft and seconds, to be consistent you need to express t in terms of seconds. You could, of course, express g in terms of "meters per minute per minute". Since there are 60 seconds to a minute there are 3600 s2 per min2 (Hey, we're using square minutes and square seconds!!) and so g= (9.81)/(3600) m/min2= 0.002725 m/min2. Use that and you are free to do all your calculation in meters and minutes!

7. Sep 14, 2005

### beginner16

I think what is confusing me the most is that by my logic if car has
acceleration

a = (10 m/s)/s then after time period of two seconds car accelerates from

v0 to v0 + 10 m/s and in one minute it accelerates from

v0 to v0 + 60*10 m/s

So by my logic acceleration a (m / min^2) should be

600 m /min^2 , but instead is 36000 m /min^2

darn it this is confusing me

8. Sep 14, 2005

### Staff: Mentor

What you are calculating is acceleration in m/sec/min, not to be confused with m/min/min. So, you are correct: the acceleration is 600 m/sec/min!

9. Sep 14, 2005

### beginner16

I think I get it now

thank you all for your help

cheers