Jump to content

Freefall acceleration: 9.8m/s/s = 9.8m/s^2


StringJunky

Recommended Posts

1 hour ago, StringJunky said:

Can someone explain how these are mathematically the same?

I think its just a different way of expressing acceleration. What the s^2 really means is [(meters per second) per second] - (m/s)/s in other words, that as every second passes, the acceleration will raise by 9.81m when free falling. 

Edited by koti
Link to comment
Share on other sites

51 minutes ago, koti said:

I think its just a different way of expressing acceleration. What the s^2 really means is [(meters per second) per second] - (m/s)/s in other words, that as every second passes, the acceleration will raise by 9.81m when free falling. 

I've read they are the same and 'm/s/s makes sense to me but m/s2 looks like to me it's saying velocity is squared with each passing second,  

Link to comment
Share on other sites

3 minutes ago, StringJunky said:

I've read they are the same and 'm/s/s makes sense to me but m/s2 looks like to me it's saying velocity is squared with each passing second,  

Its just an expression that works under the math framework. Just like c^2 or c^4 in equations which doesn’t mean that a velocity is a actually greater than c by a power of 2 or 4. 

Link to comment
Share on other sites

8 minutes ago, StringJunky said:

I've read they are the same and 'm/s/s makes sense to me but m/s2 looks like to me it's saying velocity is squared with each passing second,  

Consider what the "/s/s" means. It's "per second per second". That equates to "per second squared".

(A second is a measure of time, it's not (by itself) a measure of speed. (Velocity implies direction).)

Link to comment
Share on other sites

Don't want to push a dead topic but another thing occurred to me just before the previous post and I wanted to get it off my chest, that is, to show algebraically how the .../s/s becomes ...s^s.

The thing to remember is that A/B = A x 1/B

So m/s/s = m/s x 1/s = (m x 1)/(s x s) = m/s^2

 

Link to comment
Share on other sites

It arises because when you divide by something it is the same as multiplying by the reciprocal of that something.

Since metres per second means metres divided by seconds it also means metres multiplied by the reciprocal of seconds or 1/seconds.


[math]metres \div \sec onds = metres \times \frac{1}{{\sec onds}}[/math]


So if we divide this again by seconds we get


[math]\left[ {metres \times \frac{1}{{\sec onds}}} \right] \div \sec onds = \left[ {metres \times \frac{1}{{\sec onds}}} \right] \times \frac{1}{{\sec onds}}[/math]


or


[math] = metres \times \frac{1}{{\sec onds}} \times \frac{1}{{\sec onds}} = \frac{{metres}}{{\sec onds{}^2}}[/math]

 

Link to comment
Share on other sites

12 minutes ago, pzkpfw said:

Don't want to push a dead topic but another thing occurred to me just before the previous post and I wanted to get it off my chest, that is, to show algebraically how the .../s/s becomes ...s^s.

The thing to remember is that A/B = A x 1/B

So m/s/s = m/s x 1/s = (m x 1)/(s x s) = m/s^2

 

 

10 minutes ago, studiot said:

It arises because when you divide by something it is the same as multiplying by the reciprocal of that something.

Since metres per second means metres divided by seconds it also means metres multiplied by the reciprocal of seconds or 1/seconds.


metres÷seconds=metres×1seconds


So if we divide this again by seconds we get


[metres×1seconds]÷seconds=[metres×1seconds]×1seconds


or


=metres×1seconds×1seconds=metresseconds2

 

Thank you chaps. Great job.

31 minutes ago, Strange said:

It is confusing because, algebraically, m/s/s looks like it should re-arrange as ms/s. But that is just because the m/s/s notation is misleading ("wrong").

I appreciate now that m/s/s is the naive/wrong  way but it is intuitively self-explanatory.

Link to comment
Share on other sites

11 hours ago, StringJunky said:

I appreciate now that m/s/s is the naive/wrong  way but it is intuitively self-explanatory.

It is not wrong. You just have to use the order of operations properly.

m/s/s = (m/s)/s and not m/(s/s)

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.