StringJunky Posted April 5, 2018 Share Posted April 5, 2018 (edited) Can someone explain how these are mathematically the same? Edited April 5, 2018 by StringJunky Link to comment Share on other sites More sharing options...
koti Posted April 5, 2018 Share Posted April 5, 2018 (edited) 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 April 5, 2018 by koti Link to comment Share on other sites More sharing options...
StringJunky Posted April 5, 2018 Author Share Posted April 5, 2018 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 More sharing options...
pzkpfw Posted April 5, 2018 Share Posted April 5, 2018 It is just math. e.g. 20/2/2 = 10/2 = 5 20/2^2 = 20/4 = 5 3 Link to comment Share on other sites More sharing options...
koti Posted April 5, 2018 Share Posted April 5, 2018 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. 1 Link to comment Share on other sites More sharing options...
pzkpfw Posted April 5, 2018 Share Posted April 5, 2018 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).) 2 Link to comment Share on other sites More sharing options...
DrP Posted April 5, 2018 Share Posted April 5, 2018 I'll second that - it is just rearrangement of the same algebraic expression. s x s is s squared. m/s/s = m/ sxs 1 Link to comment Share on other sites More sharing options...
Strange Posted April 5, 2018 Share Posted April 5, 2018 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"). 1 Link to comment Share on other sites More sharing options...
StringJunky Posted April 5, 2018 Author Share Posted April 5, 2018 24 minutes ago, pzkpfw said: It is just math. e.g. 20/2/2 = 10/2 = 5 20/2^2 = 20/4 = 5 Thanks. I was just reading it wrong and couldn't see how it was mathematically equivalent. Thanks guys. Link to comment Share on other sites More sharing options...
pzkpfw Posted April 5, 2018 Share Posted April 5, 2018 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 1 Link to comment Share on other sites More sharing options...
studiot Posted April 5, 2018 Share Posted April 5, 2018 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] 2 Link to comment Share on other sites More sharing options...
StringJunky Posted April 5, 2018 Author Share Posted April 5, 2018 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 More sharing options...
Bender Posted April 5, 2018 Share Posted April 5, 2018 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 More sharing options...
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