Why is the SI unit for acceleration m/(s^2)?

AI Thread Summary
The SI unit for acceleration is expressed as meters per second squared (m/s²) because it represents the change in velocity (meters per second) over time (seconds). The confusion arises from interpreting the unit as meters per second per second, which is mathematically equivalent to meters per second squared. When broken down, acceleration can be represented as m/s divided by s, leading to m/s². This clarification resolves concerns about whether the two expressions yield different results, confirming they are indeed the same. Understanding this relationship is crucial for accurate calculations in physics.
ffleming7
Messages
25
Reaction score
0
Why is the SI unit for acceleration \frac{m}{s^2}(meters per second squared) when it is actually \frac{m}{\frac{s}{s}} (meters per second per second). Isn't the part concerning the seconds different? Wouldn't this give you different answers sometimes, or does that usually never get in the way.
 
Physics news on Phys.org
meters per second per second is the same thing as meters per second squared. If you want to do it division style, the seconds move to the denominator so you might as well write s*s as s^2.
 
ffleming7 said:
Why is the SI unit for acceleration \frac{m}{s^2}(meters per second squared) when it is actually \frac{m}{\frac{s}{s}} (meters per second per second). Isn't the part concerning the seconds different? Wouldn't this give you different answers sometimes, or does that usually never get in the way.
The way you've written it is not correct. It is \frac{\frac{m}{s}}{s} = \frac{m}{s}* \frac{1}{s}
 
Last edited:
Thank you. That makes a lot more sense now. So \frac{\frac{m}{s}}{s} = \frac{m}{s}* \frac{1}{s}=\frac{m}{s^2}.
 
Hi there, im studying nanoscience at the university in Basel. Today I looked at the topic of intertial and non-inertial reference frames and the existence of fictitious forces. I understand that you call forces real in physics if they appear in interplay. Meaning that a force is real when there is the "actio" partner to the "reactio" partner. If this condition is not satisfied the force is not real. I also understand that if you specifically look at non-inertial reference frames you can...
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/
Back
Top