Calculate Acceleration in g's from m/s²

  • Thread starter Thread starter Hemib
  • Start date Start date
  • Tags Tags
    Acceleration
AI Thread Summary
To calculate acceleration in g's from m/s², divide the acceleration value by 9.81, which represents the acceleration due to gravity (1g). For instance, if the car's acceleration is 6 m/s², the calculation would be 6 / 9.81. This yields approximately 0.61 g's. Understanding this conversion is essential for comparing different accelerations in terms of gravitational force. The formula is straightforward and can be applied to any acceleration value.
Hemib
Messages
14
Reaction score
0
Quick question:
If I have an acceleration of the car in the longitudinal direction (for example 6 m/s²), how can I calculate the acceleration in g's?

Thx!
 
Physics news on Phys.org
Acceleration due to gravity is 1g = 9.81m/s2.

So you would work out how many g's there are in your value.

For example, [your value] / 9.81 = number of g's
 
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/
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top