Discover the G Forces of Braking from 290mph in My Grand Turismo 3 Car

  • Thread starter Thread starter Gara
  • Start date Start date
  • Tags Tags
    Car
AI Thread Summary
The discussion centers on calculating the G forces experienced by a driver braking from 290 mph to 0 in 2.5 seconds. The calculations yield an acceleration of approximately -116 m/s², equating to about 11.8 Gs, while another participant calculates it as -51.9 m/s² or 5.3 Gs. Both methods confirm similar results when converted to English units, with an acceleration of -170.33 ft/sec² also resulting in 5.3 Gs. The conversation humorously concludes with a reminder about the importance of seat belts in such extreme conditions. The calculations highlight the intense forces involved in high-speed braking scenarios.
Gara
Messages
152
Reaction score
0
what kind of G forces does the driver come under when breaking from 290 mph to 0 in 2.5 seconds?
 
Physics news on Phys.org
Let's get it in meters per second...
129.642 Meters per second.

Now, our acceleration has to be cahgne in velocity over time. (FinalVelocity-InitialVelocity)/Time...

-116 m/s^2...
We are talking in absolute value...116 m/s^2. Then divide that by 9.81...11.8 G's.

I think that's right.
 
How did you get -116 m/s^2 ? Going from 129.9 m/s to 0 m/s in 2.5 second is (0- 129)/2.5= -51.9 m/s^2. That's 5.3 g.

You can, of course, do it in English units: )(290 mi/hr)/((60 min/hr)(60 sec/min)))*(5280 ft/mi)= 425.33 ft/sec.
Going from 425.33 ft/sec to 0 ft/sec in 2.5 seconds is an acceleration of (0- 425.33)/2.5= -170.33 ft/sec^2.
The acceleration due to gravity (g) is about -32.2 ft/sec^2 so
-170.33 ft/sec^2= -170.33/-32.2= 5.3 g again.
 
hope my guy is using a seat belt! lol
 
Originally posted by Gara
hope my guy is using a seat belt! lol

Hope it's not a 5-point! Unless he already has kids..
 
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