Recent content by attriola

aim, got it. After several trials with either of the two forms of the equation, I came to the realization that the initial iteration requires this form of the equation: vF=(a*t); va=(vF-vI)/2.0; t=(dx/va); And the subsequent iterations require this form: vF=sqrt((vI*vI)+((2*a)*(dx)))...

Thanks again, I came to that realization about 4:30 AM (pst), so got up to re-post but you beat me to it. Of course, it makes sense on several levels. For one thing, I can't use (a) directly anyway, it's meaningless until multiplied by time. In the code I posted, the only thing time was used...

aim, I found my error (I think): With time unknown: vF^2 = vI^2 + 2a(sF-sI) (source: http://en.wikipedia.org/wiki/Equations_of_motion) I naively oversimplified the relationship...considerably! (although it still seems intuitively correct). So now, I have to do this: vF=sqrt(vI^2 +...

My apologies, aim, I'll keep it descriptive (I hope). The term vF+=a is a short form of vF=vF+a, (a) was previously calculated (just once). Here's what the program does: Initial velocity is defined as: 230.33 (500 mph), there is no need to calculate it since it's known. Distance...

Thanks again aim, are you game for a follow-up? In trying to calculate elapsed-time (say, for a mile), I noticed that it looked too high for the velocity I was logging. (500 mph in 100 feet) Using 'average' velocity each time through the loop, (t) is essentially static. This didn't...

Thanks, your quite right, I did. Yes, I've since run into that on a thread-search and found a possible solution suggested by Mike W, but tell me if I have it right: Keeping in mind that beginning and ending velocities are known, distance is known, and, as you pointed out, we have constant...

Homework Statement Find the time it takes to traverse the distance. Given: *constant velocity *initial acceleration = 0 *ending acceleration= 11176 m/s *distance traversed = 28.96 km (18 miles) Homework Equations velocity: v^2 = v0^2 + 2a(delta x) v = sqrt(v0^2 + 2a(delta x))...