Math problem
 

Have been trying to figure this out for some time now.Say I,am running in Quick Rod. I have a perfect light and run dead on with an o. My opponent cuts a .425 light on a pro tree and runs dead on. The differance of the race would be .025. How many feet differancewould that be in inches at the finish line? Thanks

For someone to figure this out they would have to know et of both vehichles and mph of both vehicles at different points of the run. Say your car had alot more bottom end than his and he was catching you on the big end, he would be making up more ground then you making the gap he could travel in .025 bigger than if yall ran the same mph at each check point making the gap he traveled in .025 smaller. Don't know if I made any sense to you. If this was a bracket race then you would have to throw in your dial-in.

You would need the margin of victoryand the mph of the 2nd car to the line. Take mph and convert it to feet per second.

if my math is right
1 mph= 1.466 ft per sec.

so multiply the 2nd car to the finsh mph by 1.466

let's say 150 mph x 1.466 =219 feet per sec.

Take amount of time it takes the 2nd car to get there, you said .025 and multipy that by the 219 feet per sec. that comes out to 5.475 feet.

If you want it in inches, multiply that answer by 12.

It's early in the morning. Not quite awake yet. I hope my math is right.

math
 

Thanks guys. I have been trying to figure out want the win margin of each race like at a national event would of been. Just trying to get a visual of it and keep in my mind how important it is to cut an ooo light. Thanks for the input, I now better understand it.