If motorcycle #1 runs 1/4 mi in 10.49sec@133mph & #2 in 10.06@145mph how many physical feet would it be won by

Question

im trying to figure out how many feet the faster bike wins by, thanks

thanks guys for being so helpful I was trying to determine the "gap" between the slowest 600cc race bike and the fastest 1000cc race bike, not much difference in a short distance it seems.

Answer 1

It depends on the velocity profile of each vehicle.
If you model the velocity profiles as the following:

V(t) = Vm (1 - exp(-at))
t >= 0

Where Vm is the terminal or asymptotic velocity
and "a" is the exponential acceleration constant,

then
bike #1 has Vm=158 mph and a=630
bike #2 has Vm=189 mph and a=520

then displacement is

S(t) = Vm [t - (1 - exp(-at))/a]
t >= 0

displacement lag of bike #1 at 10.06 sec is 83.26 ft

If you discover that the velocity profile is something other that the guess above, then the displacement lag will be something else, but probably something close.

Answer 2

When the first hits the 1/4 mile mark, the second, traveling approximately 133 mph (195.0667 fps), arrives there exactly .43 sec later. Therefore, # 2 travels 195.0667 * .43 = 83.88 feet.

It's actually a bit less than this if you assume the bikes are accelerating throughout the race. The AVERAGE speed over the last .43 sec is a bit UNDER 133.......

Answer 3

If motorcycle two runs at 145 mph and motorcycle one runs at 133 mph, then motorcycle two runs at 145-133 or 12 mph faster. 12 mph is 5280/3600 seconds or 17.6 feet per second. So multiply 10.06 second times 17.6 and motorcycle two won by 177 feet.

Answer 4

it all depends on the reaction time of each rider there is less than 1/2sec there could be as little as 10ft to as much as 60ft go to the track and lin up have a good eye at the far end of the track to watch for the distance