Speeling?
Missuse?

Thats 2 in one post.

 

10-4.. I appologize for my errors, sir [img]i/expressions/face-icon-small-wink.gif[/img]

 

Cowboy, Is that your own writing on your website? That's good stuff if so.

 

Yes.. Songs or ramblings or poetry. The long one just my take on life at that time. the other ones were about a girlfriend that wasn't so nice to OFC and further caused OFC to want to punish women for the wicked they do [img]i/expressions/face-icon-small-wink.gif[/img]

 

Sorry, I just saw where someone said he would be passed at 1319' or 1' short of the finish. So I was showing that the Nascar would loose by some 18' and not a 1 ft. But I did fiqure out the answer, but it took a bit. The Nascar is easy because it has a consistant speed of 200mph or 293.3 ft/sec. The top fuel was quite harder as the top fuel is allways gaining through out the 1/4 mile. But I finally got it. If the top fuel maintains a constant speed right up to the 1/4 mile and ends at 336.15mph at the end, it would roughly loose 2.55mph every 10'. So every 10' you can fiqure how many ft per second then to 1/100 of a sec. So with all of this, the top fuel would pass the Nascar at the 1267.26' mark or 52.74' from the finish going 323.01mph and this would be done at 4.328 sec after the start. So in just .109 more sec. the top fuel would pass the finish in 4.437 thus beating the Nascar by 18.48'[img]i/expressions/face-icon-small-happy.gif[/img]

 

GE4x4 is the pretty much right on.

The dragster accelerates non-linearly for the first 1/8 mile. The last 1/8th mile we assume he accelerates at a constant rate. He only gains 68 mph in the last 1/8 mile out of 336 mph, so this is probably a safe assumption. Furthermore, he is probably in top gear so acceleration will not increase much, it may even fall some.

Given that we assume a constant acceleration, we can multipy average speed by time to get the distance traveled.
The average speed on the last 1/8 mile is (336.15+268)/2=302.075 mph or 0.0839097 mi/sec

Distance that dragster travels when it passes = 1/8 mile + AverageSpeedOverLast8th * (TIME - 1/8 mile time)
Distance that Nascar travels when it gets passed = 0.055555 mi/sec * TIME

the 1/8 mile time is given by 4.437 sec - 1/8 mi / averagespeed = 4.437 - 0.125/0.0839097 = 2.947303 sec

In simpler form:

S = 0.125 + 0.0839097 * (t - 2.947303)
S = 0.05555 * t

Two equations, two unknowns:

t = 4.31348 seconds is when the dragster passes the Nascar
S = 0.05555 * t = 0.2396 mile = 1265 feet

Bryce

 

Ace, now appologize to me. I am so offended.


Actually most ADD people are "extreme" in everything they do.......like RACE the 1/4 mile with red buttons and skydive (not me). The biggest problem is its hard to hold a job, a wife, and money because you react with so much impulse.

 

Are you a math teacher?

 

My last calculation was a first order approximation using average speed. It seemed to me that the average speed could be in error by about 2%.

So it seemed that we must calculate the values using acceleration instead of average speed. So the distance traveled by the NASCAR is still given by:

S = Speed * TIME
S = 0.05555 * t

The distance traveled by the dragster is given by 1/8 mile plus the distance traveled at 268 mph, plus the additional distance traveled due to the acceleration from 268mph. The acceleration in the last 1/8 mile is the difference in speeds divided by the time it takes. The 1.48974 seconds is the time from my previous post.

Acceleration = (336.15 mph - 268 mph)/1.48974 sec = (0.093375-0.074440/1.48974 = 0.01271

That equation is given as:

S = 1/8 + 268 * ( TIME - 1/8time ) + 1/2 * Acceleration * (TIME - 1/8time)^2
S = 0.125 + 0.07444 * (t-2.94726) + 1/2 * 0.006351 * (t -2.94726)^2

Once again two equations, two unknowns:

t = 4.342 seconds is when the dragster passes the NASCAR
S = 1273 feet is when the dragster passes the NASCAR

The velocity of the dragster is given by the derivative of the distance equation above or:
v = 0.01271 * t + 0.036975

At t=4.342 seconds, the dragster is going 0.092162 miles/sec
v = 331.7 MPH the speed that the dragster is going when he passes the NASCAR

Interesting thing is that my time in the first order solution was only off by about 0.03 seconds, but it made
a big difference in the distance calculation.

Bryce

 

Not only is no one "right on", no one can be. Most of the guesses are relatively close, but given the information in the scenario, the exact answer can not be extracted.

First off, the acceleration is non-linear through the entire pass. The fact that a top fuel dragster gains far fewer MPH in the second 1/8th mile is a clear indicator that the acceleration rate has dropped off dramaticly, but again, that drop off is non-linear and the curve can not be calculated with the information given.

Second, a top fuel dragster is in "top gear" right from the launch. They do not run a geared transmision of any type. The motor is linked to the rear end through a clutch, and this clutch set up is what determines acceleration curves. Rarely does this clutch ever reach a lock up that links the engine speed to tire speed directly. Top fuel cars will spike to peak RPM at launch, and stay there through the entire run. The rate that the clutch applies preasure is what detirmines how quickly the car will go. The variance from one clutch to another, or from one set up to another is part of what make these calculations impossible to make exactly.

Two cars can run the same MPH and have differing E.T.'s because of this clutch based curve of acceleration. Without knowing the E.T. of the dragster, all the other math is guess work.

Now even if we had the E.T., we couldn't be sure about the distance unless we knew that the dragster was in front of the stock car. We wouldn't know if the dragster was in front, even if it had a lower E.T., without knowing reaction times though. So, if we knew the reaction time for the dragster, and it's E.T., and those two things combined were lower than the calculated E.T. of the stock car(easily calculated due to a constant value), then we could calculate the exact distance that the stock car was behind the dragster.

If the stock car finished first, the variable acceleration curve would prevent an exact calculation of distance that the dragster was behind; even if we knew the E.T. and reaction time. A relatively close guess could be made in that case, but it couldn't be "right on" by any means unless we had full telemetry.

Although some of you show great math skills, there is much more to it than simple calculations based on the info given.