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  #46  
Old 07-14-2011, 09:10 AM
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Evan,
I didnt intend to start a ping pong match !
You could time your car to a few different speeds.
Then draw a graph of speed (y axis) v/s time (x axis)
Then calculate the area under the graph up to the speed you are interested in. That gives you the distance traveled to get to that speed.
Sorry to be a PITA but thats the correct way to do it.

The previous comment about the shape of the graph is about right.
it would be of a general form Y= kX^(1/2)
Using BC's converted amounts ~ that is time (X) in sec & your speed in ft/sec, you can work out K.
Then integrating for zero to a given time (20 in your case?)

S=(2/3)xkx(20)^(3/2)

That should give you an answer close to the actual value.

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  #47  
Old 07-14-2011, 09:52 AM
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Quote:
Originally Posted by Kuan View Post
Jeezus. Get a flexible curve and bend it to where it feels right.

Way too sensible.
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  #48  
Old 07-14-2011, 12:21 PM
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Leathermang and BC are correct.


...except that speed is different than velocity.
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  #49  
Old 07-14-2011, 01:00 PM
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Quote:
Originally Posted by jt20 View Post
Leathermang and BC are correct...
Only if acceleration is constant.
Quote:
...except that speed is different than velocity.
Only relevant if the car does not travel in a straight line.
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  #50  
Old 07-14-2011, 01:01 PM
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Quote:
Originally Posted by Honus View Post
Really? You did calculus before high school?Of course it is.
Actually there is NO right or wrong answer. If you assume infinite acceleration or infinite braking power then you can pick a number, any number, will be the right answer. In real life, it would be ..., I give up.

I suppose every one flunk their test.
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  #51  
Old 07-14-2011, 01:05 PM
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Originally Posted by Honus View Post
Only if acceleration is constant.Only relevant if the car does not travel in a straight line.

all correct, but we were given simple parameters and are left to assume kinematics with 2-3 significant figures.

I didn't read the whole thread... sorry if you already hashed this out.
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  #52  
Old 07-14-2011, 01:18 PM
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Originally Posted by ah-kay View Post
Actually there is NO right or wrong answer...
That's not true. The right answer is that he provided insufficient information to make the calculation. The wrong answer is that acceleration has nothing to do with it. With a bit of research about the acceleration characteristics of similar vehicles, one could write an equation that would give a pretty good estimate. To use the average velocity probably underestimates the distance traveled because I suspect that the acceleration is greater at the beginning of the trip than it is at the end.
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  #53  
Old 07-14-2011, 01:29 PM
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Okay, now we've done it . . . we've made math a hot topic! Well done gentlemen, well done!
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  #54  
Old 07-14-2011, 01:30 PM
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Quote:
Originally Posted by Honus View Post
That's not true. The right answer is that he provided insufficient information to make the calculation. The wrong answer is that acceleration has nothing to do with it. With a bit of research about the acceleration characteristics of similar vehicles, one could write an equation that would give a pretty good estimate. To use the average velocity probably underestimates the distance traveled because I suspect that the acceleration is greater at the beginning of the trip than it is at the end.
"pretty good estimate" = not a precise answer
"Insufficient info" = use your imagination
"acceleration has nothing to do with it" = assumption, assume to be constant

If you ask a grade school student, he/she will come up with 1466.67' as they use the most simple logic to get to their answer. They do not dwell into an ivory tower like most members here. There are better things in life to do then trying to fit an answer to a question.
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Not MBZ nor A/C trained professional but a die-hard DIY and green engineer. Use the info at your own peril. Picked up 2 Infractions because of disagreements. NOW reversed.

W124 Keyless remote, PM for details. http://www.peachparts.com/shopforum/mercedes-used-parts-sale-wanted/334620-fs-w124-chasis-keyless-remote-%2450-shipped.html

1 X 2006 CDI
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1 x 83 300D
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  #55  
Old 07-14-2011, 02:13 PM
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would have been faster just to drive it.......
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  #56  
Old 07-14-2011, 02:33 PM
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Originally Posted by lutzTD View Post
would have been faster just to drive it.......
And the only truly accurate way. The figure would vary from car to car - there is no perfect equation for it.

Rocket science is indeed rocket science but at least they have one accelerating force that is known to be constant: gravity. During liftoff, air resistance and uneven acceleration from thrust must be contended with but I'm guessing that the thrust from rockets is more constant. No gear changes.
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  #57  
Old 07-14-2011, 06:14 PM
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Originally Posted by cmac2012 View Post
... No gear changes....
When those booster rockets quit and drop off... I would call that a ' gear change '.....LOL
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  #58  
Old 07-14-2011, 06:25 PM
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Originally Posted by KarTek View Post
Guys, this has kind of been on my mind for a while and I can't seem to figure the logic of it.

Lets say a car can accelerate from 0-100 in 20 seconds. How far does it travel?

Is it an average rate of acceleration for the given number of seconds?

I just can't get my head around how it's figured...
Here ya go:

D = 1/2 a t2 + ViT where D = distance, a = acceleration, Vi= initial velocity and t= time


D= 1/2 32 ft/sec 2 (20sec) 2 + 0 (20)

D = 16 ft/sec2 (400sec 2 ) = 6400 feet
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  #59  
Old 07-14-2011, 06:36 PM
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Quote:
Originally Posted by strelnik View Post
Here ya go:

D = 1/2 a t2 + ViT where D = distance, a = acceleration, Vi= initial velocity and t= time


D= 1/2 32 ft/sec 2 (20sec) 2 + 0 (20)

D = 16 ft/sec2 (400sec 2 ) = 6400 feet

so, you dropped the car off a cliff and it takes 1.21 miles to get reach 100mph?

whats terminal velocity for this vehicle, or CD?
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  #60  
Old 07-14-2011, 07:22 PM
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How many cattle can we fit in a boxcar?

First we assume a square cow.

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