Math genius' help me solve this automotive math problem! - Page 3 - PeachParts Mercedes-Benz Forum

Chas H · #31 07-14-2011, 01:34 AM

Quote:

Originally Posted by layback40

. Not taking one from the other, that only works when one of the numbers is 0.

Which is the exact situation of the case posed in the OP.
Thank you

layback40 · #32 07-14-2011, 01:38 AM

Quote:

Originally Posted by Chas H

Which is the exact situation of the case posed in the OP.
Thank you

I didnt say average speed, just average of 2 numbers.

Now come on & tell us about your education in maths & physics !!

Chas H · #33 07-14-2011, 01:41 AM

Quote:

Originally Posted by layback40

I didnt say average speed, just average of 2 numbers.

Are you claiming that zero can not be a speed?

layback40 · #34 07-14-2011, 01:49 AM

Quote:

Originally Posted by Chas H

Are you claiming that zero can not be a speed?

I am not. Nice try at trying to change the subject.

Best you go get a high school physics book & read it !! Then you may understand what is going on with simple problems of linear motion.

It would appear that some are here just for the sake of being argumentative with little science to back them up.

Chas H · #35 07-14-2011, 01:53 AM

Quote:

Originally Posted by layback40

I am not. Nice try at trying to change the subject.

Best you go get a high school physics book & read it !! Then you may understand what is going on with simple problems of linear motion.

It would appear that some are here just for the sake of being argumentative with little science to back them up.

So give a number instead of jive.

ah-kay · #36 07-14-2011, 02:18 AM

This is grade school arithmetic, not even HS. If you treat it as Doctorate thesis then you are THE problem.

If you integrate it under the curve, it is not distance traveled. Let me know if come up with a different answer.

ah-kay · #37 07-14-2011, 02:20 AM

Quote:

Originally Posted by leathermang

So , at any level was I correct ?

At grade school level, same as me.

cmac2012 · #38 07-14-2011, 02:33 AM

Quote:

Originally Posted by Honus

Insufficient information. You need to know how acceleration varies with time. If acceleration is constant, then, the acceleration is given by:

a = (100 mile/hour * 5280 ft/mile) / (20 sec * 3600 sec/hr) = 7.33 ft/sec squared

Velocity is acceleration integrated over time, with an initial value of zero:

v=7.33 ft/sec squared * t

Distance is velocity integrated over time, again with an initial value of zero:

d = 7.33 ft/sec squared * t squared/2

At t= 20 sec, d = 1466.67 ft.

That's my story and I'm sticking to it.

If acceleration is not constant (and it wouldn't be constant in the real world), then the calculations become more complicated.

Something as imprecise as an auto accelerating could never be accurately stated with a math formula but i suspect Brian's comes close.

To get it more accurate, seems like calculus might be involved. But coming up with the right equation would be tough. Accurately measuring acceleration would be dicey. We know that wind resistance increases with the square of wind speed so I'm guessing acceleration would be some sort of sideways parabola.

**EDIT** Ah, I see you addressed that in post #20. Well done.

KarTek · #39 07-14-2011, 07:05 AM

Thanks for the input!

To address some issues:

This was intended to be a real world situation. One thing I was trying to figure out was if the area under the acceleration curve could be expressed as the average acceleration and it appears that it can not.

So, that leads to the question of how you get an acceleration curve for a particular vehicle and then how do you integrate it into a calculation?

Apparently, there's no "formula" that can be applied to this?

Botnst · #40 07-14-2011, 07:50 AM

s = 1/2 at^2 = distance traveled under acceleration.
s= dist
a = acceleration
t = time

Is it constant acceleration? Can't tell from the provided information. Be nice to have interval data. Absent that, you have to assume constant acceleration. Or Make Something Up!

Anyway, acceleration = terminal velocity - initial velocity/change in time (Newtonian)
= a = (v1-v0)/(t1-t0)
a = (100-0)/(20-0) = 5 velocity units/sec

since s = 1/2 at^2 we merely substitute a and t into the equation and do the maff.

= 1/2 (5) (20^2)
= 1,000

http://en.wikipedia.org/wiki/Equations_of_motion

Rafi · #41 07-14-2011, 08:12 AM

you did not provide enough enformation to calculate correctly the distance traveled. Acceleration is not constant. You have seen acceleration plots for car reviews in car magazines. Acceleration is a curve with sharp small spikes for gear changes. You need to integrate the area under that curve (speed vs times) to give you the precise distance traveled.

Honus · #42 07-14-2011, 08:36 AM

Quote:

Originally Posted by ah-kay

This is grade school arithmetic, not even HS...

Really? You did calculus before high school?

Quote:

... If you integrate it under the curve, it is not distance traveled...

Of course it is.

Honus · #43 07-14-2011, 08:37 AM

Quote:

Originally Posted by Chas H

That agrees with the other answers. But you later claim they are wrong.

Right. That issue has been addressed a couple of times. His calculation works if the acceleration is constant.

Honus · #44 07-14-2011, 08:41 AM

Quote:

Originally Posted by KarTek

...So, that leads to the question of how you get an acceleration curve for a particular vehicle and then how do you integrate it into a calculation?

Apparently, there's no "formula" that can be applied to this?

I think Rafi is on the right track. If you can find an acceleration curve for a similar vehicle, you can probably come up with an equation that approximates the shape of that curve. Then you tweak the equation so that it meets your conditions (0-100 in 20 seconds). Then you integrate that equation twice to get from acceleration to distance. The good news is that the small errors in your initial equation will tend to get canceled out when you integrate. Going the other direction, from a distance equation to an acceleration curve, would be more problematic in terms of accuracy.

Kuan · #45 07-14-2011, 08:47 AM

Jeezus. Get a flexible curve and bend it to where it feels right.

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