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

KarTek · #1 07-13-2011, 10:50 PM

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...

leathermang · #2 07-13-2011, 10:59 PM

I think it would be the average speed per second times the number of seconds.

Brian Carlton · #3 07-13-2011, 11:14 PM

1460 feet.

The acceleration is 7.3 ft/sec squared.

In reality, the acceleration is not linear. The figure of 7.3 is the average over the distance. The figure is higher at slower speeds and much lower as the wind resists the vehicle close to the maximum speed.

tonkovich · #4 07-13-2011, 11:42 PM

Quote:

Originally Posted by Brian Carlton

1460 feet.

The acceleration is 7.3 ft/sec squared.

In reality, the acceleration is not linear. The figure of 7.3 is the average over the distance. The figure is higher at slower speeds and much lower as the wind resists the vehicle close to the maximum speed.

nice to see a certain intelligent person around here.

(it's been thirty years since those four years of "higher math" - i have no idea, but i think mr. b.c. would be your best hope.)

Honus · #5 07-13-2011, 11:49 PM

Quote:

Originally Posted by KarTek

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...

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.

cmac2012 · #6 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.

Botnst · #7 07-13-2011, 11:53 PM

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

ah-kay · #8 07-14-2011, 12:12 AM

1466.67' is correct but it has NOTHING to do acceleration. The terminal speed and the time taken is enough to calculate the distance travel.

The average speed is (100-0)/2 = 50mph. How it gets there is not important. 50mph for 20 sec = ((50*1760*3)/3600)*20 = 1466.67'

BTW: I assume it is mph.

layback40 · #9 07-14-2011, 12:42 AM

Quote:

Originally Posted by ah-kay

1466.67' is correct but it has NOTHING to do acceleration. The terminal speed and the time taken is enough to calculate the distance travel.

The average speed is (100-0)/2 = 50mph. How it gets there is not important. 50mph for 20 sec = ((50*1760*3)/3600)*20 = 1466.67'

BTW: I assume it is mph.

Not true !!
You have not calculated the average speed !!
If we assume constant acceleration, we can use one of the equations Botnst links to ;
Distance = initial velocity X time + 1/2 X acceleration X time squared.

If acceleration varies, the area under a velocity/ time graph would give you the answer.

BC is correct in converting to compatible units.
velocity in ft/sec , time in sec , distance in ft.

Chas H · #10 07-14-2011, 12:55 AM

Quote:

Originally Posted by layback40

Not true !!
You have not calculated the average speed !!
If we assume constant acceleration, we can use one of the equations Botnst links to ;
Distance = initial velocity X time + 1/2 X acceleration X time squared.

If acceleration varies, the area under a velocity/ time graph would give you the answer.

BC is correct in converting to compatible units.
velocity in ft/sec , time in sec , distance in ft.

If Ah-Kay made a mistake, how did he come to the correct answer?
There is no graph of any sort to consider just the simple case posed by the OP. Averaging the speed is perfectly valid in this case.

layback40 · #11 07-14-2011, 01:03 AM

Quote:

Originally Posted by Chas H

If Ah-Kay made a mistake, how did he come to the correct answer?
There is no graph of any sort to consider just the simple case posed by the OP. Averaging the speed is perfectly valid in this case.

Best you get yourself a high school physics book.

A question of "averages" for you;

2 towns are 50 miles apart, you wish to average 50mph for a return journey.
At the start of the trip you get stuck behind a funeral procession & so maintain 25 mph all the way to the town. What speed must you come back at to average 50 mph for the whole trip there & back?

According to the above logic it would be 75mph. do you agree?

I dont.

Chas H · #12 07-14-2011, 01:04 AM

Quote:

Originally Posted by layback40

Best you get yourself a high school physics book.

A question of "averages" for you;

2 towns are 50 miles apart, you wish to average 50mph for a return journey.
At the start of the trip you get stuck behind a funeral procession & so maintain 25 mph all the way to the town. What speed must you come back at to average 50 mph for the whole trip there & back?

According to the above logic it would be 75mph. do you agree?

I dont.

I don't care.

leathermang · #13 07-14-2011, 01:26 AM

Quote:

Originally Posted by layback40

Best you get yourself a high school physics book.

A question of "averages" for you;

2 towns are 50 miles apart, you wish to average 50mph for a return journey.
At the start of the trip you get stuck behind a funeral procession & so maintain 25 mph all the way to the town. What speed must you come back at to average 50 mph for the whole trip there & back?

According to the above logic it would be 75mph. do you agree?

I dont.

ACTUALLY.....
you simply drive 50 mph on the way back...
because you said ' you wish to average 50 mph for a return journey' ..
thus that stupid funeral you got stuck behind had nothing to do with the return trip... except to move its start time to later than you wanted...

Honus · #14 07-14-2011, 01:19 AM

Quote:

Originally Posted by Chas H

If Ah-Kay made a mistake, how did he come to the correct answer?
There is no graph of any sort to consider just the simple case posed by the OP.

His answer is only correct if the vehicle has constant acceleration. In that case, a graph of velocity versus time is a straight line with a slope equal to the acceleration. The distance traveled is the area under the curve (i.e. the integral with respect to time.) The area under the linear curve is equal to the area under a curve that has a constant velocity of 50. Draw it and you will see. Then draw a velocity curve that is steep at the beginning and tapering off at the end. The area under that curve will be greater than the area under the linear curve. That result stands to reason because the car with the quicker acceleration at the beginning will spend more of its time traveling at high speed, even though it will not continue to gain speed as quickly as the other car will at the end of the trip.

Quote:

...Averaging the speed is perfectly valid in this case.

That's true only if acceleration is constant.

Chas H · #15 07-14-2011, 01:20 AM

Quote:

Originally Posted by Honus

His answer is only correct if the vehicle has constant acceleration. In that case a graph of velocity versus time is a straight line with a slope equal to the acceleration. The distance traveled is the area under the curve (ie the integral with respect to time.) The area under the linear curve is equal to the area under a curve that has a constant velocity of 50. Draw it and you will see. Then draw a velocity curve that is steep at the beginning and tapering off at the end. The area under that curve will be greater than the area under the linear curve. That result stands to reason because the car with the quicker acceleration at the beginning will spend more of its time traveling at high speed, even though it will not continue to gain speed as quickly as the other car will at the end of the trip.That's true only if acceleration is constant.

You know it's quite late to put on the pedantic act.