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  #1  
Old 07-13-2011, 10:50 PM
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Math genius' help me solve this automotive math problem!

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...
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  #3  
Old 07-13-2011, 11:14 PM
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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.
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  #4  
Old 07-13-2011, 11:42 PM
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Originally Posted by Brian Carlton View Post
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.)
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  #5  
Old 07-13-2011, 11:49 PM
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Quote:
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...
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.
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  #6  
Old 07-13-2011, 11:53 PM
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http://en.wikipedia.org/wiki/Equations_of_motion
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  #7  
Old 07-14-2011, 12:12 AM
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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.
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Old 07-14-2011, 12:42 AM
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Originally Posted by ah-kay View Post
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.
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  #10  
Old 07-14-2011, 12:55 AM
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Originally Posted by layback40 View Post
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.
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  #12  
Old 07-14-2011, 01:03 AM
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Originally Posted by Chas H View Post
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.
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I no longer question authority, I annoy authority. More effect, less effort....

1967 230-6 auto parts car. rust bucket.
1980 300D now parts car 800k miles
1984 300D 500k miles
1987 250td 160k miles English import
2001 jeep turbo diesel 130k miles
1998 jeep tdi ~ followed me home. Needs a turbo.
1968 Ford F750 truck. 6-354 diesel conversion.
Other toys ~J.D.,Cat & GM ~ mainly earth moving
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  #13  
Old 07-14-2011, 01:04 AM
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Originally Posted by layback40 View Post
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.
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  #14  
Old 07-14-2011, 01:04 AM
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Are these two towns in Australia or the US ?
How did the guy die ?
Is he going to be cremated or fed to the crocs ?
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  #15  
Old 07-14-2011, 01:10 AM
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Originally Posted by leathermang View Post
Are these two towns in Australia or the US ?
How did the guy die ?
Is he going to be cremated or fed to the crocs ?


The towns are in Europe.
He died from being on PP OD too much.
Neither.
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Grumpy Old Diesel Owners Club group

I no longer question authority, I annoy authority. More effect, less effort....

1967 230-6 auto parts car. rust bucket.
1980 300D now parts car 800k miles
1984 300D 500k miles
1987 250td 160k miles English import
2001 jeep turbo diesel 130k miles
1998 jeep tdi ~ followed me home. Needs a turbo.
1968 Ford F750 truck. 6-354 diesel conversion.
Other toys ~J.D.,Cat & GM ~ mainly earth moving
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