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.

Botnst · #6 07-13-2011, 11:53 PM

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

ah-kay · #7 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 · #8 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.

leathermang · #9 07-14-2011, 12:54 AM

So , at any level was I correct ?

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.

leathermang · #11 07-14-2011, 01:00 AM

So I was correct .... ?

layback40 · #12 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 · #13 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.