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What does per second per second mean?

by LT72884
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LT72884
#19
Sep14-10, 12:24 AM
P: 91
Quote Quote by mender View Post
Bingo!
ok, here is a question that might help me understnad it better.

If i were moving at a constant 32fps,or 0 acceleration, it would take me 165 seconds to travel 5280 feet(1 mile)

What if my acceleration was 32 fps and not 0, how long would it take me to go 5280?

Ok next part. 56,000lbs of thrust is how many fps? if 2lbs of thrust on a 1 lb object is 64 fps, whats the formula to find 56,000lbs on a 1lb object? is it 56,000 multiplied by 32 which is like 1.7 million. if thats correct then you would divide by 90,000 to get fps for a 747?

is my logic correct

thanx guys
mender
#20
Sep14-10, 01:02 AM
P: 563
To answer your second question first, yes, you can do it that way and you get the right answer. I find it easier to divide the thrust by the weight first, then multiply by the acceleration.

To answer your first question, if you were accelerating from zero at a constant rate of 32 ft/sec^2, how much time would it take to go 5280 feet: for this, use the equation d = 1/2 at^2 where
d = distance
a = acceleration
t = time in seconds

You know two of the three variables, so you can solve for the third. You want to isolate t so you rearrange the equation:

t = sqrt(2d/a) or the time is the square root of two times the distance divided by the acceleration rate
t = sqrt(2 x 5280/32)
t = sqrt 330
t = 18.16 seconds

Now that you know how long it took, you can figure out how fast you were going at the end of the mile.

18.16 seconds x 32ft/sec^2 = 581 ft/second = 396.3 mph.

You'll need a lot more than 775 hp to do that!
LT72884
#21
Sep15-10, 07:56 PM
P: 91
Quote Quote by mender View Post
To answer your second question first, yes, you can do it that way and you get the right answer. I find it easier to divide the thrust by the weight first, then multiply by the acceleration.

To answer your first question, if you were accelerating from zero at a constant rate of 32 ft/sec^2, how much time would it take to go 5280 feet: for this, use the equation d = 1/2 at^2 where
d = distance
a = acceleration
t = time in seconds

You know two of the three variables, so you can solve for the third. You want to isolate t so you rearrange the equation:

t = sqrt(2d/a) or the time is the square root of two times the distance divided by the acceleration rate
t = sqrt(2 x 5280/32)
t = sqrt 330
t = 18.16 seconds

Now that you know how long it took, you can figure out how fast you were going at the end of the mile.

18.16 seconds x 32ft/sec^2 = 581 ft/second = 396.3 mph.

You'll need a lot more than 775 hp to do that!
Thanx for the reply. Im usually a day late or two for my replies since i only repl;y on my work PC since i have no internet at home.

ok, so the equation would be T= D(A) or Time = Distance multiplied by (Acceleration) where acceleration is Force divided by Mas. in our case 32FPS

T = 5820(32) = A big eh number

Im lost on your equations at the top. Sorry.

Thanx
mender
#22
Sep17-10, 12:02 AM
P: 563
Quote Quote by LT72884 View Post
ok, so the equation would be T= D(A) or Time = Distance multiplied by (Acceleration) where acceleration is Force divided by Mas. in our case 32FPS

T = 5820(32) = A big eh number
Nope. Distance = velocity multiplied by time, or d = V x t

There is more than one way to get to the answer but you have to keep the basic relationships straight. Practice writing the basics in different ways as you did in algebra so you can move any one of the variables to the left side, like this:
d = V x t
t = d/V
V = d/t

Quote Quote by LT72884 View Post
Im lost on your equations at the top. Sorry.

Thanx
You'll use this equation in first year.

Being able to put together known relationships to produce usable equations is an important part of understanding how physics works. Whenever I start to get lost, I take the equation apart and try to figure out why each part is in the equation. Also, putting the units in helps me see if I have accounted for each part of the equation. Finally, putting some numbers in to check tells me if I'm doing things right.

Let's do that with d= 1/2at^2.

The question that you asked involves distance, time and acceleration. Fairly simple concepts by themselves. Here are the two major parts of what you need to figure out the answer to your question:

1. To find the average speed, you take the initial speed and the final speed and divide by two.
(V1 + V2)/2 = V average
2. To find the distance traveled at a certain speed, you multiply the average speed by the time spent at the speed.
V average x time = distance

Now let's add some units and some arbitrary numbers and do a test equation. Let's say that we start at zero and accelerate at 2 m/s every second (or 2m/s^2) for 10 seconds. Based on that, we get up to 20 m/s. So how far did we travel during that 10 seconds?

First equation:
(0 m/s + 20 m/s)/2 = average speed of 10 m/s
Second equation:
10 m/s x 10 seconds = 100 m

According to this, we travel 100 m. Now instead of using two equations, let's use the equation at the top of the page to compare.

d = 1/2 at^2
d = 1/2 (2 m/s^2 times 10 seconds squared)
d = 1/2 (2 m/s^2 times 100 s^2)
d = (200 m)/2
d = 100 m

Hey, same answer! The equation works!

If you write it for yourself (and you should when going through the same steps that I did here), you'll see how the time units cancel out and leave only metres, which of course is distance.

Let me know when you've worked through this example.
LT72884
#23
Sep19-10, 10:17 PM
P: 91
Quote Quote by mender View Post
Nope. Distance = velocity multiplied by time, or d = V x t

There is more than one way to get to the answer but you have to keep the basic relationships straight. Practice writing the basics in different ways as you did in algebra so you can move any one of the variables to the left side, like this:
d = V x t
t = d/V
V = d/t



You'll use this equation in first year.

Being able to put together known relationships to produce usable equations is an important part of understanding how physics works. Whenever I start to get lost, I take the equation apart and try to figure out why each part is in the equation. Also, putting the units in helps me see if I have accounted for each part of the equation. Finally, putting some numbers in to check tells me if I'm doing things right.

Let's do that with d= 1/2at^2.

The question that you asked involves distance, time and acceleration. Fairly simple concepts by themselves. Here are the two major parts of what you need to figure out the answer to your question:

1. To find the average speed, you take the initial speed and the final speed and divide by two.
(V1 + V2)/2 = V average
2. To find the distance traveled at a certain speed, you multiply the average speed by the time spent at the speed.
V average x time = distance

Now let's add some units and some arbitrary numbers and do a test equation. Let's say that we start at zero and accelerate at 2 m/s every second (or 2m/s^2) for 10 seconds. Based on that, we get up to 20 m/s. So how far did we travel during that 10 seconds?

First equation:
(0 m/s + 20 m/s)/2 = average speed of 10 m/s
Second equation:
10 m/s x 10 seconds = 100 m

According to this, we travel 100 m. Now instead of using two equations, let's use the equation at the top of the page to compare.

d = 1/2 at^2
d = 1/2 (2 m/s^2 times 10 seconds squared)
d = 1/2 (2 m/s^2 times 100 s^2)
d = (200 m)/2
d = 100 m

Hey, same answer! The equation works!

If you write it for yourself (and you should when going through the same steps that I did here), you'll see how the time units cancel out and leave only metres, which of course is distance.

Let me know when you've worked through this example.
Dang, thanx man. Sorry for late reply. Work was busy last night and tonight. I will print this off and read it.

Thanx
cjl
#24
Sep19-10, 10:51 PM
P: 1,018
Quote Quote by LT72884 View Post
Ok so back to thrust. If a 747 from 1970 produces 56,000 lbs of thrust, this means that it(747) is being accelerated at 1,792,000 fps? since 1 lb of thrust is 32fps. Is my thinking correct?

OR is it wrong because i have not included the weight of the plane which is 90,000lbs which is 63% of the thrust. IE 56000 is 63% of 90000

thanx
A couple quick notes about your 747 example here:

1) A 747 actually has about 56,000 pounds of thrust per engine, and it has 4 engines, making 224,000 pounds of thrust total.

2) Your weight is off by an order of magnitude - the MTOW (max takeoff weight) for a 747 is more like 900,000 pounds

Also, you're being very loose with your units here, which I think is part of the reason that you are confused. Everything here is based on the equation F = MA. If you're looking for the acceleration, you can rearrange the equation to read A = F/M. If you have 1 pound of force accelerating 1 pound of mass, you have an acceleration of 32 feet per second per second. If you swap out the mass so you have 2 pounds of mass, the new acceleration will be 16 feet per second per second, since the acceleration is the force divided by the mass (and the mass was doubled). So, you can see that what really matters is the ratio of the force to the mass.

In the case of a 747, the force is 224,000 pounds, and the mass is 900,000 pounds. So, the acceleration will be 32*(224,000/900,000), which is equal to 7.96 feet per second per second (note: the 32 is simply because of our choice of units here - if we were using the metric system, with force in newtons and mass in kilograms, then the acceleration in meters per second squared would simply be the force divided by the mass, without the additional conversion factor).
cjl
#25
Sep19-10, 11:00 PM
P: 1,018
Quote Quote by LT72884 View Post
ok, here is a question that might help me understnad it better.

If i were moving at a constant 32fps,or 0 acceleration, it would take me 165 seconds to travel 5280 feet(1 mile)
Correct so far.

Quote Quote by LT72884 View Post
What if my acceleration was 32 fps and not 0, how long would it take me to go 5280?
As I said in my last post, part of your problem is mixing units. You keep using fps interchangeably for both acceleration and velocity. Feet per second (fps) is a unit of velocity, and it doesn't make any sense to say that you're accelerating at 32 fps. What you probably mean is that you're accelerating at 32 ft/s2. This may seem pedantic, but it really helps to keep track of the differences.

Now, to answer the question itself, it's fairly easy to find the time. To do this, you can use the formula D = 1/2*a*t2. In this case, D = 5280, a = 32 ft/s2, and t is unknown. Rearranging the equation a bit, you get t = sqrt(2D/a), so t = 18.17 seconds.

Quote Quote by LT72884 View Post
Ok next part. 56,000lbs of thrust is how many fps? if 2lbs of thrust on a 1 lb object is 64 fps, whats the formula to find 56,000lbs on a 1lb object? is it 56,000 multiplied by 32 which is like 1.7 million. if thats correct then you would divide by 90,000 to get fps for a 747?

is my logic correct

thanx guys
Again, you're mixing units and formulas a bit. You can't just say that 56,000 pounds of thrust is a certain number of feet per second (or feet per second squared). Again, use F = MA (or A = F/M). If you have 56000 pounds of thrust and the object weighs one pound, then you have A = 32*(56,000/1) (once again, the 32 shows up because of our choice of units), which means A = 1.792 million ft/s2. If you have a 747, you use 224,000 pounds of thrust (4 engines) and 900,000 pounds, as I did in the post above. To use your numbers just as an example though, you could say that you have a force of 56,000 pounds and a mass of 90,000 pounds, so A = 32*(56000/90000) = 19.9 ft/s2.
LT72884
#26
Sep20-10, 07:25 PM
P: 91
Quote Quote by cjl View Post
Correct so far.


As I said in my last post, part of your problem is mixing units. You keep using fps interchangeably for both acceleration and velocity. Feet per second (fps) is a unit of velocity, and it doesn't make any sense to say that you're accelerating at 32 fps. What you probably mean is that you're accelerating at 32 ft/s2. This may seem pedantic, but it really helps to keep track of the differences.

Now, to answer the question itself, it's fairly easy to find the time. To do this, you can use the formula D = 1/2*a*t2. In this case, D = 5280, a = 32 ft/s2, and t is unknown. Rearranging the equation a bit, you get t = sqrt(2D/a), so t = 18.17 seconds.


Again, you're mixing units and formulas a bit. You can't just say that 56,000 pounds of thrust is a certain number of feet per second (or feet per second squared). Again, use F = MA (or A = F/M). If you have 56000 pounds of thrust and the object weighs one pound, then you have A = 32*(56,000/1) (once again, the 32 shows up because of our choice of units), which means A = 1.792 million ft/s2. If you have a 747, you use 224,000 pounds of thrust (4 engines) and 900,000 pounds, as I did in the post above. To use your numbers just as an example though, you could say that you have a force of 56,000 pounds and a mass of 90,000 pounds, so A = 32*(56000/90000) = 19.9 ft/s2.
I swear wiki answers said 90,000 pounds HAHA. I must have mis read it. thanx for pointing that out.

how do you get A=32 multiplied by 56,000/90,000? if the formula is A=F/M that would be A= 56,000/90,000 which = .6220... if i do F=MA and plug .622 for A, i get like 55,000 as my force so close to it so really A is .62fps rather than 32 fps.
LT72884
#27
Sep20-10, 07:37 PM
P: 91
Quote Quote by cjl View Post
A couple quick notes about your 747 example here:

1) A 747 actually has about 56,000 pounds of thrust per engine, and it has 4 engines, making 224,000 pounds of thrust total.

2) Your weight is off by an order of magnitude - the MTOW (max takeoff weight) for a 747 is more like 900,000 pounds

Also, you're being very loose with your units here, which I think is part of the reason that you are confused. Everything here is based on the equation F = MA. If you're looking for the acceleration, you can rearrange the equation to read A = F/M. If you have 1 pound of force accelerating 1 pound of mass, you have an acceleration of 32 feet per second per second. If you swap out the mass so you have 2 pounds of mass, the new acceleration will be 16 feet per second per second, since the acceleration is the force divided by the mass (and the mass was doubled). So, you can see that what really matters is the ratio of the force to the mass.

In the case of a 747, the force is 224,000 pounds, and the mass is 900,000 pounds. So, the acceleration will be 32*(224,000/900,000), which is equal to 7.96 feet per second per second (note: the 32 is simply because of our choice of units here - if we were using the metric system, with force in newtons and mass in kilograms, then the acceleration in meters per second squared would simply be the force divided by the mass, without the additional conversion factor).
i didnt know that thrust and force were the same thing. haha

If you have 1 pound of force accelerating 1 pound of mass, you have an acceleration of 32 feet per second per second.
So your saying that A=F/M which in my case would be A=1/mass which is 1 pound. so A= 1pound of force divided by 1 pound of mass equals 32fps. so i dont see how 32 is a solution to A=F/M mainly because i dont know the conversion of mass to pounds and the conversion of newtons for force.

thanx
cjl
#28
Sep20-10, 08:06 PM
P: 1,018
Quote Quote by LT72884 View Post
i didnt know that thrust and force were the same thing. haha
Yep - thrust is a force, and behaves just like any other force.

Quote Quote by LT72884 View Post
So your saying that A=F/M which in my case would be A=1/mass which is 1 pound. so A= 1pound of force divided by 1 pound of mass equals 32fps. so i dont see how 32 is a solution to A=F/M mainly because i dont know the conversion of mass to pounds and the conversion of newtons for force.

thanx
32 is a solution because of the units chosen.

If we used standard metric units, it would work out a little more cleanly. In that case, if you had one newton of thrust pushing on one kilogram, you would get one meter per second per second of acceleration. Because English units are being used here though, it's not as pretty.

Basically, in this case, you actually get A = k*(F/M), in which k is a conversion factor based on your units chosen. For units of ft/s2, pounds force, and pounds mass, k is equal to 32.
LT72884
#29
Sep20-10, 08:39 PM
P: 91
Quote Quote by cjl View Post
Yep - thrust is a force, and behaves just like any other force.


32 is a solution because of the units chosen.

If we used standard metric units, it would work out a little more cleanly. In that case, if you had one newton of thrust pushing on one kilogram, you would get one meter per second per second of acceleration. Because English units are being used here though, it's not as pretty.

Basically, in this case, you actually get A = k*(F/M), in which k is a conversion factor based on your units chosen. For units of ft/s2, pounds force, and pounds mass, k is equal to 32.
forgive my ignorance, but i aint seein how 1 meter per second is the same as 32 feet per second. i cant see how 1pound of force/1pound of mass = 32 fps. to me its still 1. So how did they get 32fps for acceleration if it is measured in meters and its 3 feet to a meter?

I aint following how they got the value 32fps for K. if i follow the formula correctly using my numbers i get this A= 56,000/90,000 --> .622 meters per second which is 1.5 fps. so how they got 32fps is not making any sense at all. If i plug that .622 into F=MA i get F=90,000(.622) which gives me something close to 56,000 newtons. wait. i think i might have figured something out. The .622 is based off of the mass being in KG not pounds. basically i need the thrust and weight of the 747 in metric. ok here i go. lets see if i learned something.

249088 newtons = 56,000 pounds of thrust. 4.448newtons per pound so 4.48*56,000
mass = .45KG per pound so 90,000*.45 = 40500pounds

A= 249088/40500 = 6.1 mps * 3. = 19.5fps

or

A=4.48/.45 = 9.9 mps *3.2 = OMG 31.85fps. Holy crap. i learned something..i learned to use the metric system rather than English for physics. Now its time to do it backwards. haha. I hate how stuff works.com gives everything i standard when the formula was written for metric. Granted i hate the metric system but since the formula was written for it, then i have to use it. to bad the formula wasnt written for english folk like me. haha

ok i think i can understnad my very first post now.

Thanx
mender
#30
Sep21-10, 12:04 AM
P: 563
Quote Quote by LT72884 View Post
So your saying that A=F/M which in my case would be A=1/mass which is 1 pound. so A= 1pound of force divided by 1 pound of mass equals 32fps. so i dont see how 32 is a solution to A=F/M mainly because i dont know the conversion of mass to pounds and the conversion of newtons for force.

thanx
The reason for 32 ft/sec^2 (don't forget this part) as acceleration is that we live on the earth. Earth's gravity pulls a 1 pound mass down with enough force to accelerate it at 32 ft/sec^2 when dropped. 1 pound of force is needed to counter gravity and hold the mass from dropping, therefore a pound of force will accelerate a 1 pound mass at one g.

Using the metric system, objects accelerate at 9.8 m/s^2 when dropped. To accelerate a 1 kg mass at 9.8 m/s (one g) you need 9.8 Newtons of force.

There's nothing magic about the metric system; I use both 'cause I'm bilingual!


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