
#19
Sep1410, 12:24 AM

P: 91

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 



#20
Sep1410, 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! 



#21
Sep1510, 07:56 PM

P: 91

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 



#22
Sep1710, 12:02 AM

P: 563

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 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. 



#23
Sep1910, 10:17 PM

P: 91

Thanx 



#24
Sep1910, 10:51 PM

P: 975

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). 



#25
Sep1910, 11:00 PM

P: 975

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*t^{2}. In this case, D = 5280, a = 32 ft/s^{2}, and t is unknown. Rearranging the equation a bit, you get t = sqrt(2D/a), so t = 18.17 seconds. 



#26
Sep2010, 07:25 PM

P: 91

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. 



#27
Sep2010, 07:37 PM

P: 91

thanx 



#28
Sep2010, 08:06 PM

P: 975

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/s^{2}, pounds force, and pounds mass, k is equal to 32. 



#29
Sep2010, 08:39 PM

P: 91

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 



#30
Sep2110, 12:04 AM

P: 563

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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