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What does per second per second mean?

  1. Sep 11, 2010 #1
    thrust is the ability to move a 1 pound object at 32 feet per second per second. What does per second per second mean? Does that mean for every second, the object has moved 64 feet in a given direction?

    if air molecules are moved at 3200 feet per second per second to deliver 100 pounds of thrust, does that mean what ever device is producing the thrust(gas turbine engine) is moving the exhaust(air molecules) at 3200 feet per second per second? Thats a little over half a mile those air molecules are thrown if that is true.


  2. jcsd
  3. Sep 11, 2010 #2


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    Staff: Mentor

    Re: Thrust

    32 feet per second per second means that after 1 second the object is traveling at a speed of 32 feet per second. After 2 seconds it is traveling at 64 feet per second. Etc.
  4. Sep 12, 2010 #3


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    Re: Thrust

    "distance per second per second" is a reference to acceleration, the change in speed per unit of time. It might be easier to understand if the units are changed, 1 g of acceleration means an increase in speed by about 21.9 mph every second.
  5. Sep 12, 2010 #4
    Re: Thrust

    So at 3 seconds it is going 128 feet per second and then at 4 seconds its 256 feet per second? Or is it just a constant 32 feet per second. So at 3 seconds it has traveled 96 feet and at 4 seconds, 124 feet?

  6. Sep 12, 2010 #5
    Re: Thrust

    No, the distance traveled is a separate calculation.

    After 3 seconds the speed is 96 feet per second and after 4 seconds it's 128 feet per second.
  7. Sep 12, 2010 #6
    Re: Thrust

    From the information given there is no way to know how fast the molecules are moving, only that a specific amount of thrust is being produced.
  8. Sep 12, 2010 #7


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    Re: Thrust

    The speed after 3 seconds is 96 feet per second. The total distance is 144 feet.

    Speed is acceleration times time.

    Distance is average speed times time. With constant acceleration, the average speed is the final speed divided by two.
  9. Sep 13, 2010 #8
    Re: Thrust

    To me thats a contridictiction because 96 FEET per second is distance. feet is a measure of distance so how can the speed be 96 FEET per second but after 3 seconds, its 144 FEET.

  10. Sep 13, 2010 #9
    Re: Thrust

    No, 96 feet per second is a speed. If the speed is constant, the object covers 96 feet every second.

    Since the object is accelerating, the speed that it is traveling and and therefore the distance that it covers each second is changing. That's why the average speed needs to be figured in.

    Example: the object starts from zero and accelerates to 32 feet per second in one second. You take the initial speed and the final speed during that second, add them together and divide by two to get the average, then multiply by the time, one second. The average speed was 16 ft/sec so during the first second the object traveled 16 feet. If the object then stays at that speed, it travels 32 feet every second thereafter.

    Whenever an object is accelerating, you have to take the average speed times the amount of time accelerating to find the distance traveled during that time. If the object accelerates from zero at 32 feet per second, every second (32 ft/sec^2) for 3 seconds, it will be at 96 feet per second in speed but will only have traveled half the distance that it would if it was at that speed from the beginning. That's because the average speed during that time is 48 ft/sec. 48 feet times 3 is 144 feet.

    Now, if the object stops accelerating after three seconds and stays at 96 ft/sec for the next three seconds, it will cover 288 feet. That's because its average speed during the final three seconds is 96 ft/sec.
    Last edited: Sep 13, 2010
  11. Sep 13, 2010 #10
    Re: Thrust

    Hmm, so since it takes a certain amount of time to go from 0 to 32 feet per second, thats why you add them together then divide by two? I dont get how an object traveling at 96 feet per second can only cover half that distance. its not the problem with the math, its the theory of why. To me, if it is moving at 96 feet per second then it has traveled the same distance. IE, i am traveling at 96MPH, how many miles is that in one hour, its 96.

    I know im complicated but i need to understand this. This is my first year as a ME student and have not taken ANY physics classes except an intro.

  12. Sep 13, 2010 #11


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    Re: Thrust

    The object isn't traveling at 96 feet per second for the entire three seconds. In fact, it is only traveling at 96 feet per second at the last instant of the 3 seconds.

    If you're traveling at 96 mph for one hour, at a constant speed, you'll travel 96 miles. But then your acceleration is 0. Acceleration is what it's called when your speed is changing.

    In the example already given, the speed is 32 fps after 1 second and 64 fps after 2 seconds, so clearly it isn't moving at 96 fps for the entire 3 seconds.
    Last edited: Sep 13, 2010
  13. Sep 13, 2010 #12
    Re: Thrust

    If you travel at 96 mph for 30 minutes then stop for the rest of the hour, how far did you travel? What was your average speed?
  14. Sep 13, 2010 #13
    Re: Thrust

    haha. i set myself up for this one.

    ok. so since it took 3 seconds get up to 96 fps, its not a constant is what your gettin to. ok thats cool. So if an object is moving at a constant 32fps, it will travel a distance of 32 feet every second because its acceleration 0. But if its acceleration is 32fps, then every second, the speed is increased by 32 fps. Acceleration is not the same is "how fast are you going" like i and 50 million other people think of it. IE, when i ask someone how fast we are going in the car, they say 175mph. (mustang GT supercharged 775HP at the wheels)


  15. Sep 13, 2010 #14
    Re: Thrust

    Sounds like you're on the right track now (pun intended!).

    Be very careful with definitions in physics; I see you've already discovered that "common" meanings don't always translate well!
  16. Sep 13, 2010 #15
    Re: Thrust

    OMG, that is so true. im going to have to print this thread out so i have it in my notes for reference.

    Hopefully in a year, i will have a better understanding of how it all works.

    thanx guys.

    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

    Last edited: Sep 13, 2010
  17. Sep 13, 2010 #16
    Re: Thrust

    Back to definitions:
    "Thrust is the ability to accelerate a 1 pound object at 32 feet per second per second." How thrust are we talking about? 1 pound of thrust. So 1 pound of thrust can accelerate a 1 pound object at 32ft/sec^2. How fast will it accelerate 2 pounds?

    In other words, your second thought is correct, you have to divide the thrust (force) by the weight (mass) in order to get the acceleration. 63% of 32 ft/sec^2 is 20.2 ft/sec^2. You'll soon find that the equation F=MA (and its variations) is your friend! The equation/variation you're using here is F/M=A.

    Incidentally (or not!), a Mustang GT with that much power should also be able to accelerate at .63 gs at 96 mph!
    Last edited: Sep 13, 2010
  18. Sep 13, 2010 #17
    Re: Thrust

    So a two pound object accelerated at 1 lb of thrust or 2 lbs of thrust? if it is 1 lb of thrust, then it would be 16fps, however, if it is 2lbs of thrust accelerating a two lb object, it would be 64fps. wait... maybe still 32fps because the weight/thrust ratio has not changed. its still a 1:1 ratio. if it was 2 lbs of thrust on a 1lb object, then it would be 64fps.
  19. Sep 13, 2010 #18
    Re: Thrust

  20. Sep 14, 2010 #19
    Re: Thrust

    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
  21. Sep 14, 2010 #20
    Re: Thrust

    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!
  22. Sep 15, 2010 #21
    Re: Thrust

    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.

  23. Sep 17, 2010 #22
    Re: Thrust

    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.
    Last edited: Sep 17, 2010
  24. Sep 19, 2010 #23
    Re: Thrust

    Dang, thanx man. Sorry for late reply. Work was busy last night and tonight. I will print this off and read it.

  25. Sep 19, 2010 #24


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    Re: Thrust

    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).
  26. Sep 19, 2010 #25


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    Re: Thrust

    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.
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