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Dumb question

  1. May 10, 2006 #1
    hey all, i'm pretty ignorant when it comes to physics.. so i have a question for you physic braniacs.

    If 1 N is 1 kg m / s^2, meaning the force required to accelerate a 1 kg object at 1 m/s^2, then how much does 1 N accelerate a 1 kg object over 1 millisecond?

    I googled for 1 m/s^2 in m/ms^2 and the result is as expected 10^-6 = 1 / (1000*1000). I plugged this into a formula, and i come up short.

    at t=10 the velocity of a 1 kg object should be 10 m/s, and the distance traveled is 55 m.

    But if i calculate velocity and position at each millisecond, the resulting velocity at t=10 s is 0.01 m/ms^2 = 10 m/s (which is correct) but my distance traveled is only 50.01 m.. why is that?

    please slap around my physics deprived head a bit.

    [edit]

    I should add that the reason i want to calculate velocity and distance traveled each millisecond is that i'm making a C++ program to describe some functions, but doing it each second does not give me high enough 'resolution'.
     
    Last edited: May 10, 2006
  2. jcsd
  3. May 10, 2006 #2
    a millisecond is 10^-3 s not 10^-6
    try that see if it works:surprised
     
  4. May 10, 2006 #3

    Integral

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    Why would the acceleration be different at 1ms or 1hour?

    You have a constant force applied to a constant mass, therefore the acceleration is constant. It starts the instant the force is applied to the mass and continues until the force ends.

    The quantities of interest are the displacement and velocity. These are determined by the acceleration and are what you would be computing after 1 ms.

    we have :
    [tex] v = a t + v_0 [/tex]
    And
    [tex] d = \frac 1 2 a t^2 + v_0 t + d_0 [/tex]

    where
    v = velocity
    [itex] v_0 [/itex] = initial velocity
    d = distance traveled
    [itex] d_0 [/itex] = initial displacement
    a is the acceleration.
    t is the time. (here is where your 1ms comes in)

    another point, to avoid round off errors do not take steps in multiples of .1 use nice computer friendly steps like [itex] 2^{-n} [/itex] where n defines the step of your choice.

    Where did you get the 55m displacement? 50 is the correct answer. Perhaps you should share with us more of what you are doing.
     
    Last edited: May 10, 2006
  5. May 10, 2006 #4
  6. May 10, 2006 #5
    Hey Poposhka there are not as many dumb questions as dumb answers, so don't worry about asking. That is how we learn - hopefully.
     
  7. May 10, 2006 #6
    Yes but since it is per second squared, then it is 10^-6 per millisecond squared
     
  8. May 10, 2006 #7
    Actually i think i have a fundamental flaw in my understanding here... lets backtrack for a second...

    Since 1 Newton is the force required to accelerate a mass of 1 kg at a rate of 1 m/s/s (or m/s^2). Does that mean that a increases by 1 each second?

    So at t=0, a=0, t=1 a=1, t=2 a=2, t=3 a=3?? or should i look back at Newton's 3rd that states F = m*a, meaning that a is always 1 ( since F = 1, m = 1 )?

    Say you have a tiny spaceship with the mass of 1 kg, and a rocket booster producing thrust of 1 N, will the acceleration be a constant 1?
     
  9. May 10, 2006 #8

    Doc Al

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    No. If an object has an acceleration of 1 m/s^2, that means its velocity increases by 1 m/s each second.

    If a net force of 1 N is exerted on a mass of 1 kg, then its acceleration will be 1 m/s^2. As long as the net force and mass do not change, the acceleration will remain constant.
     
  10. May 10, 2006 #9
    Not having plugged in the numbers, I don't know how you got 50.01. But using Integrals' equations in incremental steps, you have v(i)=at + v(i-1), or v(i) = 10^-6 meters per millisecond + v(i-1). This makes your velocity v(i) at any time (since v(0) = 0) = i*10^-6 meters per millisecond or i*10^-3 meters per second, which is what you would expect since there are 10^4 increments of milliseconds (10^4*10^-3 = 10). The distance travelled during any increment is d(i) = 0.5at^2 + v(i)t +d(0), or d(i) = (0.5 + i)*10^-6 meters. When summed over 10^4 increments, this equals 50 meters.
     
  11. May 10, 2006 #10
    here's how i got 55 m from my original statement...

    t= 0.00, d= 0.00, v= 0.00, a= 1.00
    t= 1.00, d= 1.00, v= 1.00, a= 1.00
    t= 2.00, d= 3.00, v= 2.00, a= 1.00
    t= 3.00, d= 6.00, v= 3.00, a= 1.00
    t= 4.00, d= 10.00, v= 4.00, a= 1.00
    t= 5.00, d= 15.00, v= 5.00, a= 1.00
    t= 6.00, d= 21.00, v= 6.00, a= 1.00
    t= 7.00, d= 28.00, v= 7.00, a= 1.00
    t= 8.00, d= 36.00, v= 8.00, a= 1.00
    t= 9.00, d= 45.00, v= 9.00, a= 1.00
    t= 10.00, d= 55.00, v= 10.00, a= 1.00
    [edit] .. hmm should d be calculated before v?? .. let me try [edit2] no go
     
    Last edited: May 10, 2006
  12. May 10, 2006 #11

    Doc Al

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    These values are incorrect. For example:
    t= 1.00, d= 0.50, v= 1.00, a= 1.00

    To find the distance at any time (starting from 0 at t = 0), use:
    [tex]d = 1/2 a t^2[/tex]
    (As Integral explained.)

    How are you calculating your values?
     
  13. May 10, 2006 #12
    aah i think i know what the problem is here ...let me investigate

    ... :rofl:
     
    Last edited: May 10, 2006
  14. May 10, 2006 #13

    Doc Al

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    If it traveled at a constant speed of 1 m/s then it will travel 1 m each second. But you are talking about accelerated motion, so speed is not constant.

    That's the proper way to handle accelerated motion.

    Here's how to understand this: In the first second, you go from a speed of 0 to 1 m/s. The average speed during that 1 second interval is 0.5 m/s, so the distance traveled is (0.5 m/s)*(1 s) = 0.5 m.
     
  15. May 10, 2006 #14
    Ok i think i figured it out, Thanks Al, i gathered that, when i posted that last note, and realized my error.. see what you think of this.

    It's in C code, but i'll simplify it so that it might make sense... the sequence of operation goes from top to bottom.


    F = 1
    m = 1
    d = 0
    a = F/m
    t = 0
    v = 0

    while ( t <= 10 )
    {
    d = d + (0.5 * (v + v + a) )
    v = v + a
    t = t + 1
    }

    Ok let me try and explain unless it makes sense
    the symbols used are all in SI and F (force), m (mass), d (distance), a (Acceleration), t ( time ), v (velocity)

    the "while" blurb is a "loop" that will execute the operations below the '{' and above the '}' as long as t is equal or less than 10.

    inside the { and }, you see 3 statements

    the first statement d = d + (0.5 * (v + v + a) ) says that distance changes by the average of the velocity.

    the second statement v = v + a says that velocity changes by acceleration

    the third statement simply increases time by 1 second.

    This seems correct to me, and plugging the numbers in the distance formula gives me the same results... let me experiment with milliseconds now :surprised
     
  16. May 10, 2006 #15

    Doc Al

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    Gives you the same results as what? This looks correct to me. (Of course, you'll need to modify the code if your interval is milliseconds.)

    Any reason why you are writing code to calculate distance second by second, when the same formula gives you the final answer in one step?
     
  17. May 11, 2006 #16

    Integral

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    Please read my post #3. Your method is not clear. You are adding velocity and acceleration to get a distance. Since velocity and acceleration do not have the same units you cannot add them without some conversion factor. Look post #3 notice that velocity is the result of multiplying acceleration and time. To do your addition you are making a hidden multiplication factor of 1 s. Thus your difficulty with other sized steps.

    If you use the relationships I gave you in #3 you will not need to compute intermediate steps. You can compute for ANY time t.

    If you were to continue with the same algorithm and chose a step of .1 you would see serious numerical errors in a pretty short time. DO NOT take additive steps of .1 (or any multiple of .1 which is not a multiple of .5) These numbers cannot be precisely represented in a computer.

    If you feel you must take a step of .1 then use an integer counter and do a multiplication by the step size to get your "position". This way the round off error does not accumulate.
     
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