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Thread: rate of accelleration plus time = ????? ( how far?? )

  1. #1 rate of accelleration plus time = ????? ( how far?? ) 
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    hey there. hypothetical question. let say and object has a constant accaleration of 0.161 in/s.. how far would it have travelled after 10 years and.. how fast would it be going. i can figure how fast. but how far is beyond my skillset.
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    Quote Originally Posted by Mr LeClair View Post
    hey there. hypothetical question. let say and object has a constant accaleration of 0.161 in/s.. how far would it have travelled after 10 years and.. how fast would it be going. i can figure how fast. but how far is beyond my skillset.
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    Here's the thing. The object can't maintain that constant acceleration relative to the initial starting frame. If you try to calculate its final velocity on that basis you get an answer that exceeds the speed of light.
    However, you can do it if you assume a constant acceleration as measured by the object itself, then you use the either


    or



    depending on whether the ten years is measured by the object (T), or by the initial rest frame (t).
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    thx fellas but im not sure how to go about calculating those formulas. and yea i know the rate of acceleration tapers off to infinity. i just figured it wouldnt get to the speed of light but maybe travel a ly or two. rough estimates would be fine under the circumstances.
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    After going back over my calculation for the final speed I noted a decimal displacement when converting units. It turns out that you won't get that close to the speed of light after all. You'll reach roughly 0.4% of c and a distance of just a bit under 9 light days
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    not very far.

    so with that said, i could add lets say twice the amount of acceleration and then it'll get twice the speed and twice the distant? minus of course the eventual drop off in acceleration.
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    Quote Originally Posted by Mr LeClair View Post
    not very far.

    so with that said, i could add lets say twice the amount of acceleration and then it'll get twice the speed and twice the distant? minus of course the eventual drop off in acceleration.
    Is it so difficult for you to substitute values for in ? They teach this in 9-th grade.
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    yea i was a under acheaver in highschool. hense the mispelling.
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    Quote Originally Posted by Mr LeClair View Post
    yea i was a under acheaver in highschool. hense the mispelling.
    Okay, Using AndrewC's equation:



    s is the distance traveled.
    a is the value of the acceleration
    t is the time you accelerated.

    The little 2 above the t stands for "squared" and means you multiply t by itself.

    Now to use this equation, you have to make sure that everything you enter in are in comparable units. Thus since you give the acceleration as inches and seconds (Quick note: acceleration is distance per sec per sec, and here would be in/sec/sec. in/sec would be speed, and acceleration is how much the speed is changing over time.), then your time must be in seconds also (you'll have to convert 10 years to the number of seconds in ten years) and the answer you will get out will be in inches.

    Thus to use the formula, you would take the value of your acceleration (0.161 in/sec/sec), multiply it by the number of seconds in 10 years multiplied by itself, and divide by 2 to get a distance in inches.
    if you double the acceleration, this is the multiplying your answer by 2.

    This will give you a good answer as long as the combination of the acceleration and time aren't too large. If a times t is a good-sized fraction of the speed of light, then this equation becomes less and less accurate. Then you have to use one of the equations I gave, and it also matters as to whether the 10 years is being measured by a clock traveling with our object or one that stayed put.

    As you can see, there is a bit more to my equations.
    I used d for distance instead of s,
    a is still the acceleration
    T is the time if measured by the object
    t is the time if measured by the stay at home clock.
    c is the speed of light.
    cosh stands for the hyperbolic cos
    Another way to express it would be like this:

    where e is Euler's number and equal to 2.71828...

    Either way, it not something you are going to calculate easily by hand. The best bet is to use a scientific calculator or spreadsheet with a cosh function.

    In the second equation, the

    symbol stands for the square-root. This is the reverse of the when you put the little number above and to the right of a number. It is the number that when multiplied by itself results in the number under the symbol. So for example, the square-root of 4 is 2 because 2 x 2 = 4. Most calculators have this function.

    Here, c is ~11,802,852,680 in/sec. Alternatively, you can convert the acceleration to meters/sec/sec and use 299792458 m/s( or 300,000,000 if you are worried about extreme accuracy) for c.

    So for the first equation, you would multiply the acceleration by the time and then divide by the speed of light. Then you find the hyperbolic cosine of this result, subtract 1 from it and multiply this by the speed of light times itself divided by the acceleration.

    With the second equation, you multiply the acceleration by the time, divide that by the speed of light. multiply the result by itself and add 1, then find the square-root of this result and multiply this by the speed of light times itself divided by the acceleration.

    Please excuse me if I have explained things that you already knew, but from your posts I wasn't quite sure of how basic you needed things.
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