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Math Help - Obtaining deceleration from speed, distance and time

  1. #1
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    Obtaining deceleration from speed, distance and time

    Is it possible to obtain the rate of deceleration of an object, if you know it's original speed/sec, expected distance to travel, and expected time to take e.g.

    Object A is travelling at 800 units per second, towards a point (Z) which is 400 units away. A is expected to reach this point in 1 second, as it is decelerating all the way (reaching 0 units/second upon reaching Z) at a constant rate.

    I hope I've explained what I'm trying to do well enough.
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  2. #2
    Junior Member Evales's Avatar
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    Quote Originally Posted by Flare View Post
    Is it possible to obtain the rate of deceleration of an object, if you know it's original speed/sec, expected distance to travel, and expected time to take e.g.
    Why yes there is! Have you come across:
    a = (v - u)/t?

    When I do these I sometimes forget that I have v (0m/s) Because that is when it stops.

    How I haven't been too confusing.
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  3. #3
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    Are you ready for "Limits"?

    Figure it out.

    Without deceleration: 400/800 = 1/2 second We're not real interested in this result.

    Try it half-way at a time.

    200 / 800 + 200 / x = 1 ==> 1/2 + 200/x = 1 ==> x < 0. This one went badly, too, since we accidentally made the whole trip on the first piece.

    Try it 1/4 of the way at a time.

    100/800 + 100/(800-x) + 100/(800-2x) + 100/(800-3x) = 1

    Now try it 1/8 of the way at a time.

    1) Can you write the expression?
    2) Can you solve it?
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    @Evales: Would you mind explaining what each part of that is?

    I'm guessing
    a = accel?
    v = 0 (or initial speed)
    u = ???
    t = time?

    @TKHunny: I haven't a clue about what you're saying
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  5. #5
    Junior Member Evales's Avatar
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    a = acceleration
    v = is not the initial speed it is the final speed but since you are talking about deceleration it is 0 anyway.
    u = initial speed
    t = time

    Sorry I forget of course the symbols aren't standard in every curriculum.
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  6. #6
    Junior Member Evales's Avatar
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    But it almost seems as if you are given too much information. As we don't use the distance.
    Alternatively you could use

    a = (v^2 - u^2) / 2s

    Which would be using the distance traveled instead of the time. One second I'm going to check if they come up with the same answer.

    (Edit: s = distance)
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  7. #7
    Junior Member Evales's Avatar
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    They answers they give are hugely different are you sure that it decelerates fully?
    Are you sure that it takes 1 second?
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    Well, I'm doing this for a system I'm making for a Warcraft 3 mod.

    The a = (v(squared) - u(squared))/2(distance) seems to work.

    In the first test, at 600 speed, 400 distance and 1.5second time, I was only out .15 seconds on the time, and 9 units on the distance.

    In the second test, at 600 speed, 400 distance and 1second time, I was out -18 units on the distance, and had the time perfect.

    Those errors shouldn't make a big difference though (who's gonna notice 18 units when it's only about this _ long )

    Thanks very much for the help
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  9. #9
    Junior Member Evales's Avatar
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    Wow Physics can be used for something fun? Who would have thought....
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  10. #10
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    I know, it's pretty awesome
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  11. #11
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Flare View Post
    Well, I'm doing this for a system I'm making for a Warcraft 3 mod.

    The a = (v(squared) - u(squared))/2(distance) seems to work.

    In the first test, at 600 speed, 400 distance and 1.5second time, I was only out .15 seconds on the time, and 9 units on the distance.

    In the second test, at 600 speed, 400 distance and 1second time, I was out -18 units on the distance, and had the time perfect.

    Those errors shouldn't make a big difference though (who's gonna notice 18 units when it's only about this _ long )

    Thanks very much for the help
    I am rather confused about all of this. Both equations you were given will give the same result if you put the same data into it. The only way that it will not is if you have inconsistent data. Since the v^2 - u^2 equation works, I suspect you have an incorrect time interval.

    -Dan
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  12. #12
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    Quote Originally Posted by Flare View Post
    @TKHunny: I haven't a clue about what you're saying
    It's an exploration!

    For this one:

    100/800 + 100/(800-x) + 100/(800-2x) + 100/(800-3x) = 1

    I split the trip up into four pieces of equal length. The trip is 400, so this explains the 100 in each numerator.

    We start out at 800 ups. This explains the 800s in the denominator.

    You defined uniform decelleration, so I applied a deceleration of 'x' to each successive unit.

    First 100, speed = 800 -- no deceleration
    Second 100, speed = 800 - x -- one deceleration of unknown size
    Third 100, speed = 800 - 2x -- one more deceleration of the same size.
    Fourth 100, speed = 800 - 3x -- one more deceleration of the same size.

    How long did the first piece take? 100/800 seconds
    How long did the second piece take? 100/(800-x) seconds
    How long did the third piece take? 100/(800-2x) seconds
    How long did the fourth piece take? 100/(800-3x) seconds

    The last requirement is that the entire trip takes 1 second.

    100/800 + 100/(800-x) + 100/(800-2x) + 100/(800-3x) = 1

    Finally, there is a nice solution, x = 195.087. Let's check it out.

    First 100, speed = 800, time = 100/800 = 0.125
    Second 100, speed = 800 - 195.087 = 604.913, time = 100/604.913 = 0.165
    Third 100, speed = 800 - 2(195.087) = 409.826, time = 100/409.826 = 0.244
    Fourth 100, speed = 800 - 3(195.087) = 214.739, time = 100/214.739 = 0.466

    Checking: 0.125+0.165+0.244+0.466 = 1.000

    Thus, a constant decelleration of 195.087 ups every 100 units produces a good approximation of 1.95087 per unit.

    Now for 8s. There are 8 pieces of the trip, each measuring 50 units. Again, the first speed is 800 ups, the second 800-x, the third 800-2x, down to the last 800-7x. For this one, I get a constant decelleration of 87.317 ups every 50 units produces a good approximation of 87.317/50 = 1.74634 per unit.

    16s? x = 41.621 and the approximation is 41.621/25 = 1.66484 per unit

    Noting how things are going:

    4 -- 1.95087
    8 -- 1.74634
    16 -- 1.66484

    Is there a limit to this process? If we tried 1,000 pieces, what woud we get? Maybe 1,000,000? Does it just keep going down or will it level off?

    In four steps, from 4 to 8, the decrease averaged 0.051133. In eight steps, from 8 to 16, the decrease averaged only 0.010187. It's decreasing more slowly. I'm beginning to suspect that there is a limit to how low it will go.

    Note: (That is, if you managed to get this far.) Like I said, this is an exploration. I'm just playing around with the situation a bit and seeing what there is to see and pointing out what there is to point out. If you're making a Warcraft Mod, this suggests you may have sufficient time on your hands to be able to enjoy such an exploration. Try to remember that the learning of mathematics can be FUN!!
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  13. #13
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    Here's the 5 formula that are useful in calculating linear motion:

    v = u + at
    s = \frac12(u+v)t
    s = ut + \frac12 at^2
    s = vt - \frac12at^2
    v^2 = u^2 + 2as

    Where:
    v = Final Velocity
    u = Initial (Start) Velocity
    a = Acceleration
    s = Displacement (Distance travelled in a given direction)
    t = Time of motion

    Combining the first and the second equation allowed the bottom three equations to be derived.
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