View Full Version : Trouble with an equation

Jim Starluck

2004-Mar-23, 08:41 AM

I'm trying to figure this out, but so far it's stumped me...(knew I should've studied harder in math class...)

An object is moving through space, not only accelerationg, but INCREASING its acceleration. I want a formula that lets me plug in how fast it's increasing its acceleration (in meters/second³) and how long it's doing it (in seconds). I want to be able to figure out how fast it's traveling at a given time, and how far it has traveled from its point of origin.

Help? :-?

tusenfem

2004-Mar-23, 10:39 AM

Well, I guess I should know the answer. We can just write the following:

an object starte moving at location x0, travelling with velocity v, accelerating with a and hyperating (huh what? :o ) with h, now lets write down x(t)

x(t) = x0 + v * t + a * t^2 / 2 + h * t^3 / (2 * 3)

then we would find

v(t) = d/dt x(t) = v + a * t + h * t^2 / 2

and

a(t) = d/dt v(t) = a + h * t

Seems like the derivatives work, so I guess now you can put in your values. Naturally you would have to do it correctly in a relativistic manner, and not start as I did here, but start with the Equotion of Motion :wink:

Okeliedokelie neighborino

JohnOwens

2004-Mar-23, 10:48 AM

An object is moving through space, not only accelerationg, but INCREASING its acceleration. I want a formula that lets me plug in how fast it's increasing its acceleration (in meters/second³) and how long it's doing it (in seconds).

And just FYI, the name for that increase of acceleration is "jerk". No, I am not making this up. It's the third derivative of displacement over time.

And if you thought that was fun, wait until you get into cases where the acceleration varies as a function of location instead of time.... *grumble*, *grumble* :x

swansont

2004-Mar-23, 11:25 AM

And if you thought that was fun, wait until you get into cases where the acceleration varies as a function of location instead of time.... *grumble*, *grumble* :x

Then you get to use the chain rule: dv/dt = dv/dx * dx/dt = v* dv/dx

(I'm not disagreeing that such problems are a pain in the asterix)

Jim Starluck

2004-Mar-23, 01:34 PM

Okay...I got as far as the variables, but you lost me pretty quick after that. >.<

Remind me what x(t), a(t) and v(t) stand for again? :-?

x(t) is the position of the object as a function of time.

v(t) = dx/dt is the velocity of the object as a f'n of time.

a(t) = dv/dt is the acceleration as a f'n of time.

x(0), v(0), and a(0) are the initial conditions.

AGN Fuel

2004-Mar-23, 11:42 PM

Okay...I got as far as the variables, but you lost me pretty quick after that. >.<

Remind me what x(t), a(t) and v(t) stand for again? :-?

What you are doing here Jim, is just using a bit of calculus. The first derivative here is the rate of change of position (ie velocity). The second derivative is the rate of change of velocity (ie acceleration). What you are looking for is the rate of change of acceleration. In other words, you are looking for the third derivative of your initial equation.

kelly

2004-Mar-26, 01:27 AM

I want a formula that lets me plug in how fast it's increasing its acceleration (in meters/second³) and...

acceleration is m/s^2 for what u need, not cubed.. is this a typo?

Wiley

2004-Mar-26, 01:40 AM

I want a formula that lets me plug in how fast it's increasing its acceleration (in meters/second³) and...

acceleration is m/s^2 for what u need, not cubed.. is this a typo?

No, it's not a typo. The derivative of acceleration, or the increase in acceleration, would have units of m/s^3.

If j is the increase in acceleration:

a(t) = j*t + a(0)

v(t) = j*t^2/2 + a(0)*t + v(0)

x(t) = j*t^3/6 + a(0)*t^2/2 + v(0)*t + x(0)

where a(0), v(0), and x(0), are the initial acceleration, initial velocity, and initial position.

tracer

2004-Mar-26, 01:44 AM

And if phpBB allowed HTML tags in our posts, we could write those equations with real subscripts and exponents, the way God and Leibnitz intended.

AstroSmurf

2004-Mar-26, 09:11 AM

You can; x(t) = j*t³/6 + a(0)*t²/2 + v(0)*t + x(0)

Anyway, this is all according to Newtonian physics. Doing it relativistically is ... complicated.

pi is exactly 3

2004-Mar-26, 01:01 PM

So if the initial a,v,x are 0 then the equation could simply be put

x(t)= j*t^3/6 Correct? and then you take the first and second derivatives to find the v(t) and a(t) correct? JohnOwens mention that increase in acceleration is called a jerk. So how is that used in a sentence? is it the same as acceleration? " The acceleration of that object is 10m/s^2", "The jerk of that object is 10m/s^3"

Mr. X

2004-Mar-26, 03:48 PM

Jerk is also called impulse.

Worthy of note: The equatons given are true only for constant jerk. If you have jerk as a function of time J(t), then integrate it once for acceleration a(t) and add a constant, integrate J(t) twice for speed and add contants each time, and thrice (I love that word!) for position.

Worthy of note also is that this process might or might not be possible depending on what type of function J(t) is. You might have to do some acrobatics, such as using series to get an approx. solution.

If you have x(t) (position) as a function of time, then you can always find the jerk by derivating until you find the third derivative.

More notes:

J(t) can, and a lot of times IS a function of its lower derivatives.

For example, J(t) could be:

J(t) = g(t)a(t)+ h(t)v(t) + k(t)x(t) + C

That would mean that you have a differential equation of third order.

J(t) = x'''(t)

a(t) = x''(t)

v(t) = x'(t)

x(t) = x(t)

You'd have x'''(t) = g(t)x''(t) + h(t)x'(t) + k(t)x(t) + C

And g(t), h(t), k(t) and x(t) can be anything and everything in the whole universe. And you're going to be in deep doo doo because a lot of these problems are not really solvable analytically, and you need numerical methods.

I'm dropping a few constants and the function of J(t) because it is usually possible to bring it to that form. C is a constant, but if you had to bring it from a form where the third derivative x'''(t) had a function, C would become C(t), a function of t also.

At any rate the process from J(t) upwards to x(t) can be easy as pi (pie... hmmm pie) or very very very complex.

The other way around is always possible (from x(t) to J(t) unless your x(t) function happens to be not differentiable... but why would you do that?)

At any rate I hope I could help.

tracer

2004-Mar-26, 04:17 PM

You can; x(t) = j*t³/6 + a(0)*t²/2 + v(0)*t + x(0)

That's cheating! [-X You're just using the ² and ³ characters from the ANSI character set. Let's see you write an equation with an "n" or "x" (or even a "4") as an exponent! Or an equation with subscripts.

Seriously, it would be really nice if BBCode supported [sub] and [super] directives, like HTML does.

tracer

2004-Mar-26, 04:26 PM

Jerk is also called impulse.

Not where I come from, it isn't! "Impulse," as used in every physics book I've seen, means a force applied over a length of time (equivalent to change-in-momentum), and is in units of Newton-seconds or kg*m/s. Jerk, by contrast, is in units of m/s^3.

Mr. X

2004-Mar-26, 04:29 PM

In my language it's called impulse, since jerk has no equivalent.

I don't remember what I used to call what you describe...

Eroica

2004-Mar-26, 04:48 PM

So if the initial a,v,x are 0 then the equation could simply be put

x(t)= j*t^3/6 Correct?

No. If the initial v and the initial a are both zero, then you're not going to get anywhere! :)

The jerk is the rate of change of the acceleration, da/dt. If the initial acceleration is zero, then j = 0. Initial acceleration must be some positive value.

Mr. X

2004-Mar-26, 04:48 PM

Ah, there. It's in the so-so serious ways of referring to it.

http://math.ucr.edu/home/baez/physics/General/jerk.html

Eroica

2004-Mar-26, 04:51 PM

Anyway, this is all according to Newtonian physics. Doing it relativistically is ... complicated.

Anyone care to have a go? I'm not sure how to proceed. :-k

Mr. X

2004-Mar-26, 04:55 PM

Anyone care to have a go? I'm not sure how to proceed. :-k

Pfff... ask Phil :).

AstroSmurf

2004-Mar-26, 05:02 PM

So if the initial a,v,x are 0 then the equation could simply be put

x(t)= j*t^3/6 Correct?

Quite right - acceleration will increase linearly and velocity quadratically.

No. If the initial v and the initial a are both zero, then you're not going to get anywhere! :)

The jerk is the rate of change of the acceleration, da/dt. If the initial acceleration is zero, then j = 0. Initial acceleration must be some positive value.

Hate to bring it to you, but you're wrong. Nothing in the math requires the acceleration to have any particular value. You'd also have trouble starting your car, otherwise...

If you start dealing with this relativistically, you also need to know in what frame a and j are determined (I would recommend the moving frame).

Eroica

2004-Mar-26, 05:36 PM

The jerk is the rate of change of the acceleration, da/dt. If the initial acceleration is zero, then j = 0. Initial acceleration must be some positive value.

Hate to bring it to you, but you're wrong. Nothing in the math requires the acceleration to have any particular value. You'd also have trouble starting your car, otherwise...

Ah, yes. I get it now. Thanks. I stand corrected.

Wiley

2004-Mar-27, 12:32 AM

If you start dealing with this relativistically, you also need to know in what frame a and j are determined (I would recommend the moving frame).

I am gonna recommend using the observer's frame. The object's frame, or the moving frame, would not be an inertial frame. The Lorentz transformations are technically only valid between two inertial frames. I know you can get around this requirement for special cases, such as when the acceleration is "zero almost everywhere" (my favorite, precise mathematical phrase :)).

Or were you planning on using GR?

The Watcher

2004-Mar-27, 12:46 AM

Or how about-

**Hands in pockets and whistles** Look at that mother increase acceleration through space.

Silence

More Silence

I'll log off...

AGN Fuel

2004-Mar-27, 04:35 AM

Bet you're glad you asked, huh Jim?! :lol:

Eroica

2004-Mar-27, 08:42 AM

To make up for my unforgivable lapse earlier, I've done the donkey work for anyone who still cares.

Let us assume that when time t = 0: velocity v = v°; acceleration a = a°; displacement s = s°; and jerk j = j (a constant).

By definition, Jerk is equal to the rate of change of acceleration:

da/dt = j

=> da = jdt

Integrating both sides gives:

a = jt + k

Set the boundary condition: when t =0, a = a°. This gives k = a°

=> a = jt + a° .... Equation 1

Now, acceleration, by definition, is equal to the rate of change of velocity. So, replacing a in Equation 1 with dv/dt, we get:

dv/dt = jt + a°

=> dv = jtdt + a°dt

Integrating both sides gives:

v = ½jt² + a°t + k

Set the boundary condition: when t = 0, v = v°. This gives k = v°

=> v = ½jt² + a°t + v° .... Equation 2

Now, velocity, by definition, is equal to the rate of change of displacement. So, replacing v in Equation 2 with ds/dt, we get:

ds/dt = ½jt² + a°t + v°

=> ds = ½jt²dt + a°tdt + v°dt

Integrating both sides gives:

s = (1/6)jt³ + ½a°t² + v°t + k

Set the bounday condition: when t = 0, s = s°. This gives k = s°

=> s = (1/6)jt³ + ½a°t² + v°t + s° .... Equation 3

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