What is an equation for distance which relates the jerk and snap?

[warfreak131]

Homework Statement

Well I was perusing Wikipedia on physics and came across the "jerk" of an object, the rate of change of acceleration, as well as the "snap", the rate of change of the jerk. The units for these were $$\frac{m}{s^3}$$ and $$\frac{m}{s^4}$$ respectively.

Is there some way to relate these to distance in the equation:

$$d = x + v_{i}t + \frac{1}{2}at^2$$ ?

Homework Equations

$$d = x + v_{i}t + \frac{1}{2}at^2$$

The Attempt at a Solution

The only good guess I have is that the coefficient before each variable is equal to $$\frac{1}{n!}$$ where $$n$$ is the power that $$t$$ is raised to.

So according to my rule, the equation including both snap and jerk would be:

$$d = \frac{1}{0!}xt^{0} + \frac{1}{1!}v_{i}t^{1} + \frac{1}{2!}at^2 + \frac{1}{3!}jt^3 + \frac{1}{4!}st^4$$

simplifying to:

$$d = x + v_{i}t + \frac{1}{2}at^2 + \frac{1}{6}jt^3 + \frac{1}{24}st^4$$

Am I right?

 

[rock.freak667]

As much as I can say about it is

$$jerk= \frac{da}{dt} = \frac{d}{dt}(\frac{d^2x}{dt^2})= \frac{d^3x}{dt^3}$$


though I've never learned about jerk and snap .

 

[tomcue]

Your attempt at a solution is close, but not quite correct. The equation you have written does include the jerk and snap, but it does not directly relate them to distance. The equation you have written is essentially just a polynomial expansion of the original equation, with the addition of the jerk and snap terms.

To directly relate the jerk and snap to distance, we can use the fundamental theorem of calculus. Since jerk is the rate of change of acceleration, we can integrate it to get the acceleration:

a = \int j dt

Similarly, we can integrate snap to get the jerk:

j = \int s dt

We can then substitute these expressions into the original equation for distance:

d = x + v_{i}t + \frac{1}{2}\int a dt^2

Expanding the integral, we get:

d = x + v_{i}t + \frac{1}{2} \bigg( \frac{1}{2} \int j dt^3 + \frac{1}{6} \int s dt^4 \bigg)

Using the fundamental theorem of calculus again, we can integrate the jerk and snap to get:

d = x + v_{i}t + \frac{1}{4}jt^4 + \frac{1}{24}st^5

This is the equation that directly relates distance to jerk and snap. Your attempt at a solution was on the right track, but it did not take into account the integration of the jerk and snap terms.