# Kinematics - Jerk Problem

1. Nov 29, 2011

### GreenPrint

1. The problem statement, all variables and given/known data

In another universe, the force of gravity causes not a constant acceleration, but a constant "jerk" (third derivative of position with respect to time) with a value of j in the downward direction. An object is launched from the ground with an initial velocity of v_0 at an angle of theta and the initial acceleration is zero. Derive an expression for the horizontal distance the object will travel before it hits the ground. Answer in terms of theta, J, and v_0.

2. Relevant equations

3. The attempt at a solution

I know that

x = Ʃ[k=0,n] (x^(k)t^k)/k!
note that x^(k) is the kth derivative not kth power
so to come up with the equation were we consider the first three derivatives of x we got
x = x_0 + (dx/dt) t + 1/2 (d^2x/dt^2) t^2 + 1/6 (d^3x/dt^3)t^3
or
x = x_0 + v_0 t + 1/2 at^2 + 1/6 jt^3
in this case we want the horizontal component of the velocity so
x = x_0 + v_0 cos(θ) t + 1/2 at^2 + 1/6 jt^3
we are given that the initial acceleration is zero... so can I just ignore it completely from this problem sense it doesn't say to include air friction? This is what I assumed
x = x_0 + v_0 cos(θ) t + 1/6 jt^3
does this look correct?

2. Nov 29, 2011

### Staff: Mentor

You'll want to solve for the time that the projectile spends in the air, then use that to find the horizontal range (presumably the horizontal velocity remains constant in this universe, as it does in ours). When does the projectile strike the ground?

3. Nov 30, 2011

### cepheid

Staff Emeritus
There are a few issues with what you've done. First of all, let me just point out that it's not necessary in this case to use this fancy general formula. You can just start with what you've been given (which is a third derivative of position) and integrate it three times to recover position as a function of time. You're told that the jerk is constant and in the negative vertical (y) direction. The jerk is the derivative of the acceleration. Therefore, you could express what you've been given mathematically as:$$\frac{d^3y}{dt^3} = \frac{da(t)}{dt} = -J$$You can simply integrate this equation once with respect to time to get:$$\frac{d^2y}{dt^2} = a(t) = \int (-J)\,dt = -J\int\,dt = -Jt + C_1$$To solve for the constant of integration C1, consider the initial condition i.e. consider the acceleration at t = 0:$$a(0) = 0 = -J(0) + C_1 = C_1$$Therefore:$$a(t) = -Jt$$Now we can integrate this equation a second time in order to get the vertical velocity:$$\frac{dy}{dt} = v_y(t) = \int a(t)\,dt = -J\int t\,dt = -\frac{J}{2}t^2 + C_2$$To solve for the constant of integration C2, consider the initial condition i.e. consider the vertical velocity at t = 0:$$v_y(0) = v_0 \sin\theta = -\frac{J}{2}(0)^2 + C_2 = C_2$$Therefore:$$v_y(t) = v_0 \sin\theta - \frac{J}{2}t^2$$Now we can integrate this equation a third time in order to get the vertical position as a function of time:$$y(t) = \int v_y(t)\,dt = \int v_0\sin\theta\,dt - \frac{J}{2}\int t^2\,dt = (v_0\sin\theta)t - \frac{J}{6}t^3 + C_3$$The constant of integration C3 is just 0 because it is equal to the initial vertical position y(0) = 0. Therefore, the result is:$$y(t) = (v_0\sin\theta)t - \frac{J}{6}t^3$$So you can see that since you were given the third derivative of the vertical position w.r.t. time in this problem, simply integrating it three times was easy and a lot simpler than applying the general formula that you included in your original post. However, if you wanted to use the general formula, then I have worked out that it is actually as follows:

If $y^n(t) \equiv d^ny/dt^n$is a constant, then the function y(t) is given by the expression$$y(t) = \sum_{k=0}^n \left.\frac{1}{k!}\frac{d^ky}{dt^k}\right|_{t=0} \!t^k \ \ = \ \ \sum_{k=0}^n \frac{1}{k!}y^k(0)t^k$$where the right hand version is just if you prefer the "superscript k" notation for derivatives rather than the dk/dtk notation.

Notice that the coefficients of the resulting polynomial are the derivatives of y(t) evaluated at t = 0. This is something that you neglected to include in your original post. The fact that the derivatives are evaluated at t = 0 explains, for instance, why the k = 2 term went to 0 in your original post. It's because the k = 2 coefficient was a(0) = 0. It was not because you have to "assume" something about the acceleration.

Anyway, if you apply this formula for n = 3, you'll of course arrive at the result that we arrived at much more easily just by straight integration.

Last edited: Nov 30, 2011