Equation for position, moving with a value J of the third derivative of position.

Question

Q. An object moves in one dimension (described by an x-value) with a constant value J of the third derivative of position with respect to time. Write an equation for the position $x_0$ and an initial velocity $v_0$ (the sign of $v_0$ indicates the initial direction of motion).

A. $x_0 + v_0t + \frac{1}{2}a_0 t^2 + \frac{1}{6}J t^3$

My thinking. I'm familiar with the formula: $s = v_0t + \frac{1}{2}a t^2$ and if the starting point is x0 and not the origin then: $s = x_0 + v_0t + \frac{1}{2}a t^2$

I found something about the third derivative of position being called a Jerk and seen this $\frac{1}{6}J t^3$ in a few places but can't seem to get my head around this question. What is the explanation?

Answer 1

we have that $\frac{d^3x}{dt^3} = J$ where x is the position of the particle and that at $ t = 0, \frac{d^2x}{dt^2} =a_0, \frac{dx}{dt} = v_0, x = x_0. $ (I'm assuming the $a_0$ in your answer is initial acceleration.)

We integrate the first equation with respect to time: $$ \frac{d^2x}{dt^2} = Jt + c_1$$ for some constant $c_1$. But at t = 0, $\frac{d^2x}{dt^2} =a_0$ so $c_1 = a_0$. You integrate twice more in the same way, finding the constant given at each step using the conditions at time t = 0 and that gives an expression for $x$ in terms of t.

Answer 2

For a constant acceleration $a$ we get $ x(t) = x_0 + v_0t+\frac{1}{2}at^2$ by integrating: $ \frac{d^2x}{dt^2} = a$ twice.

Similar for a constant Jerk we intergate $ \frac{d^3x}{dt^3} = J$ three times to get: $ x(t) = x_0 + v_0t+\frac{1}{2}a_0t^2+\frac{1}{6}Jt^3 $