Curve in osculating plane if acceleration is constant

Question

The book Second Year Calculus: From Celestial Mechanics to Special Relativity makes this comment that if acceleration vector is constant, then the curve lies in the osculating plane spanned by unit normal and unit tangent. The book generally proves its claims but this is just given as comment. So I think there is something very easy to see here but I am missing it. Can someone prove that if acceleration is a constant vector, than curve is planar and lies in the osculating plane? Since the book is full of applications, an explanation with physical intuition would also be welcomed.

Answer 1

Here's my intuitive understanding.

If there were no acceleration, the path of the particle/body would be a straight line. Call that line $\ell.$

But now we suppose there is a constant acceleration. That means constant magnitude, but also constant direction. If the particle started at rest, it would move along a line parallel to the acceleration vector.

The motion of a particle with some initial velocity under this acceleration is the composition of the displacement along the line $\ell$ that the particle would have at time $t$ (due to its initial velocity) if there were no acceleration plus the displacement the acceleration would have caused at time $t$ to a particle that started at rest.

Since the displacements due to acceleration are all parallel and all have "starting points" on the line $\ell,$ they are all parallel and all lie in the same plane, which also contains $\ell.$ The position of the particle at time $t$ is at the "end point" of one of these displacements, so it also is in the same plane.

Once we have determined that a curve lies in a single plane, we know it has tangent and normal vectors in the same plane.

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An alternative way to look at it is to consider a particular event along the space-time path of the particle, that is, a location where the particle was at some instant of time $t_0$ according to its own timeline. We then take the inertial frame of reference $F_1$ in which this event occurs at time $t_0$ and the particle is at rest at that time. In this frame, an acceleration in a constant direction moves the particle along a straight line.

Now we go back to the inertial frame of reference in which we originally wanted to describe the particle's motion. Since the origin of the inertial frame $F_1$ is moving at a constant velocity in our selected inertial frame, any line in the frame $F_1$ (in particular, the line in that frame along which the particle moves) will sweep out a plane in our chosen frame of reference. Since the particle is always found on that line in $F_1,$ it is always found in that plane in our chosen frame of reference.

Answer 2

If the acceleration is a constant vector, the trajectory follows the well-known equation

$$\mathbb r=\mathbb a\frac{t^2}2+\mathbb v_0t+\mathbb r_0$$

which is a parabola lying in the affine plane spanned by $\mathbb a$ and $\mathbb v_0$.