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This question already has an answer here:

How do we determine the direction of instantaneous acceleration when the body is moving in a plane (or a 3D space)? This question has been truly bothering me for nearly two weeks. I looked it up, found a similar post, but that didn't really clear up my doubts, so I decided to put it up.

Let's get to the point. I do understand the direction of acceleration (average or instantaneous) is along the direction of "change in velocity" over a time interval t. And it's relatively much more easy, to find the direction of that "change in velocity" (vector addition/subtraction), if the time interval over which the change takes place is significantly larger, say 5 seconds, 10 seconds...etc. But it gets much more challenging to determine this direction when the time interval becomes infinitesimally small, i.e when it approaches zero. Let's say for example, a body is moving along a curve, and it's trajectory equation is $y = x²$. It means the body is moving along a parabolic path. What I know is, the body's instantaneous velocity at any point, is along the tangent to the curve at that point. But,

  1. How do we find the direction of its instantaneous acceleration at that point, if all we're given is its trajectory equation?

And

  1. If we differentiate its trajectory equation partially wrt time, and plot its Vy vs Vx relation, what does the tangent at any point to this Vy vs Vx curve give? Does the slope of a Vy vs Vx curve have any physical meaning?
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marked as duplicate by ja72, user191954, Jon Custer, Kyle Kanos, Aaron Stevens Oct 11 '18 at 0:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I think the equation should be written in the parametric form with time as the parameter.Then differentiate it. $\endgroup$ – Mohan Oct 8 '18 at 13:28
  • $\begingroup$ Can you please elaborate it? $\endgroup$ – π times e Oct 8 '18 at 13:38
  • $\begingroup$ There isn't such a unique point or line where accelerations are zero in general like instantaneous velocities have. $\endgroup$ – ja72 Oct 8 '18 at 16:59
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The easiest way is to get your position as a function of time - instead of defining your trajectory as a curve $y(x)$, use the separate equations $x(t)$ and $y(t)$. Then, the acceleration as a function of time will just be the vector $\langle \frac{d^2 x}{dt^2}, \frac{d^2 y}{dt^2}\rangle$.

If for some reason that's not an option, given the curve $y(x)$, we can differentiate it to get

$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

Since the velocity vector is $\langle \frac{dx}{dt},\frac{dy}{dt}\rangle$, the angle $\theta_v$ that the velocity vector makes with the horizontal axis is given by

$$\theta_v = \tan^{-1}\left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right)=\tan^{-1}\left(\frac{dy}{dx}\right)$$

But finding the direction of the velocity vector is easy, because the velocity vector always points along the tangent to your curve $y(x)$, so you can get that information directly from the curve. However, the acceleration vector is not subject to the same restrictions. It can have both a tangential component and a normal component. The normal component:

$$a_n = \frac{|v|^2}{R}$$

changes only the direction of the velocity vector. As you can see, it does have some dependence on the instantaneous radius of curvature $R$ of the curve $y(x)$, but it also depends on the magnitude of the velocity vector $v$, which we do not have because we aren't given the functions $x(t)$ or $y(t)$. So we can't calculate the normal component. The tangential component changes only the magnitude of the velocity vector (and whether it's moving "forward" or "backward" along the curve), which entirely depends on how quickly the particle moves along the curve and not on the shape of the curve itself. So we can't calculate either component separately. What we want is the angle between the velocity and acceleration vectors

$$\theta_{va} = \tan^{-1}\left(\frac{a_n}{a_t}\right) = \tan^{-1}\left(\frac{\frac{v^2}{R}}{\frac{dv}{dt}}\right)$$

from which we cannot, in general, eliminate the dependence on the magnitude of the velocity. So, in general, you cannot find the direction of the acceleration vector from the shape of the path $y(x)$ alone. You must have some information about the motion as a function of time.

That said, there are two special cases in which you can get the direction of the acceleration vector based only on the shape of the curve:

  • If the motion is at constant speed, then the magnitude of the velocity vector $v$ is fixed, and so $a_t=0$. In that case, the acceleration, if it is nonzero, will always point in the direction perpendicular to the curve (and by extension, perpendicular to the velocity vector).

  • If the motion is in a straight line, then the direction of the velocity vector is fixed, and so $a_n=0$. In that case, the acceleration will either be parallel or antiparallel to the velocity vector (depending on if the speed is increasing or decreasing at a particular moment in time).

And just as a final note, taking the second derivative of the curve $y(x)$ gives you

$$\frac{d^2 y}{dx^2} = \frac{\frac{d^2 y}{dt^2}\frac{dx}{dt}-\frac{d^2x}{dt^2}\frac{dy}{dt}}{\left(\frac{dx}{dt}\right)^3}$$

which doesn't give you any way of separating the two second time-derivatives to get the ratio $\frac{\frac{d^2y}{dt^2}}{\frac{d^2x}{dt^2}}$ that you would need to calculate the direction of the acceleration vector. So you can't use the same trick as we used to determine the direction of the velocity vector, either.

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If you think of $$ (a_x,a_y)=\frac{d}{dt}(v_x,v_y) $$ then you can find $$ \frac{a_y}{a_x}=\frac{dv_y/dt}{dv_x/dt}=\frac{dv_y}{dv_x} =\frac{\Delta v_y}{\Delta v_x} $$ In other words, the ratio of accelerations at time $t$, from which you can get the direction of $\vec a$, is just the slope of the $(v_x,v_y)$ graph at that point.

You are not going to get information about the magnitude of the acceleration from the $v_y$ vs $v_x$ graph as this parametric curve eliminates $t$. The simplest example of why this is so would be two curves $(v_x,v_y)=(t,2t)$ and $(2t,4t)$ which completely overlap but for which the accelerations are different. You would need locate $(v_x,v_y)$ for a sequence of different $t$ on the same plot, and indicate values of $t$ on your curve so you can recover the time interval between two values of $\vec v$.

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I think what you are asking is how to decompose an acceleration vector $\vec{a}$ into tangential and normal components and find the center of rotation of the normal (centrifugal) component.

  • A velocity vector $\vec{v}$ is always tangent to the path, with a tangent vector $\hat{e}$ and magnitude (speed) $v$ $$ \vec{v} = v \,\hat{e} $$
  • An acceleration vector $\vec{a}$ has both a tangent component with magnitude $\dot{v}$ and a normal component along $\hat{n}$ with magnitude $v^2/r$ $$ \vec{a} = \dot{v} \hat{e} + \frac{v^2}{r} \hat{n} $$ where $r$ is the radius of curvature of the path.

See the linked Wikipedia article on the calculation of the radius of curvature from the path definition:

$$ r = \frac{ \left(1+ \left( \frac{{\rm d}y}{{\rm d}x} \right)^2 \right)^{3/2} }{ \frac{{\rm d}^2 y}{{\rm d}x^2} } $$

$$ r = \frac{ \left( \dot{x}^2 + \dot{y}^2 \right)^{3/2} }{ \dot{y} \ddot{x} - \ddot{y} \dot{x} } $$

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