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My book says that when a mass travels in a curved path, like a circle for example, the instantaneous velocity and displacement vectors are both tangent to the path.

enter image description here

I agree that velocity vector is. But no displacement vector. In the previous chapter I learned that vectors like velocity and force is different from displacement in what their vectors mean.

Displacement vector indicates by its head and tail the actual location the object will arrive at.

But velocity does not do this. An object may never arrive at what the arrow head indicates.

For example, in this pic., the vectors are velocity, the one to the very right goes into the ground. But obviously, the object won't ever go there, the place indicated by the arrow head.

enter image description here

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  • $\begingroup$ Think about how velocity and displacement are related. $\endgroup$ – HDE 226868 Jul 22 '15 at 21:45
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As HDE 226868 says in his comment, the relationship between velocity and displacement is important.

This problem is about an infinitesimal time period. The average, the total, and the instantaneous velocity are identical during this infinitesimal.

Average velocity over any time period equals the total change in displacement divided by the time. For an infinitesimal time period, the average and total velocities are the same.

Therefore the ∆ displacement vector must be identical to the velocity vector during this infinitesimal period, in order for the relationship between average velocity and total change in displacement to hold.

So far as the direction of the ∆ displacement vector is concerned, during an infinitesimal time period, the object doesn't really go anywhere, so you approximate its path by a tangent to the circle. As K7PEH says in his 2nd comment, if you sum all the infinitesimal tangent ∆ displacement vectors, your object arrives at the head of the final displacement vector.

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  • $\begingroup$ Is it acceleration then that I mix up with velocity? $\endgroup$ – most venerable sir Jul 22 '15 at 22:16
  • $\begingroup$ But for velocity (or maybe acceleration if I am mistaken) vector , you don't arrive at the end? $\endgroup$ – most venerable sir Jul 22 '15 at 22:18
  • $\begingroup$ I apologize if I may have confused you. Acceleration doesn't enter into this problem. In the case of uniform circular motion, the acceleration vector (unlike the displacement vector) can not have a component in the direction of the velocity vector, or it would cause a change of speed. So the instantaneous acceleration vector is perpendicular to the velocity vector and points toward the origin of the circle. It's called centripetal acceleration. The 1st paragraph of this link is a good explanation: labman.phys.utk.edu/phys135/modules/m3/… $\endgroup$ – Ernie Jul 22 '15 at 22:32
  • $\begingroup$ I don't think understand my second question. The difference between displacement and velocity vectors, as I perceive it, is that displacement vector is drawn from where an object starts and where it will arrive like in actuality. But for velocity, it only indicates the direction. I edited my answer, $\endgroup$ – most venerable sir Jul 22 '15 at 22:38
  • $\begingroup$ @Doeser I think it is fair to use the term displacement for both the final position vector as well as the instantaneous (or, infinitesimal) vector. The key difference though in these two usages is that one is a static result where there is no longer any motion (velocity or acceleration) and the other (infinitesimal) is computed for the instance of the position itself where motion (velocities and accelerations) are continuously changing the infinitesimal displacement vector. "Displacement" is not a precise term meaning only one thing. $\endgroup$ – K7PEH Jul 22 '15 at 22:56
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The acceleration, $\frac{d}{dx} v$, is always tangent to the velocity vector. You could imagine a circular orbit where the velocity vector always is tangent to the ground itself and where the acceleration is pointing downwards (since the force is pointing down and $a = \frac{F}{m} = \frac{1}{m} F$). Since the constant $\frac{1}{m}$ is just a scalar, the acceleration keeps the same direction as the force, but it changes magnitude with $\frac{1}{m}$.

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