Relation between the velocity vector and the acceleration vector of an object

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In uniform circular motion, the position vector is expressed as u=cos(wt)i +sin(wt)j, with the velocity and acceleration vectors being constant in magnitude and perpendicular at each instant. Knowing the velocity vector at a specific time allows for the calculation of angular frequency (ω), which can then be used to predict the position vector at any future time. This relationship holds true because the motion is circular and uniform, providing a consistent framework for analysis. Additionally, the acceleration vector can also be derived similarly, reinforcing the predictability of the motion. Thus, both vectors provide essential information for determining future positions in circular motion.
Clockclocle
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Homework Statement
velocity vector and acceleration vector
Relevant Equations
u=cos(wt)i +sin(wt)j
A uniform circular motion of a point always yield an equation u=cos(wt)i +sin(wt)j of position vector. Which we deduce the acceleration and velocity vector with constant magnitude and they are perpendicular at each instant. Can I use the information of them at one instant to predict the position vector at any instant later? Or they are only provide the information that the point is moving in the direction of velocity vector and also is affected by a centripetal force point toward the center of the circle?
 
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Clockclocle said:
Homework Statement: velocity vector and acceleration vector
Relevant Equations: u=cos(wt)i +sin(wt)j

A uniform circular motion of a point always yield an equation u=cos(wt)i +sin(wt)j of position vector. Which we deduce the acceleration and velocity vector with constant magnitude and they are perpendicular at each instant. Can I use the information of them at one instant to predict the position vector at any instant later? Or they are only provide the information that the point is moving in the direction of velocity vector and also is affected by a centripetal force point toward the center of the circle?

Somewhat. If you already know the motion is circular around the origin with radius 1 and uniform, you can write $$\vec u = \cos(\omega t)\hat \imath+\sin(\omega t)\hat \jmath)\tag{1}$$
for the position vector, as you did.

Taking the derivative ##\vec v \equiv {d\vec u\over dt}## gives $$\vec u = \omega\left (-\sin(\omega t)\hat \imath+\cos(\omega t)\hat \jmath)\right )\tag {2}$$

Knowing the velocity vector at some instant ##t_0## would give you ##\omega## and ##t_0##, which you can use in ##(1)## for any ##t##.

And you could also do something similar from ##\vec a##, the second derivative.

##\ ##
 
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