how do you find position vector?

Question

So at time zero a particle is at x= 4 m and y= 3 m and has a velocity of $${ v= \left( 2.0 \ \boldsymbol{\hat{\imath}}-9.0 \ \boldsymbol{\hat{\jmath}} \right) \text{m/s}}$$

The acceleration of the particle is constant and given by

$${a= \left(4.0 \ \boldsymbol{\hat{\imath}}+3.0 \ \boldsymbol{\hat{\jmath}} \right) \text{m/s}^2}$$

Express the position vector at t= 4.0 sec in terms of $\boldsymbol{\hat{\imath}}$ and $\boldsymbol{\hat{\jmath}}$. Also give the magnitude and direction of the position vector at this time.

I believe I have to use the velocity formula and divide the two numbers by ${t= 4}$ sec. From there I think I have the $x$ and add it to the x and y when ${t= 0}$ sec. But I don't know what to do from there.

Answer 1

Formally you would have to integrate the acceleration in order to find the velocity, and then integrate the velocity to find the position as a function of time. But given that this is a uniformly accelerated motion, the solution is well known, and of the form $$\mathbf x = \mathbf x_0 +\mathbf v_0t + \frac12\mathbf at^2,$$ which is just a special case of the more general procedure that I have mentioned above.

Answer 2

A Position vector is a vector that describes position of a point in space. Often the position of such a point is nothing but the x,y, and z coordinates of the point with origin as the reference point. The direction of such a vector is the arrow originating from the origin towards the point and magnitude would be the length of this arrow.

So imagine a point in space with co-ordinates (2,3,6). If you draw a line starting from origin (tail) and ending in the point (head) then this arrow so made would be the position vector for the co-ordinate (2,3,6). You can do a bit of math and algebra to establish its length and direction

You can watch this video that I have made to understand this better

Position Vectors