Question

Consider a physical situation in which a particle moves from point A to point B. This process is described from two coordinate systems that are identical except that they have different origins.The __________ of the particle at point A differ(s) as expressed in one coordinate system compared to the other, but the __________ from A to B is/are the same as expressed in both coordinate systems.

Answer 1

Answer:

The POSITION of the particle at point A differ(s) as expressed in one coordinate system compared to the other, but the VELOCITY from A to B is/are the same as expressed in both coordinate systems.

Explanation:

Imagine that you got 2 coordinate systems; A and B. Now, this coordinate systems start at different points.

Measured from A, the starting point of the A coordinate system its, of course, at $(0,0,0)_A$. But B its somewhere else. Lets put the origin of the system B MEASURED from the system A, at $(b_x,b_y,b_z)_A$.

This position, in the system B will be at $(0,0,0)_B$, cause its it origin, be careful, we could write:

$(b_x,b_y,b_z)_A = (0,0,0)_B$

as this two vectors are the same, the starting position for the B system.

Lets imagine that in thei got something displaced from the starting point of the system B by (1,0,0). Of course, in the system B this is located at $(1,0,0)_B$, What is the position for the system A? Well, it will be at

$(b_x,b_y,b_z)_A + (1,0,0)_A = (b_x +1,b_y,b_z)_A$.

The distance is measured the same way in both system.

Any position

$(x,y,z)_B$

in the system B, will be at

$(b_x,b_y,b_z)_A + (x,y,z)_A = (b_x + x,b_y + y,b_z + z)_A$

at system A, cause this positions ARE THE SAME, even though they are represented with different vector in this systems.

Now, why the velocity its the same? Well, the velocity its the derivative of the position with respect time.

So, for a position $R=(x,y,z)_B$, in the system B we got:

$\frac{d}{dt} R = \frac{d}{dt} (x,y,z)_B = ( \frac{dx}{dt} , \frac{dy}{dt} , \frac{dz}{dt} )_B$

R in the system A takes the form:

$R=(b_x,b_y,b_z)_A + (x,y,z)_A$

and the derivative its:

$\frac{d}{dt} R = \frac{d}{dt} [(b_x,b_y,b_z)+ (x,y,z)]_A =\frac{d}{dt} (b_x,b_y,b_z)_A + \frac{d}{dt} (x,y,z)_A$

now, the $(b_x,b_y,b_z)_A$ its constant, so:

$\frac{d}{dt} (b_x,b_y,b_z)_A = 0$

which leave us with:

$\frac{d}{dt} R =\frac{d}{dt} (x,y,z)_A$

$\frac{d}{dt} R = (\frac{d}{dt} x,\frac{d}{dt} y,\frac{d}{dt} z)_A$

This is the same for both systems!