Newtonian Mechanics — Question 2

Question 2

A translating observer changes velocity but not acceleration

In laboratory coordinates a body has $x=1+4t+2t^2$ m. A second origin moves as $X=5+3t$ m, with parallel axes and the same clock. Numerical t is in seconds.

1. Find the second observer’s position and velocity for the body.

2. Compare the two accelerations.

3. Explain why a different measured velocity need not require a different physical interaction.

Question 2 - Solution

A translating observer changes velocity but not acceleration

(a) Subtract the origin history.

$$x'=x-X=-4+t+2t^2,\quad \boxed{v'=\dot x-\dot X=1+4t}$$ $x^{\prime}=x-X=-4+t+2t^2$ m, so $v^{\prime}=1+4t$ m/s. The laboratory velocity is $4+4t$ m/s.

(b) Differentiate again.

$$a'=\ddot x-\ddot X=4-0=\boxed{4\,\mathrm{m/s^2}}$$ Both accelerations are $4$ m/s$^2$ because the relative origin velocity is constant. The fixed initial offset also disappears on differentiation.

(c) Identify the shared physical event.

$$\Delta v=-3\,\mathrm{m/s},\quad\Delta a=0,\quad m a'=m a$$ The observers assign different positions and velocities to the same body. Constant translation does not change its acceleration. Newtonian dynamics is formulated so that inertial observers related in this way can describe the same interactions consistently.