Relative Motion in One Dimension — Question 9

Question 9

A constant-speed frame preserves acceleration, not velocity

In ground coordinates, $x_A=11+3t+t^2$ m and the observer origin is $x_B=5+3t$ m, with numerical t in seconds. Their axes have the same orientation.

1. Find position, velocity and acceleration of A relative to B.

2. At t=0, is A at the observer’s origin, at rest relative to the observer, both, or neither?

3. Explain why ground and observer accelerations agree even though their velocities differ.

Question 9 - Solution

A constant-speed frame preserves acceleration, not velocity

(a) Subtract the moving-origin history.

$$x_{A/B}=(11+3t+t^2)-(5+3t)=6+t^2,\quad\boxed{v_{A/B}=2t,\quad a_{A/B}=2}\quad\text{(SI)}$$ $x_{A/B}=6+t^2$ m, $v_{A/B}=2t$ m/s and $a_{A/B}=2$ m/s$^2$. Ground velocity of A is $3+2t$ m/s and ground acceleration is also 2 m/s$^2$.

(b) Separate position from motion.

$$x_{A/B}(0)=6\,\mathrm m\ne0,\quad v_{A/B}(0)=0,\quad a_{A/B}(0)=2\,\mathrm{m/s^2}$$ At zero the particle is 6 m east of B and has zero relative velocity. It is momentarily at rest relative to the observer but not located at the observer’s origin. The positive relative acceleration makes it begin moving east in that frame.

(c) Differentiate the transformation twice.

$$a_{A/B}=\frac{d}{dt}(v_A-v_B)=a_A-a_B=2-0=2\,\mathrm{m/s^2}$$ $v_{A/B}=v_{A/G}-v_{B/G}$ and $a_{A/B}=a_{A/G}-a_{B/G}$. Since B has constant velocity, $a_{B/G}=0$. If the observer accelerated, this equality of accelerations would generally fail; parallel axes alone are not enough.