Relative Motion in One Dimension — Question 2

Question 2

A moving origin requires its initial offset as well as its speed

In ground coordinates, an observer’s origin is $x_B(t)=5\ \mathrm m+(3\ \mathrm{m/s})t$. A particle has $x_A(t)=20\ \mathrm m-(2\ \mathrm{m/s})t$. The observer’s axis is parallel and east-positive.

1. Find the particle’s position $x_{A/B}=x_A-x_B$ and velocity relative to B.

2. Find when A passes B and the ground location of that event.

3. What error results from using only $x_{A/B}=x_A-3t$ with SI numerical values?

Question 2 - Solution

A moving origin requires its initial offset as well as its speed

(a) Subtract the whole origin history.

$$x_{A/B}=(20-2t)-(5+3t)=15-5t,\quad\boxed{v_{A/B}=-5\,\mathrm{m/s}}$$ $x_{A/B}=15\ \mathrm m-(5\ \mathrm{m/s})t$, and $v_{A/B}=-5$ m/s. The 15 m initial separation and the closing velocity come from different parts of the transformation.

(b) Set relative position to zero.

$$15-5t=0\ \Rightarrow\ t=3\,\mathrm s,\quad x_A=20-6=14\,\mathrm m=x_B$$ $15-5t=0$ gives t=3 s. Both ground positions are then 14 m: $x_A=20-6$ and $x_B=5+9$. Agreement in relative coordinates matches one common physical event.

(c) An omitted offset corrupts positions but not this velocity.

$$x'_{\mathrm{wrong}}=20-5t\ \Rightarrow\ t'_{\mathrm{encounter}}=4\,\mathrm s,\quad\Delta t=1\,\mathrm s$$ The incomplete expression would give $20-5t$ m and predict crossing at 4 s, one second late. Its derivative is still $-5$ m/s because a constant offset differentiates to zero. Correct relative velocity alone does not guarantee a correct encounter calculation.