Relative Motion in One Dimension — Question 10

Question 10

Relative-velocity records must close around a frame cycle

Three observers A, B and C move along one east-positive line. Their reported pairwise velocities are $v_{A/B}=10$ m/s, $v_{B/C}=-4$ m/s and $v_{C/A}=-5$ m/s. The records refer to the same instant.

1. Derive the condition the three records must satisfy and test it.

2. If the first two entries are trusted, repair the third.

3. If C moves at +1 m/s relative to ground, find the ground velocities of A and B and verify the repaired records.

Question 10 - Solution

Relative-velocity records must close around a frame cycle

(a) Sum a complete reference-frame cycle.

$$v_{A/B}+v_{B/C}+v_{C/A}=(v_A-v_B)+(v_B-v_C)+(v_C-v_A)=0$$ Using $v_{A/B}=v_A-v_B$ and analogous expressions, the sum $v_{A/B}+v_{B/C}+v_{C/A}$ must be zero. The reported sum is $10-4-5=1$ m/s, so the exact records are inconsistent.

(b) Use transitivity before reversing a label.

$$10-4-5=1\,\mathrm{m/s}\ne0,\quad\boxed{v_{C/A}=-(10-4)=-6\,\mathrm{m/s}}$$ $v_{A/C}=v_{A/B}+v_{B/C}=6$ m/s. Reversing object and reference gives $v_{C/A}=-6$ m/s. The repaired value differs from the report by $-1$ m/s.

(c) Anchor the relative data to one frame.

$$v_B=-4+1=-3,\quad v_A=10-3=7\quad\mathrm{m/s};\qquad7-(-3)=10,\ -3-1=-4,\ 1-7=-6$$ $v_{B/G}=v_{B/C}+v_{C/G}=-4+1=-3$ m/s. Then $v_{A/G}=10+(-3)=7$ m/s. The differences are $7-(-3)=10$, $-3-1=-4$, and $1-7=-6$ m/s, closing the cycle.