Relative Motion in Two Dimensions — Question 7

Question 7

Transform an entire two-dimensional history

For numerical t in seconds, a particle has ground position $\vec r_A=(10+4t,3+2t-t^2)$ m. A translating observer has $\vec r_B=(1+t,-2+3t)$ m. The axes remain parallel and do not rotate.

1. Find the relative position, velocity and acceleration.

2. Evaluate the relative position and velocity at 1 s and compare with the ground velocity.

3. Explain which equality of accelerations depends on B’s constant velocity.

Question 7 - Solution

Transform an entire two-dimensional history

(a) Transform both coordinate components.

$$\vec r_{A/B}=(9+3t,5-t-t^2),\quad\vec v_{A/B}=(3,-1-2t),\quad\boxed{\vec a_{A/B}=(0,-2)}\quad\text{(SI)}$$ $\vec r_{A/B}=(9+3t,5-t-t^2)$ m. Differentiating gives $\vec v_{A/B}=(3,-1-2t)$ m/s and $\vec a_{A/B}=(0,-2)$ m/s$^2$. The initial offsets remain in position but disappear from derivatives.

(b) Compare simultaneous records.

$$\vec r_{A/B}(1)=(12,3)\,\mathrm m,\quad\vec v_{A/B}(1)=(4,0)-(1,3)=(3,-3)\,\mathrm{m/s}$$ At 1 s, relative position is $(12,3)$ m and relative velocity is $(3,-3)$ m/s. Ground velocity is $(4,0)$ m/s. Subtracting observer velocity $(1,3)$ m/s gives the same relative velocity. Horizontal motion on the ground can appear diagonal in a moving frame.

(c) Use the acceleration transformation.

$$\vec a_{A/B}=\vec a_{A/G}-\vec a_{B/G}=(0,-2)-(0,0)=(0,-2)\,\mathrm{m/s^2}$$ $\vec a_{A/B}=\vec a_{A/G}-\vec a_{B/G}$. Here B’s velocity is constant, so its acceleration is zero and the two descriptions give the same particle acceleration. If B accelerated, its acceleration would have to be subtracted as well; nonrotating axes alone do not ensure equality.