Position and Displacement — Question 6

Question 6

Changing the origin changes positions, not displacement

A cart starts at $\vec r_i=(7,-2)$ m and ends at $\vec r_f=(10,2)$ m. A new origin is at old coordinates $\vec a=(5,-4)$ m, with axes parallel to the originals. New coordinates satisfy $\vec r\,{}^{\prime}=\vec r-\vec a$.

1. Find both new position vectors and the displacement in each system.

2. Compare the initial distance from the origin in the two systems.

3. A student transforms only the final position before subtracting the old initial position. Find that result and explain the error.

Question 6 - Solution

Changing the origin changes positions, not displacement

(a) Translate both endpoints consistently.

$$\vec r_i'=(2,2),\quad\vec r_f'=(5,6),\quad\boxed{\Delta\vec r'=(3,4)=\Delta\vec r}\quad\mathrm m$$ $\vec r_i\,{}^{\prime}=(2,2)$ m and $\vec r_f\,{}^{\prime}=(5,6)$ m. Both coordinate systems give $\Delta\vec r=(3,4)$ m, magnitude 5 m, because the common translation cancels in the subtraction.

(b) Position depends on a reference point.

$$|\vec r_i|=\sqrt{7^2+(-2)^2}=\sqrt{53}\,\mathrm m,\quad|\vec r_i'|=\sqrt{2^2+2^2}=\sqrt8\,\mathrm m$$ The original initial distance is $\sqrt{53}$ m; the new one is $\sqrt8$ m. These distances refer to different origins and need not agree. The physical cart position is unchanged, but its position vector has a different tail.

(c) Do not mix reference origins.

$$\vec r_f'-\vec r_i=(-2,8)\,\mathrm m,\quad\text{error}=(-2,8)-(3,4)=(-5,4)\,\mathrm m=-\vec a$$ The mixed subtraction gives $(5,6)-(7,-2)=(-2,8)$ m. Its error relative to the true $(3,4)$ m is $(-5,4)$ m, exactly $-\vec a$. It combines two coordinate descriptions as though they had the same reference point.