Position and Displacement — Question 7

Question 7

A closed path can have nonzero partial displacements

A particle follows three straight legs from O $(0,0)$ m to P $(6,0)$ m to Q $(6,8)$ m and then back to O.

1. Find each leg displacement and the complete-trip displacement.

2. Find the traveled distance and the displacement from the midpoint of OP to the midpoint of QO.

3. Explain why a zero complete-trip displacement does not mean that position was constant.

Question 7 - Solution

A closed path can have nonzero partial displacements

(a) Add directed legs.

$$\boxed{\Delta\vec r=(6,0)+(0,8)+(-6,-8)=(0,0)\,\mathrm m}$$ The three displacements are $(6,0)$ m, $(0,8)$ m and $(-6,-8)$ m. Their sum is zero. The last leg has magnitude 10 m and returns the particle to its initial position.

(b) Use the specified subinterval endpoints.

$$s=6+8+\sqrt{6^2+8^2}=24\,\mathrm m,\quad\Delta\vec r_{\mathrm{mid}}=(3,4)-(3,0)=(0,4)\,\mathrm m$$ Total distance is $6+8+10=24$ m. The midpoint of OP is $(3,0)$ m and that of QO is $(3,4)$ m. Their displacement is therefore $(0,4)$ m, magnitude 4 m. It is not half of the full-trip displacement.

(c) A global cancellation does not describe each instant.

$$\Delta\vec r=\int\vec v\,dt=\vec0,\qquad s=\int|\vec v|\,dt=24\,\mathrm m\ne0$$ The position changes throughout each leg even though its final value equals its initial value. A zero interval displacement constrains only the two selected endpoints. It says nothing about the distance traveled or whether the particle was ever at rest between them.