Position and Displacement — Question 1

Question 1

The displacement arrow does not start at the origin

A robot moves from P at $(3,-2)$ m to Q at $(-1,5)$ m in fixed perpendicular floor coordinates. Its position vector points from the origin to its location; its displacement is $\Delta\vec r=\vec r_Q-\vec r_P$.

1. Find both position magnitudes and the displacement vector.

2. Find the displacement magnitude and direction counterclockwise from positive x.

3. Explain why subtracting position magnitudes does not give the displacement magnitude. Is the traveled distance determined?

Question 1 - Solution

The displacement arrow does not start at the origin

(a) Subtract coordinates, not lengths.

$$|\vec r_P|=\sqrt{3^2+(-2)^2}=\sqrt{13},\quad|\vec r_Q|=\sqrt{26},\quad\boxed{\Delta\vec r=(-4,7)\,\mathrm m}$$ $|\vec r_P|=\sqrt{13}$ m and $|\vec r_Q|=\sqrt{26}$ m. The displacement is $\Delta\vec r=(-4\hat\imath+7\hat\jmath)$ m, pointing from P to Q rather than from O to Q.

(b) Use the correct quadrant.

$$|\Delta\vec r|=\sqrt{65}=8.062\,\mathrm m,\quad\theta=180^\circ-\tan^{-1}(7/4)=119.745^\circ$$ $|\Delta\vec r|=\sqrt{65}\approx8.062$ m. Negative x and positive y put the arrow in quadrant II: $\theta=180^\circ-\tan^{-1}(7/4)\approx119.745^\circ$.

(c) Keep endpoint geometry separate from the route.

$$\big||\vec r_Q|-|\vec r_P|\big|\le|\vec r_Q-\vec r_P|\le s,\quad1.493<8.062\le s\quad\mathrm m$$ $\sqrt{26}-\sqrt{13}\approx1.493$ m is merely a change in distance from the origin. It discards the change of direction. The route length is not specified; it is at least $\sqrt{65}$ m, with equality for a straight direct path. Detours can make it arbitrarily longer.