Position and Displacement — Question 5

Question 5

An endpoint constraint can leave two destinations

A particle starts at $(2,1)$ m. Its final x coordinate is 8 m, and the magnitude of its displacement is exactly 10 m.

1. Find every possible final y coordinate.

2. Find both corresponding displacement directions.

3. An added observation says the final point lies below the x axis. Select the endpoint and distinguish this condition from merely moving downward initially.

Question 5 - Solution

An endpoint constraint can leave two destinations

(a) Apply the length condition to coordinate changes.

$$(8-2)^2+(y_f-1)^2=10^2\ \Rightarrow\ y_f=1\pm8\ \Rightarrow\ \boxed{y_f=9\text{ or }-7\,\mathrm m}$$ $\Delta x=8-2=6$ m. Thus $(y_f-1\ \mathrm m)^2=100-36=64\ \mathrm{m^2}$. The two values are $y_f=9$ m and $y_f=-7$ m. Using $y_f$ itself as the vertical displacement would incorrectly ignore the starting height.

(b) Use the sign of the vertical change.

$$\theta_+=\tan^{-1}(8/6)=53.130^\circ,\quad\theta_-=360^\circ-53.130^\circ=306.870^\circ$$ The displacements are $(6,8)$ m and $(6,-8)$ m. Their directions are $53.130^\circ$ and $306.870^\circ$ from positive x, respectively; each has magnitude 10 m.

(c) Use an endpoint observation to select an endpoint.

$$y_f<0\ \Rightarrow\ \vec r_f=(8,-7)\,\mathrm m;\qquad v_y(0)<0\ \not\Rightarrow\ y_f<0$$ Below the x axis means $y_f<0$, selecting $(8,-7)$ m. Initial downward motion alone would not select the final point: the particle could later turn and finish above the axis. Displacement data constrain endpoints, not every part of the intervening motion.