Position and Displacement — Question 9

Question 9

A destination region gives a displacement region

A particle starts at $(2,3)$ m. Its destination may be any point with $x_f\in[5,7]$ m and $y_f\in[-1,2]$ m. All combinations are allowed hard bounds.

1. Find the complete allowed set of displacement components.

2. Find the minimum and maximum displacement magnitudes and the endpoints attaining them.

3. Can the displacement point due east or due south? Explain using component signs.

Question 9 - Solution

A destination region gives a displacement region

(a) Subtract the fixed start from the whole region.

$$\Delta x\in[5-2,7-2]=[3,5],\quad\Delta y\in[-1-3,2-3]=[-4,-1]\quad\mathrm m$$ $\Delta x\in[3,5]$ m and $\Delta y\in[-4,-1]$ m. The allowed displacement set is a rectangle entirely in quadrant IV. The bounds are not probabilities and are not combined statistically.

(b) Locate nearest and farthest corners.

$$\boxed{d_{\min}=\sqrt{3^2+1^2}=\sqrt{10}\,\mathrm m},\quad d_{\max}=\sqrt{5^2+4^2}=\sqrt{41}\,\mathrm m$$ The smallest length is $\sqrt{3^2+(-1)^2}=\sqrt{10}$ m at destination $(5,2)$ m. The largest is $\sqrt{5^2+(-4)^2}=\sqrt{41}$ m at $(7,-1)$ m. Both are attained because every endpoint in the rectangle is allowed.

(c) Check zero-component requirements.

$$\Delta x\ge3\,\mathrm m>0,\quad\Delta y\le-1\,\mathrm m<0\ \Rightarrow\ \Delta x\ne0,\ \Delta y\ne0$$ Due east would require $\Delta y=0$, excluded by the negative vertical interval. Due south would require $\Delta x=0$, excluded by the positive horizontal interval. Every allowed displacement has both eastward and southward components.