Question

1. (5 points) At 6pm, ghost A is 5 kilometers due west of ghost B. Ghost A is flying west

Answer 1

Answer:

The distance between the ghost changes at 10 pm approximately at a rate of 24.981 kilometers per hour.

Step-by-step explanation:

At first we assume that north and east directions both represent positive quantities. Let suppose that $\vec r_{A,o} = (0\,km,0\,km)$ and $\vec r_{B,o} = (5\,km, 0\,km)$. If both ghosts moves at constant velocity such that $\vec v_{A} = \left(-15\,\frac{km}{h}, 0\,\frac{km}{h} \right)$ and $\vec v_{B} = \left(0\,\frac{km}{h},20\,\frac{km}{h} \right)$, then the final positions of both ghosts are, respectively:

Ghost A

$\vec r_{A} = \vec r_{A,o}+t\cdot \vec v_{A}$ (Eq. 1)

Ghost B

$\vec r_{B} = \vec r_{B,o}+t\cdot \vec v_{B}$ (Eq. 2)

Where $t$ is the time, measured in hours.

Then, the equations of motion of each ghost are, respectively:

Ghost A

$\vec r_{A} = (0\,km,0\,km)+t\cdot \left(-15\,\frac{km}{h}, 0\,\frac{km}{h} \right)$

$\vec r_{A} = \left(-15\cdot t, 0)\,\,\,\left[km \right]$

Ghost B

$\vec r_{B} = (5\,km, 0\,km)+t\cdot \left(0\,\frac{km}{h}, 20\,\frac{km}{h} \right)$

$\vec r_{B} = (5, 20\cdot t)\,\,\,\left[km\right]$

Then, the distance between both ghosts is:

$\vec r_{B/A} = (5,20\cdot t)-(-15\cdot t, 0)\,\,\,[km]$

$\vec r_{B/A} =(5+15\cdot t, 20\cdot t)\,\,\,[km]$ (Eq. 3)

The magnitude of the relative is represented by the following Pythagorean identity:

$r^{2}_{B/A} = (5+15\cdot t)^{2}+(20\cdot t)^{2}$

Then, we find the rate of change of the relative distance ($\dot r_{B/A}$), measured in kilometers per hour, by implicit differentiation:

$2\cdot r_{B/A}\cdot \dot r_{B/A} = 2\cdot (5+15\cdot t)\cdot 15+2\cdot (20\cdot t)\cdot 20$

$r_{B/A}\cdot \dot r_{B/A} = 15\cdot (5+15\cdot t)+20\cdot (20\cdot t)$

$\dot r_{B/A} = \frac{75+625\cdot t}{r_{B/A}}$

$\dot r_{B/A} = \frac{75+625\cdot t}{\sqrt{(5+15\cdot t)^{2}+(20\cdot t)^{2}}}$ (Eq. 4)

If we know that $t = 4\,h$, then the rate of change of the relative distance at 10 PM is:

$\dot r_{B/A} = \frac{75+625\cdot (4)}{\sqrt{[5+15\cdot (4)]^{2}+[20\cdot (4)]^{2}}}$

$\dot r_{B/A} \approx 24.981\,\frac{km}{h}$

The distance between the ghost changes at 10 pm approximately at a rate of 24.981 kilometers per hour.