Question

One afternoon, a couple walks three-fourths of the way around a circular lake, the radius of which is 2.51 km. they start at the west side of the lake and head due south to begin with. (a) what is the distance they travel? (b) what is the magnitude of the couple's displacement? (c) what is the direction (relative to due east) of the couple's displacement?

Answer 1

(a) Since the lake has a circular shape, the distance they traveled is exactly 3/4 of a circumference. The radius of the lake is 2.51 km, and the circumference is given by $2\pi r$, therefore the distance covered is
$d= \frac{3}{4} (2 \pi r) = \frac{3}{2} \pi (2.51 km)=11.82 km$

(b) We can consider the lake to be on a xy-plane with the origin of the axes being at the center of the lake. In this system of coordinates, the starting point of the motion is at the west side of the lake, so at coordinates (-2.51 km,0). The final point is after 3/4 of circumference, therefore at the north side, at coordinates (0, 2.51 km).
So we can calculate the magnitude of the displacement as
$d= \sqrt{(x_f-x_i)^2+(y_f-y_i)^2} = \sqrt{(0-(-2.51))^2+(2.51-0)^2}=$
$=3.55 km$

(c) Considering only the initial and final point of the motion, the couple moved 2.51 km north (on the x-axis) and 2.51 km east (on the y-axis). Therefore, we can calculate the angle of the displacement with respect to the east direction:
$\tan \alpha = \frac{\Delta y}{\Delta x} = \frac{2.51 km}{2.51 km}=1$
from which
$\alpha=45^{\circ}$