Question

12 kelsey swims 1.5km east, then bikes 40km north, and then runs 10km west. based on the properties that define vector quantities in two - dimensional motion, which choice gives the correct solution for the magnitude of kelseys displacement? clear all c²=(40²)+(10²)+(1.5²) c²=(40²)+(8.5²) c²=(40²)+(10²) c²=(1.5²)+(40²)

Explanation:

Step1: Define the displacements as vectors

Let the east - west direction be along the x - axis (east is positive x and west is negative x) and north - south direction be along the y - axis (north is positive y). The east - ward displacement $d_1=1.5$ km, the north - ward displacement $d_2 = 40$ km and the west - ward displacement $d_3=- 10$ km. The net x - component of the displacement $x=d_1 + d_3=1.5-10=-8.5$ km and the y - component of the displacement $y = 40$ km.

Step2: Use the Pythagorean theorem for magnitude of displacement

The magnitude of the displacement vector $C$ in two - dimensional space is given by $C=\sqrt{x^{2}+y^{2}}$. Substituting $x=-8.5$ km and $y = 40$ km, we get $C^{2}=x^{2}+y^{2}=(40)^{2}+(8.5)^{2}$.

Answer:

$C^{2}=(40)^{2}+(8.5)^{2}$