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1. kelsey swims 1.5km east, then bikes 40km north, and then runs 10km w…

Question

  1. 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 o c²=(40²)+(8.5²) o c²=(40²)+(10²)+(1.5²) o c²=(1.5²)+(40²) o c²=(40²)+(10²)

Explanation:

Step1: Define vector components

Let the east - west direction be the x - axis (east is positive x, west is negative x) and north - south direction be the y - axis (north is positive y). The east - ward displacement $x_1 = 15$ km, the north - ward displacement $y_1=40$ km and the west - ward displacement $x_2=- 10$ km. The net x - component of the displacement $x=x_1 + x_2=15-10 = 5$ km and the y - component of the displacement $y = 40$ km.

Step2: Use the Pythagorean theorem for magnitude

The magnitude of the displacement vector $d$ in two - dimensions is given by $d=\sqrt{x^{2}+y^{2}}$. Substituting $x = 5$ km and $y = 40$ km, we can also calculate it using the individual displacements directly. The magnitude of the displacement $d$ of the motion is given by the Pythagorean theorem for vector addition. If we consider the three displacements, the magnitude of the net displacement $d$ of the motion is calculated as $d=\sqrt{(15 - 10)^2+(40)^2}=\sqrt{(5)^2+(40)^2}$. In terms of the original displacements without calculating the net x - component first, we know that for a right - angled triangle formed by the net displacement vector, the magnitude $d$ of the displacement is given by $d=\sqrt{(15)^2+(40)^2+( - 10)^2}=\sqrt{(15)^2+(40)^2+(10)^2}$. The formula for the magnitude of the displacement vector when adding vectors in two - dimensions is $d=\sqrt{d_x^{2}+d_y^{2}}$, where $d_x$ and $d_y$ are the net x and y components of the displacement. The correct formula for the magnitude of the displacement $C$ (using the Pythagorean theorem for vector addition) is $C=\sqrt{(15)^2+(40)^2+(10)^2}$, which is equivalent to $C^{2}=(15)^{2}+(40)^{2}+(10)^{2}$.

Answer:

The correct option is the one with $C^{2}=(15)^{2}+(40)^{2}+(10)^{2}$