Question

1. an athlete from your schools cross - country team goes for a training run. she runs 3.5 km south, 4.5 km north, then another 2.4 km east.

a. draw the vectors that represent each portion of the students run.
b. draw the vector from the students initial position to her final position (the resultant).
c. at the end of her run, how far is the athlete from her starting position?

1. resolve the two - dimensional vector below into horizontal and vertical vectors.
2. resolve the two - dimensional vector below into horizontal and vertical vectors.

Explanation:

Step1: Identify the displacements as vectors

The athlete runs 3.5 km south (a vector $\vec{A}$ with magnitude 3.5 km in the -y direction), 4.3 km north ($\vec{B}$ with magnitude 4.3 km in the +y direction) and 2.4 km east ($\vec{C}$ with magnitude 2.4 km in the +x direction).

Step2: Calculate the net displacement in the y - direction

The net displacement in the y - direction $y$ is $y=4.3 - 3.5=0.8$ km (north).

Step3: Calculate the net displacement in the x - direction

The net displacement in the x - direction $x = 2.4$ km (east).

Step4: Use the Pythagorean theorem to find the magnitude of the resultant vector

The magnitude of the resultant vector $R$ from the starting point to the ending point is given by $R=\sqrt{x^{2}+y^{2}}$. Substitute $x = 2.4$ km and $y = 0.8$ km into the formula: $R=\sqrt{(2.4)^{2}+(0.8)^{2}}=\sqrt{5.76 + 0.64}=\sqrt{6.4}\approx 2.53$ km.

Answer:

Approximately 2.53 km