Question

problem 1: a person walks to the store and does a displacement of 200 meters to the right and 300 meters up, (200, 300) m. then the person walks to the gym with another displacement of (-500, 150) m. find the net displacement of the person.

Explanation:

Step1: Identify the two displacement vectors

The first displacement vector is $\vec{d_1} = (200, 300)$ m and the second is $\vec{d_2} = (-500, 150)$ m.

Step2: Add the corresponding components of the vectors

To find the net displacement $\vec{d_{net}}$, we add the x - components and the y - components separately.
For the x - component: $d_{netx}=200+(-500)=200 - 500=-300$
For the y - component: $d_{nety}=300 + 150 = 450$
So the net displacement vector is $(-300,450)$ m.

Step3: (Optional) Find the magnitude of the net displacement (if needed)

The magnitude of a vector $(x,y)$ is given by $|\vec{v}|=\sqrt{x^{2}+y^{2}}$.
For $\vec{d_{net}}=(-300,450)$, the magnitude $|\vec{d_{net}}|=\sqrt{(-300)^{2}+450^{2}}=\sqrt{90000 + 202500}=\sqrt{292500}\approx540.83$ m. But the question asks for the net displacement (the vector), so the vector form is more appropriate.

Answer:

The net displacement of the person is $(-300, 450)$ meters (or approximately $540.83$ meters at an angle, but the vector form $(-300,450)$ m is the direct result of adding the displacements). If we consider the vector, the answer is $\boldsymbol{(-300, 450)\text{ m}}$.