Question

A student walks to home from school. They walk 65m north and the. 40m East. What is their total displacement?

Answer 1

Let the north be towards the positive of the y-axis and the east be towards the positive of the x-axis.

Let i be the unit vector along the x-axis and j be the unit vector along the y-axis.

The distance traveled by the student towards the east is d_1 = 40 m.

The distance traveled by the student towards the north is d_2 = 65 m.

The displacement of the student is,

`\begin{gathered} d=d_1i+d_2j \\ d=40\text{ i+65 j} \end{gathered}`

The magnitude of the displacement is,

`\begin{gathered} |d|=\sqrt[]{d^2_1+d^2_2} \\ |d|=\sqrt[]{40^2+65^2} \\ |d|=\sqrt[]{1600+4225} \\ |d|=76.32\text{ m} \end{gathered}`

Thus, the total displacement of the student is 76.32 meters.