Question

A disoriented physics professor drives 3.28 km north, then 4.79 km west, and then 1.50 km south. Use components to find the magnitude of the resultant displacement of this professor. Use components to find the direction of the resultant displacement of this professor. θ = 148.9 148.9 ∘ west of north

Use components to find the magnitude of the resultant displacement of this professor.

Use components to find the direction of the resultant displacement of this professor.

Answer 1

Given that :

$\vec{A}$ = 3.28 km   (in direction of north)

$\vec{B}$ = 4.79 km   (in direction of west)

$\vec{C}$ = 1.50 km   (in direction of south)

The x and y coordinate of displacement which will be given below as -

$\vec{A}$ = (0 km, 3.28 km)

$\vec{B}$ = (-4.79 km, 0 km)

$\vec{C}$ = (0 km, -1.50 km)

The net resultant displacement of this professor which will be given as -

we know that, $\vec{R}$ = $\vec{A}$ + $\vec{B}$ + $\vec{C}$

$\vec{R}$ = [(0 km, 3.28 km) + (-4.79 km, 0 km) + (0 km, -1.50 km)]

$\vec{R}$ = (-4.79 km, 1.78 km)

Magnitude of the resultant displacement of this professor which will be given by -

| $\vec{R}$ | = _$\sqrt{}$ [(-4.79 km)² + (1.78 km)²]

| $\vec{R}$ | = _$\sqrt{}$ 26.1125 km²

| $\vec{R}$ | = 5.11 km

Direction of the resultant displacement of this professor which will be given by -

$\theta$ = tan^-1 [(1.78 km) / (-4.79 km)]

$\theta$ = tan^-1 (0.3716)

$\theta$ = - 20.4 degree

OR

$\theta$ = - 20.4⁰ + 180⁰

$\theta$ = 159.6 degree