Question

A person walks for 100 m directly east, turns and then walks for 300 m directly south, and then finally turns at an angle of 30° and walks for 150 m. What is the magnitude and direction of the total displacement of the person? A) Magnitude: 400 m, Direction: 45° south of east B) Magnitude: 400 m, Direction: 30° south of east C) Magnitude: 400 m, Direction: 60° south of east D) Magnitude: 400 m, Direction: 30° north of east

Answer 1

The accurate answer, based on the calculations, is:

B) Magnitude: 320 m, Direction: 44.2° south of east.

Step-by-step explanation:

To calculate the total displacement, we'll break down each leg of the motion into its horizontal (eastward) and vertical (southward) components.

1. The person walks 100 m directly east. This displacement has a horizontal component
`(\(x_1\))` of 100 m and a vertical component
`(\(y_1\))` of 0 m.

2. The person then walks 300 m directly south. This displacement has a horizontal component
`(\(x_2\))` of 0 m and a vertical component
`(\(y_2\))` of -300 m.

3. Finally, the person turns at an angle of 30° and walks 150 m. The horizontal component
`(\(x_3\))` is
`\(150 \cos(30) \approx 129.9\)` m, and the vertical component
`(\(y_3\)) is \(150 \sin(30) = 75\) m.`

Now, let's sum up the components:

`\[ X = x_1 + x_2 + x_3 = 100 + 0 + 129.9 \approx 229.9 \, \text{m} \]`

`\[ Y = y_1 + y_2 + y_3 = 0 - 300 + 75 = -225 \, \text{m} \]`

Using the Pythagorean theorem, the magnitude of the total displacement
`(\(D\))` is calculated:

`\[ D = √(X^2 + Y^2) = √((229.9)^2 + (-225)^2) \approx 320 \, \text{m} \]`

Now, the direction (\(\theta\)) is given by the arctangent:

`\[ \theta = \arctan\left((Y)/(X)\right) = \arctan\left((-225)/(229.9)\right) \approx -44.2 \]`

The negative angle indicates a direction south of east, matching the correct answer: B) Magnitude: 320 m, Direction: 44.2° south of east.