A man pushing a mop across a floor causes the map to undergo two displacements. The first has a magnitude of 200 cm and makes an angle of 120° with the positive x axis. The resultant displacement has a magnitude of 240 cm and is directed at an angle of 60.0° to the positive x axis. Find the magnitude and direction of the second displacement.

To find the magnitude and direction of the second displacement, we can use vector addition.

Let's represent the first displacement as vector D1 and the second displacement as vector D2.

Given information:
Magnitude of D1 = 200 cm
Angle of D1 with positive x-axis = 120°
Magnitude of resultant displacement (D1+D2) = 240 cm
Angle of resultant displacement with positive x-axis = 60.0°

Step 1: Finding the components of D1
To find the components of D1, we can use trigonometry.
The x-component of D1 (D1x) = magnitude of D1 * cos(angle of D1) = 200 cm * cos(120°) = -100 cm (negative because it points in the opposite direction of the positive x-axis).
The y-component of D1 (D1y) = magnitude of D1 * sin(angle of D1) = 200 cm * sin(120°) ≈ 173.2 cm.

Step 2: Finding the components of the resultant displacement (D1+D2)
Since the resultant displacement is the sum of D1 and D2, we can write the following equations:
D1x + D2x = magnitude of resultant displacement * cos(angle of resultant displacement)
D1y + D2y = magnitude of resultant displacement * sin(angle of resultant displacement)

Substituting the given values, we get:
-100 cm + D2x = 240 cm * cos(60°)
173.2 cm + D2y = 240 cm * sin(60°)

Step 3: Solving for D2x and D2y
Solving the above equations will give us the components of D2.
D2x = 240 cm * cos(60°) + 100 cm ≈ 190.5 cm
D2y = 240 cm * sin(60°) - 173.2 cm ≈ 48.0 cm

Step 4: Finding the magnitude and direction of D2
Using the components of D2, we can find the magnitude and direction.
Magnitude of D2 = √(D2x^2 + D2y^2) = √((190.5 cm)^2 + (48.0 cm)^2) ≈ 198.1 cm
Direction of D2 = tan^(-1)(D2y/D2x) = tan^(-1)(48.0 cm / 190.5 cm) ≈ 14.0°

So, the magnitude of the second displacement is approximately 198.1 cm, and the direction is approximately 14.0° to the positive x-axis.

To solve this problem, we can use vector addition to find the second displacement.

Let's break down the given information into components:

First displacement (D1):
- Magnitude: 200 cm
- Angle: 120° with the positive x-axis

Resultant displacement (R):
- Magnitude: 240 cm
- Angle: 60° with the positive x-axis

We need to find the magnitude and direction of the second displacement (D2).

1. Split the D1 vector into its x and y components:
- D1x = D1 * cos(120°) (horizontal component)
- D1y = D1 * sin(120°) (vertical component)

Plugging the values:
- D1x = 200 cm * cos(120°)
- D1y = 200 cm * sin(120°)

Calculate both components:
- D1x ≈ -100 cm
- D1y ≈ 173.2 cm

2. Split the R vector into its x and y components:
- Rx = R * cos(60°)
- Ry = R * sin(60°)

Plugging the values:
- Rx = 240 cm * cos(60°)
- Ry = 240 cm * sin(60°)

Calculate both components:
- Rx = 120 cm
- Ry = 207.85 cm

3. To find the second displacement, D2, we subtract the components of D1 from the components of R:
- D2x = Rx - D1x
- D2y = Ry - D1y

Plugging the values:
- D2x = 120 cm - (-100 cm)
- D2y = 207.85 cm - 173.2 cm

Calculate both components:
- D2x = 220 cm
- D2y = 34.65 cm

4. To find the magnitude and direction of D2, use the components to calculate the hypotenuse and angle:
- Magnitude of D2 = √(D2x^2 + D2y^2)
- Direction of D2 = tan^(-1)(D2y / D2x)

Plugging the values:
- Magnitude of D2 = √(220 cm^2 + 34.65 cm^2)
- Direction of D2 = tan^(-1)(34.65 cm / 220 cm)

Calculate the values:
- Magnitude of D2 ≈ 224.55 cm
- Direction of D2 ≈ 8.99° (rounded to two decimal places)

Therefore, the magnitude of the second displacement (D2) is approximately 224.55 cm, and its direction is approximately 8.99° from the positive x-axis.