Vectors Addition Worksheet PDF Form - FormsPal - Free Printable

Problem Analysis:

The task involves breaking down two vectors into their respective x- and y-components, then summing these components to find the total displacement in the x- and y-directions. The vectors are given as follows:

1. A vector of magnitude 50 m at an angle of 135°.
2. A vector of magnitude 30 m at an angle of 90°.

We need to:
- Determine the x- and y-components of each vector.
- Sum the components to find the total displacement in the x- and y-directions.

Step-by-Step Solution:

#### Vector 1:
- Magnitude: \( 50 \, \text{m} \)
- Angle: \( 135^\circ \)

The x- and y-components of a vector can be calculated using trigonometry:
\[
x_1 = r \cos(\theta)
\]
\[
y_1 = r \sin(\theta)
\]

For Vector 1:
\[
x_1 = 50 \cos(135^\circ)
\]
\[
y_1 = 50 \sin(135^\circ)
\]

Using the values of trigonometric functions:
\[
\cos(135^\circ) = -\frac{\sqrt{2}}{2}
\]
\[
\sin(135^\circ) = \frac{\sqrt{2}}{2}
\]

Substitute these values:
\[
x_1 = 50 \left(-\frac{\sqrt{2}}{2}\right) = -25\sqrt{2} \approx -35.36 \, \text{m}
\]
\[
y_1 = 50 \left(\frac{\sqrt{2}}{2}\right) = 25\sqrt{2} \approx 35.36 \, \text{m}
\]

So, for Vector 1:
\[
x_1 = -25\sqrt{2} \, \text{m}, \quad y_1 = 25\sqrt{2} \, \text{m}
\]

#### Vector 2:
- Magnitude: \( 30 \, \text{m} \)
- Angle: \( 90^\circ \)

Using the same formulas:
\[
x_2 = r \cos(\theta)
\]
\[
y_2 = r \sin(\theta)
\]

For Vector 2:
\[
x_2 = 30 \cos(90^\circ)
\]
\[
y_2 = 30 \sin(90^\circ)
\]

Using the values of trigonometric functions:
\[
\cos(90^\circ) = 0
\]
\[
\sin(90^\circ) = 1
\]

Substitute these values:
\[
x_2 = 30 \cdot 0 = 0 \, \text{m}
\]
\[
y_2 = 30 \cdot 1 = 30 \, \text{m}
\]

So, for Vector 2:
\[
x_2 = 0 \, \text{m}, \quad y_2 = 30 \, \text{m}
\]

#### Total Displacement:
To find the total displacement in the x- and y-directions, we sum the components:
\[
x_{\text{tot}} = x_1 + x_2
\]
\[
y_{\text{tot}} = y_1 + y_2
\]

Substitute the values:
\[
x_{\text{tot}} = -25\sqrt{2} + 0 = -25\sqrt{2} \approx -35.36 \, \text{m}
\]
\[
y_{\text{tot}} = 25\sqrt{2} + 30 \approx 35.36 + 30 = 65.36 \, \text{m}
\]

Final Answer:

\[
\boxed{
\begin{aligned}
x_1 &= -25\sqrt{2} \, \text{m}, & x_2 &= 0 \, \text{m}, & x_{\text{tot}} &= -25\sqrt{2} \, \text{m} \\
y_1 &= 25\sqrt{2} \, \text{m}, & y_2 &= 30 \, \text{m}, & y_{\text{tot}} &= 25\sqrt{2} + 30 \, \text{m}
\end{aligned}
}
\]

In approximate numerical form:
\[
\boxed{
\begin{aligned}
x_1 &\approx -35.36 \, \text{m}, & x_2 &= 0 \, \text{m}, & x_{\text{tot}} &\approx -35.36 \, \text{m} \\
y_1 &\approx 35.36 \, \text{m}, & y_2 &= 30 \, \text{m}, & y_{\text{tot}} &\approx 65.36 \, \text{m}
\end{aligned}
}
\]