Vector Addition Physics Worksheet - Free Printable

Problem Analysis:

The problem involves vector addition to determine the resultant displacement when two displacements are given:
1. A displacement of 15 m East.
2. A displacement of 25 m at 30° North of East.

We need to:
- Sketch the vector diagram.
- Break each vector into its perpendicular components (East and North).
- Add the components to find the resultant displacement.
- Determine the magnitude and direction of the resultant vector.

---

Step-by-Step Solution:

#### Part a: Sketching the Vector Diagram
1. Draw a horizontal line representing the first displacement of 15 m East.
2. From the endpoint of the first vector, draw a second vector of 25 m at an angle of 30° North of East.
3. The resultant displacement will be the vector from the starting point to the endpoint of the second vector.

#### Part b: Calculating Perpendicular Components
We break each vector into its East (x) and North (y) components.

##### Vector 1: 15 m East
- This vector is entirely in the East direction.
- \( x_1 = 15 \, \text{m} \)
- \( y_1 = 0 \, \text{m} \)

##### Vector 2: 25 m at 30° North of East
- Use trigonometry to find the components:
- The East component (\( x_2 \)) is given by:
\[
x_2 = 25 \cos(30^\circ)
\]
Using \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[
x_2 = 25 \times \frac{\sqrt{3}}{2} = 25 \times 0.866 = 21.65 \, \text{m}
\]
- The North component (\( y_2 \)) is given by:
\[
y_2 = 25 \sin(30^\circ)
\]
Using \( \sin(30^\circ) = \frac{1}{2} \):
\[
y_2 = 25 \times \frac{1}{2} = 25 \times 0.5 = 12.5 \, \text{m}
\]

So, the components are:
- \( x_2 = 21.65 \, \text{m} \)
- \( y_2 = 12.5 \, \text{m} \)

#### Part c: Calculating the Resultant Displacement
To find the resultant displacement, we add the components of the two vectors:
- Total East component (\( R_x \)):
\[
R_x = x_1 + x_2 = 15 + 21.65 = 36.65 \, \text{m}
\]
- Total North component (\( R_y \)):
\[
R_y = y_1 + y_2 = 0 + 12.5 = 12.5 \, \text{m}
\]

##### Magnitude of the Resultant Vector
The magnitude \( R \) is given by the Pythagorean theorem:
\[
R = \sqrt{R_x^2 + R_y^2}
\]
Substitute the values:
\[
R = \sqrt{(36.65)^2 + (12.5)^2}
\]
\[
R = \sqrt{1343.4225 + 156.25}
\]
\[
R = \sqrt{1509.6725}
\]
\[
R \approx 38.85 \, \text{m}
\]

##### Direction of the Resultant Vector
The direction \( \theta \) (angle North of East) is given by:
\[
\tan(\theta) = \frac{R_y}{R_x}
\]
Substitute the values:
\[
\tan(\theta) = \frac{12.5}{36.65}
\]
\[
\tan(\theta) \approx 0.341
\]
\[
\theta = \tan^{-1}(0.341)
\]
\[
\theta \approx 18.8^\circ
\]

---

Final Answer:

The resultant displacement is:
\[
\boxed{38.85 \, \text{m}, \, 18.8^\circ \, \text{North of East}}
\]