Question

find the magnituide of the vector sum -->a+-->b
vector a is 17.4m long in a 303 degrees direction.
vector b is 25.6m long in a 22.0 degrees direction.

Answer 1

To find the magnitude of the vector sum \(\vec{a} + \vec{b}\), we'll break down each vector into its components, sum those components, and then find the magnitude of the resulting vector.

1. **Convert each vector into components:**
- Vector \(\vec{a}\):
- Magnitude: \(17.4 \, \text{m}\)
- Direction: \(303^\circ\)
- Components:
\[
a_x = 17.4 \cos(303^\circ)
\]
\[
a_y = 17.4 \sin(303^\circ)
\]
- Since \(303^\circ\) is in the fourth quadrant, \(\cos(303^\circ)\) is positive and \(\sin(303^\circ)\) is negative.
- Vector \(\vec{b}\):
- Magnitude: \(25.6 \, \text{m}\)
- Direction: \(22.0^\circ\)
- Components:
\[
b_x = 25.6 \cos(22.0^\circ)
\]
\[
b_y = 25.6 \sin(22.0^\circ)
\]

2. **Calculate the components:**

For \(\vec{a}\):
\[
a_x = 17.4 \cos(303^\circ) = 17.4 \cos(360^\circ - 57^\circ) = 17.4 \cos(-57^\circ) = 17.4 \cos(57^\circ) \approx 17.4 \times 0.5446 \approx 9.48 \, \text{m}
\]
\[
a_y = 17.4 \sin(303^\circ) = 17.4 \sin(360^\circ - 57^\circ) = 17.4 \sin(-57^\circ) = 17.4 (-\sin(57^\circ)) \approx 17.4 \times (-0.8387) \approx -14.59 \, \text{m}
\]

For \(\vec{b}\):
\[
b_x = 25.6 \cos(22.0^\circ) \approx 25.6 \times 0.9272 \approx 23.73 \, \text{m}
\]
\[
b_y = 25.6 \sin(22.0^\circ) \approx 25.6 \times 0.3746 \approx 9.60 \, \text{m}
\]

3. **Sum the components:**
\[
\vec{R}_x = a_x + b_x \approx 9.48 + 23.73 \approx 33.21 \, \text{m}
\]
\[
\vec{R}_y = a_y + b_y \approx -14.59 + 9.60 \approx -4.99 \, \text{m}
\]

4. **Calculate the magnitude of the resultant vector:**
\[
|\vec{R}| = \sqrt{(\vec{R}_x)^2 + (\vec{R}_y)^2} \approx \sqrt{(33.21)^2 + (-4.99)^2} \approx \sqrt{1102.94 + 24.90} \approx \sqrt{1127.84} \approx 33.58 \, \text{m}
\]

So, the magnitude of the vector sum \(\vec{a} + \vec{b}\) is approximately \(33.58 \, \text{m}\).