Question
find the direction 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.
vector a is 17.4m long in a 303 degrees direction.
vector b is 25.6m long in a 22.0 degrees direction.
Answers
GPT-4o
To find the direction of the vector sum \(\vec{a} + \vec{b}\), we first need to break each vector into its components and then add those components together.
### Step 1: Break vectors into components
#### Vector \(\vec{a}\)
- Magnitude \( |\vec{a}| = 17.4 \) m
- Direction \( \theta_a = 303^\circ \)
Since 303 degrees is measured counterclockwise from the positive x-axis, it translates to an angle of \( 360^\circ - 303^\circ = 57^\circ \) below the positive x-axis (towards the fourth quadrant).
The components of \(\vec{a}\) are:
- \( a_x = 17.4 \cos(303^\circ) \)
- \( a_y = 17.4 \sin(303^\circ) \)
Using the angle in the fourth quadrant:
- \( a_x = 17.4 \cos(57^\circ) \)
- \( a_y = 17.4 \sin(57^\circ) \)
Given that \( \cos(303^\circ) = \cos(57^\circ) \) and \( \sin(303^\circ) = -\sin(57^\circ) \):
\[
a_x = 17.4 \cos(57^\circ)
\]
\[
a_y = 17.4 (-\sin(57^\circ))
\]
#### Vector \(\vec{b}\)
- Magnitude \( |\vec{b}| = 25.6 \) m
- Direction \( \theta_b = 22^\circ \)
The components of \(\vec{b}\) are:
- \( b_x = 25.6 \cos(22^\circ) \)
- \( b_y = 25.6 \sin(22^\circ) \)
### Step 2: Calculate the components
First, we need to find the trigonometric values.
For \( \theta = 57^\circ \):
\[
\cos(57^\circ) \approx 0.5446,\quad \sin(57^\circ) \approx 0.8387
\]
For \( \theta = 22^\circ \):
\[
\cos(22^\circ) \approx 0.9272,\quad \sin(22^\circ) \approx 0.3746
\]
Substituting these into the component equations:
\[
a_x = 17.4 \times 0.5446 \approx 9.48 \, \text{m}
\]
\[
a_y = 17.4 \times (-0.8387) \approx -14.59 \, \text{m}
\]
\[
b_x = 25.6 \times 0.9272 \approx 23.75 \, \text{m}
\]
\[
b_y = 25.6 \times 0.3746 \approx 9.59 \, \text{m}
\]
### Step 3: Sum the components
\[
\vec{R_x} = a_x + b_x = 9.48 + 23.75 \approx 33.23 \, \text{m}
\]
\[
\vec{R_y} = a_y + b_y = -14.59 + 9.59 \approx -5.00 \, \text{m}
\]
### Step 4: Find the direction of \(\vec{R}\)
To find the direction, we use the tangent function:
\[
\tan \theta_R = \frac{R_y}{R_x} = \frac{-5.00}{33.23} \approx -0.1505
\]
\[
\theta_R = \tan^{-1}(-0.1505) \approx -8.58^\circ
\]
This angle is negative, indicating it's below the x-axis. To express it in standard position (counterclockwise from the positive x-axis):
\[
\theta_\text{standard} = 360^\circ - 8.58^\circ \approx 351.42^\circ
\]
### Final Answer:
The direction of the vector sum \(\vec{a} + \vec{b}\) is approximately \(351.42^\circ\) from the positive x-axis.
### Step 1: Break vectors into components
#### Vector \(\vec{a}\)
- Magnitude \( |\vec{a}| = 17.4 \) m
- Direction \( \theta_a = 303^\circ \)
Since 303 degrees is measured counterclockwise from the positive x-axis, it translates to an angle of \( 360^\circ - 303^\circ = 57^\circ \) below the positive x-axis (towards the fourth quadrant).
The components of \(\vec{a}\) are:
- \( a_x = 17.4 \cos(303^\circ) \)
- \( a_y = 17.4 \sin(303^\circ) \)
Using the angle in the fourth quadrant:
- \( a_x = 17.4 \cos(57^\circ) \)
- \( a_y = 17.4 \sin(57^\circ) \)
Given that \( \cos(303^\circ) = \cos(57^\circ) \) and \( \sin(303^\circ) = -\sin(57^\circ) \):
\[
a_x = 17.4 \cos(57^\circ)
\]
\[
a_y = 17.4 (-\sin(57^\circ))
\]
#### Vector \(\vec{b}\)
- Magnitude \( |\vec{b}| = 25.6 \) m
- Direction \( \theta_b = 22^\circ \)
The components of \(\vec{b}\) are:
- \( b_x = 25.6 \cos(22^\circ) \)
- \( b_y = 25.6 \sin(22^\circ) \)
### Step 2: Calculate the components
First, we need to find the trigonometric values.
For \( \theta = 57^\circ \):
\[
\cos(57^\circ) \approx 0.5446,\quad \sin(57^\circ) \approx 0.8387
\]
For \( \theta = 22^\circ \):
\[
\cos(22^\circ) \approx 0.9272,\quad \sin(22^\circ) \approx 0.3746
\]
Substituting these into the component equations:
\[
a_x = 17.4 \times 0.5446 \approx 9.48 \, \text{m}
\]
\[
a_y = 17.4 \times (-0.8387) \approx -14.59 \, \text{m}
\]
\[
b_x = 25.6 \times 0.9272 \approx 23.75 \, \text{m}
\]
\[
b_y = 25.6 \times 0.3746 \approx 9.59 \, \text{m}
\]
### Step 3: Sum the components
\[
\vec{R_x} = a_x + b_x = 9.48 + 23.75 \approx 33.23 \, \text{m}
\]
\[
\vec{R_y} = a_y + b_y = -14.59 + 9.59 \approx -5.00 \, \text{m}
\]
### Step 4: Find the direction of \(\vec{R}\)
To find the direction, we use the tangent function:
\[
\tan \theta_R = \frac{R_y}{R_x} = \frac{-5.00}{33.23} \approx -0.1505
\]
\[
\theta_R = \tan^{-1}(-0.1505) \approx -8.58^\circ
\]
This angle is negative, indicating it's below the x-axis. To express it in standard position (counterclockwise from the positive x-axis):
\[
\theta_\text{standard} = 360^\circ - 8.58^\circ \approx 351.42^\circ
\]
### Final Answer:
The direction of the vector sum \(\vec{a} + \vec{b}\) is approximately \(351.42^\circ\) from the positive x-axis.