Question

homework (1): vectors & relativity

1. which quadrant are each of the following vectors in:

$vec{a}=5m$ @ 30 degrees
$vec{b}=8m/s$ @ 170 degrees
$vec{c}=20m/s^{2}$ @ southwest
$vec{d}=4m$ @ 15 degrees north of west
$vec{e}=7m$ @ 40 west of south
$vec{f}=23km$ @ - 120 degrees
problems 2 - 3

1. what is the difference of vector a and vector b? $vec{a}-vec{b}$

(a) a (b) b (c) c (d) d (e) e

1. to satisfy this equation: $vec{a}+vec{b}+vec{c}=0$ vector c would represent:

(a) a (b) b (c) c (d) d (e) e

1. find the resultant vector of each sum graphically and algebraically

$vec{a}+vec{b}-vec{c}=$ $vec{a}=5m/s n$, $vec{b}=7m/s$ @ - 25 degrees, $vec{c}=3m/s n$ of $w$ @ 45 degrees

Explanation:

Step1: Determine quadrants for vectors in 1)

• For $\vec{A}=5m@30$ degrees, angles between $0$ and $90$ degrees are in the first - quadrant. So $\vec{A}$ is in the first quadrant.
• For $\vec{B}=8m/s@170$ degrees, angles between $90$ and $180$ degrees are in the second - quadrant. So $\vec{B}$ is in the second quadrant.
• For $\vec{C}=20m/s^{2}@$ Southwest, Southwest corresponds to an angle between $180$ and $270$ degrees, so $\vec{C}$ is in the third quadrant.
• For $\vec{D}=4m@15$ degrees North of West, this is an angle between $90$ and $180$ degrees, so $\vec{D}$ is in the second quadrant.
• For $\vec{E}=7m@40$ West of South, this is an angle between $180$ and $270$ degrees, so $\vec{E}$ is in the third quadrant.
• For $\vec{F}=23km@ - 120$ degrees, $-120$ degrees is equivalent to $240$ degrees ($360 - 120$), and angles between $180$ and $270$ degrees are in the third quadrant. So $\vec{F}$ is in the third quadrant.

Step2: Solve 2)

To find $\vec{A}-\vec{B}=\vec{A}+(-\vec{B})$. Geometrically, when we subtract $\vec{B}$ from $\vec{A}$, we reverse the direction of $\vec{B}$ and then add it to $\vec{A}$. Without specific coordinate - based values, we can use vector addition rules. If we consider the directions of $\vec{A}$ and $\vec{B}$ in the given figure, $\vec{A}-\vec{B}$ will be a vector in the fourth quadrant. Among the given options, assume a proper vector subtraction visualization, and we find the answer by comparing the direction of the resultant vector with the options.

Step3: Solve 3)

Given $\vec{A}+\vec{B}+\vec{C}=0$, then $\vec{C}=-(\vec{A}+\vec{B})$. Geometrically, we first add $\vec{A}$ and $\vec{B}$ using the parallelogram or triangle law of vector addition, and then reverse the direction of the resultant of $\vec{A}+\vec{B}$ to get $\vec{C}$. By comparing with the given vectors in the figure, we can determine the correct option.

Step4: Solve 4)

Algebraic method:

1. Resolve the vectors into components:

• For $\vec{A}=5m/s\ N$, in a two - dimensional coordinate system (assuming $x$ - axis as East - West and $y$ - axis as North - South), $A_x = 0$ and $A_y=5$.
• For $\vec{B}=7m/s@ - 25$ degrees, $B_x=7\cos(-25^{\circ})\approx7\times0.9063 = 6.3441$ and $B_y = 7\sin(-25^{\circ})\approx7\times(- 0.4226)=-2.9582$.
• For $\vec{C}=3m/s\ N$ of $W@45$ degrees, $C_x=-3\cos45^{\circ}\approx - 2.1213$ and $C_y = 3\sin45^{\circ}\approx2.1213$.

1. Calculate the $x$ and $y$ components of the resultant vector $\vec{R}=\vec{A}+\vec{B}-\vec{C}$:

• $R_x=A_x + B_x-C_x=0 + 6.3441-(-2.1213)=8.4654$.
• $R_y=A_y + B_y - C_y=5-2.9582 - 2.1213= - 0.0795$.

1. Find the magnitude of $\vec{R}$ using $R=\sqrt{R_x^{2}+R_y^{2}}=\sqrt{(8.4654)^{2}+(-0.0795)^{2}}\approx8.4654$.
2. Find the direction $\theta$ using $\tan\theta=\frac{R_y}{R_x}=\frac{-0.0795}{8.4654}\approx - 0.0094$, so $\theta\approx - 0.54$ degrees.

Graphical method:

1. Draw $\vec{A}$ as a vector pointing North with a length of 5 units.
2. Draw $\vec{B}$ with an angle of $-25$ degrees and length of 7 units.
3. Reverse the direction of $\vec{C}$ (since we are subtracting it) and draw it.
4. Use the head - to - tail method to add $\vec{A}$, $\vec{B}$ and the reversed $\vec{C}$. Measure the magnitude and direction of the resultant vector.

Answer:

1. $\vec{A}$: First quadrant; $\vec{B}$: Second quadrant; $\vec{C}$: Third quadrant; $\vec{D}$: Second quadrant; $\vec{E}$: Third quadrant; $\vec{F}$: Third quadrant.
2. (Answer depends on vector visualization in the figure, assume correct option based on vector subtraction rules).
3. (Answer depends on vector visualization in the figure, assume correct option based on $\vec{C}=-(\vec{A}+\vec{B})$).
4. Algebraic: Magnitude $\approx8.4654m/s$, Direction $\approx - 0.54$ degrees. Graphical: Follow the steps of vector addition and subtraction as described above to measure the magnitude and direction.