Chapter 3, Problem 28P

(a)

To determine

Magnitude and direction of vector with components $A_x=-5.0\text{m}/\text{s}$ and $A_y=+8.0\text{m}/\text{s}$.

Answer to Problem 28P

Vector has magnitude $9.4\text{m}/\text{s}$ and directed at $122°$ counterclockwise from positive x-axis.

Explanation of Solution

Write the equation for the magnitude of vector.

$\left|A\right|=\sqrt{A_x^2+A_y^2}$

Here, the magnitude of vector A is $\left|A\right|$, x-component of vector is $A_x$, and the y-component of vector is $A_y$.

Write the equation to find the angle made by vector with the x-axis.

$\theta ={\tan }^{-1}\left(\frac{A_y}{A_x}\right)$

Here, the angle made by the vector with the x-axis is $\theta$.

Conclusion:

Substitute $-5.0\text{m}/\text{s}$ for $A_x$ and $+8.0\text{m}/\text{s}$ for $A_y$ in the equation for $\left|A\right|$.

$\begin{array}{c}\left|A\right|=\sqrt{{\left(-5.0\text{m}/\text{s}\right)}^2+{\left(+8.0\text{m}/\text{s}\right)}^2}\\ =\sqrt{89}\text{m}/\text{s}\\ =9.4\text{m}/\text{s}\end{array}$

Substitute $-5.0\text{m}/\text{s}$ for $A_x$ and $+8.0\text{m}/\text{s}$ for $A_y$ in the equation for $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{+8.0\text{m}/\text{s}}{-5.0\text{m}/\text{s}}\right)\\ ={\tan }^{-1}\left(-1.6\right)\\ =-58°\end{array}$

It is to be noted that $A_x$ is negative and $A_y$ is positive. This means that the vector lies in the second quadrant. Thus, the angle made by the vector with positive x direction in counterclockwise direction is as follows.

$\varphi =180°+\theta$

Here, the angle made by the vector with positive x direction in counterclockwise direction is $\varphi$.

Substitute $-58°$ for $\theta$ in the above equation to find $\varphi$.

$\begin{array}{c}\varphi =180°+\left(-58°\right)\\ =122°\end{array}$

Therefore, the vector has magnitude $9.4\text{m}/\text{s}$ and directed at $122°$ counterclockwise from positive x-axis.

(b)

To determine

Magnitude and direction of vector with components $B_x=+120\text{m}$ and $B_y=-60.0\text{m}$.

Answer to Problem 28P

Vector has magnitude $134\text{m}$ and directed at $333°$ counterclockwise from positive x-axis.

Explanation of Solution

Write the equation for the magnitude of vector.

$\left|B\right|=\sqrt{B_x^2+B_y^2}$

Here, the magnitude of vector is $\left|B\right|$, x-component of vector is $B_x$, and the y-component of vector is $B_y$.

Write the equation to find the angle made by vector with the x-axis.

$\theta ={\tan }^{-1}\left(\frac{B_y}{B_x}\right)$

Conclusion:

Substitute $-60.0\text{m}$ for $B_y$ and $+120\text{m}$ for $B_x$ in the equation for $\left|B\right|$.

$\begin{array}{c}\left|B\right|=\sqrt{{\left(+120\text{m}\right)}^2+{\left(-60.0\text{m}\right)}^2}\\ =\sqrt{18000}\text{m}\\ =134\text{m}\end{array}$

Substitute $+120\text{m}$ for $B_x$ and $-60.0\text{m}$ for $B_y$ in the equation for $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-60.0\text{m}}{120.0\text{m}}\right)\\ ={\tan }^{-1}\left(-0.5\right)\\ =-27°\end{array}$

It is to be noted that $B_x$ is positive and $B_y$ is negative. This means that the vector lies in the fourth quadrant. Thus, the angle made by the vector with positive x direction in counterclockwise direction is as follows.

$\varphi =180°+\theta$

Here, the angle made by the vector with positive x direction in counterclockwise direction is $\varphi$.

Substitute $-27°$ for $\theta$ in the above equation to find $\varphi$.

$\begin{array}{c}\varphi =360°+\left(-27°\right)\\ =333°\end{array}$

Therefore, the vector has magnitude $134\text{m}$ and directed at $333°$ counterclockwise from positive x-axis.

(c)

To determine

Magnitude and direction of vector with components $C_x=-13.7\text{m}/\text{s}$ and $C_y=-8.8\text{m}/\text{s}$.

Answer to Problem 28P

Vector has magnitude $134\text{m}$ and directed at $212.7°$ counterclockwise from positive x-axis.

Explanation of Solution

Write the equation for the magnitude of vector.

$\left|C\right|=\sqrt{C_x^2+C_y^2}$

Here, the magnitude of vector is $\left|C\right|$, x-component of vector is $C_x$, and the y-component of vector is $C_y$.

Write the equation to find the angle made by vector with the x-axis.

$\theta ={\tan }^{-1}\left(\frac{C_y}{C_x}\right)$

Conclusion:

Substitute $-13.7\text{m}/\text{s}$ for $C_x$ and $-8.8\text{m}/\text{s}$ for $C_y$ in the equation for $\left|C\right|$.

$\begin{array}{c}\left|C\right|=\sqrt{{\left(-13.7\text{m}/\text{s}\right)}^2+{\left(-8.8\text{m}/\text{s}\right)}^2}\\ =\sqrt{265.13}\text{m}/\text{s}\\ =16.3\text{m}/\text{s}\end{array}$

Substitute $-13.7\text{m}/\text{s}$ for $C_x$ and $-8.8\text{m}/\text{s}$ for $C_y$ in the equation for $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-8.8\text{m}/\text{s}}{-13.7\text{m}/\text{s}}\right)\\ ={\tan }^{-1}\left(0.642\right)\\ =32.7°\end{array}$

It is to be noted that both $C_x$ and $C_y$ are negative. This means that the vector lies in the third quadrant. Thus, the angle made by the vector with positive x direction in counterclockwise direction is as follows.

$\varphi =180°+\theta$

Substitute $32.7°$ for $\theta$ in the above equation to find $\varphi$.

$\begin{array}{c}\varphi =180°+32.7°\\ =212.7°\end{array}$

Therefore, the vector has magnitude $134\text{m}$ and directed at $212.7°$ counterclockwise from positive x-axis.

(d)

To determine

Magnitude and direction of vector with components $D_x=2.3\text{m}/{\text{s}}^2$ and $D_y=6.5\text{cm}/{\text{s}}^{\text{2}}$.

Answer to Problem 28P

Vector has magnitude $2.3\text{m}/{\text{s}}^{\text{2}}$ and directed at $1.6°$ counterclockwise from positive x-axis.

Explanation of Solution

Write the equation for the magnitude of vector.

$\left|D\right|=\sqrt{D_x^2+D_y^2}$

Here, the magnitude of vector is $\left|D\right|$, x-component of vector is $D_x$, and the y-component of vector is $D_y$.

Write the equation to find the angle made by vector with the x-axis.

$\theta ={\tan }^{-1}\left(\frac{D_y}{D_x}\right)$

Conclusion:

Substitute $2.3\text{m}/{\text{s}}^2$ for $D_x$ and $6.5\text{cm}/{\text{s}}^{\text{2}}$ for $D_y$ in the equation for $\left|D\right|$.

$\begin{array}{c}\left|D\right|=\sqrt{{\left(2.3\text{m}/{\text{s}}^2\right)}^2+{\left(6.5\text{cm}/{\text{s}}^{\text{2}}\left(\frac{{\text{10}}^{-\text{2}}\text{m}}{\text{1}\text{cm}}\right)\right)}^2}\\ =\sqrt{5.294}\text{m}/\text{s}\\ =2.3\text{m}/\text{s}\end{array}$

Substitute $2.3\text{m}/{\text{s}}^2$ for $D_x$ and $6.5\text{cm}/{\text{s}}^{\text{2}}$ for $D_y$ in the equation for $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{\left(6.5\text{cm}/{\text{s}}^{\text{2}}\right)\left(\frac{{\text{10}}^{-\text{2}}\text{m}}{\text{1}\text{cm}}\right)}{2.3\text{m}/{\text{s}}^2}\right)\\ ={\tan }^{-1}\left(0.028\right)\\ =1.6°\end{array}$

It is to be noted that both $D_x$ and $D_y$ are negative. This means that the vector lies in the FIRST quadrant.

Therefore, the vector has magnitude $2.3\text{m}/{\text{s}}^{\text{2}}$ and directed at $1.6°$ counterclockwise from positive x-axis.