Chapter 3, Problem 39QAP

To determine

(a)

The magnitude and direction of the resultant velocity vector $\overrightarrow{A}+\overrightarrow{B}$

Answer to Problem 39QAP

The resultant velocity vector $\overrightarrow{A}+\overrightarrow{B}$ has a magnitude 64.8 m/s and is directed 70.9^oN of E.

Explanation of Solution

Given:

The magnitude and the directions of the velocity vectors $\overrightarrow{A}\text{ and }\overrightarrow{B}$.

  $\begin{array}{c}A=30\text{ m/s at 45}°\text{ North of East}\\ B=40\text{ m/s along North}\end{array}$

Formula used:

If $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$, then the x and y components of the resultant velocity vector $\overrightarrow{C}$ are given by,

  $\begin{array}{c}{C}_x=A_x+B_x\\ C_y=A_y+B_y\end{array}$

The magnitude of the vector $\overrightarrow{C}$ is given by,

  $C=\sqrt{C_x^2+C_y^2}$

The angle the vector makes with the x axis is given by,

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

Calculation:

Assume the +x axis to be directed along the East and the +yaxis along the North.

Draw a vector diagram for the velocities.

The vector $\overrightarrow{A}$ makes an angle 45^o with the x axis (East). Its components along the x axis (East) and the y axis (North) are calculated as follows:

  $\begin{array}{c}{A}_x=A\cos 45°\\ =\left(30\text{ m/s}\right)\left(\cos 45°\right)\\ =21.2\text{ m/s}\\ A_y=A\sin 45°\\ =\left(30\text{ m/s}\right)\left(\sin 45°\right)\\ =21.2\text{ m/s}\end{array}$

The velocity vector $\overrightarrow{B}$ is directed along the North ( y axis), hence it makes an angle 90^o with the x axis (East). Its components along the x axis (East) and the y axis (North) are calculated as follows:

  $\begin{array}{c}{B}_x=B\cos 90°\\ =\left(40\text{ m/s}\right)\left(0\right)\\ =0\\ B_y=B\sin 90°\\ =\left(40\text{ m/s}\right)\left(1\right)\\ =40\text{ m/s}\end{array}$

Calculate the x and y components of the vector $\overrightarrow{C}$.

  $\begin{array}{c}{C}_x=A_x+B_x\\ =21.2\text{ m/s}+0\\ =21.2\text{ m/s}\\ C_y=A_y+B_y\\ =21.2\text{ m/s}+40\text{ m/s}\\ =61.2\text{ m/s}\end{array}$

Calculate the magnitude of the resultant velocity vector $\overrightarrow{C}$.

  $\begin{array}{c}C=\sqrt{C_x^2+C_y^2}\\ =\sqrt{{\left(21.2\text{ m/s}\right)}^2+{\left(61.2\text{ m/s}\right)}^2}\\ =64.8\text{ m/s}\end{array}$

Calculate the angle made by the vector with the +x axis (East).

  $\begin{array}{c}\alpha ={\tan }^{-1}\left(\frac{C_y}{C_x}\right)\\ ={\tan }^{-1}\left(\frac{61.2\text{ m/s}}{21.2\text{ m/s}}\right)\\ =70.9°\end{array}$

Conclusion:

The resultant velocity vector $\overrightarrow{A}+\overrightarrow{B}$ has a magnitude 64.8 m/s and is directed 70.9^oN of E.

To determine

(b)

The magnitude and direction of the resultant velocity vector $\overrightarrow{A}-\overrightarrow{B}$

Answer to Problem 39QAP

The resultant velocity vector $\overrightarrow{A}-\overrightarrow{B}$ has a magnitude 28.3 m/s and is directed at an angle 41.6^oSouth of East.

Explanation of Solution

Given:

From part (a), the components of the vectors $\overrightarrow{A}\text{ and }\overrightarrow{B}$ along the x and y directions.

  $\begin{array}{c}{A}_x=21.2\text{ m/s}\\ A_y=21.2\text{ m/s}\\ B_x=0\\ B_y=40\text{ m/s}\end{array}$

Formula used:

If $\overrightarrow{D}=\overrightarrow{A}-\overrightarrow{B}$, then the x and y components of the vector $\overrightarrow{D}$ are given by,

  $D_x=A_x-B_x$

  $D_y=A_y-B_y$

The magnitude of the vector $\overrightarrow{D}$ is given by,

  $D=\sqrt{D_x^2+D_y^2}$

The angle made by the vector with the x axis is given by,

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

Calculation:

Calculate the x and y components of the vector $\overrightarrow{D}$.

  $\begin{array}{c}{D}_x=A_x-B_x\\ =21.2\text{ m/s}-0\\ =21.2\text{ m/s}\\ D_y=A_y-B_y\\ =21.2\text{ m/s}-40\text{ m/s}\\ =-18.8\text{ m/s}\end{array}$

Calculate the magnitude of the vector $\overrightarrow{D}$.

  $\begin{array}{c}D=\sqrt{D_x^2+D_y^2}\\ =\sqrt{{\left(21.2\text{ m/s}\right)}^2+{\left(-18.8\text{ m/s}\right)}^2}\\ =28.3\text{ m/s}\end{array}$

Calculate the angle made by the vector $\overrightarrow{D}$ with East.

  $\begin{array}{c}\beta ={\tan }^{-1}\left(\frac{D_y}{D_x}\right)\\ ={\tan }^{-1}\left(\frac{-18.8\text{ m/s}}{21.2\text{ m/s}}\right)\\ =-41.6°\end{array}$

Since the y component of $\overrightarrow{D}$ is negative, it is directed along South. Therefore, the vector is directed at an angle 41.6^o South of East.

Conclusion:

The resultant velocity vector $\overrightarrow{A}-\overrightarrow{B}$ has a magnitude 28.3 m/s and is directed at an angle 41.6^oSouth of East.

To determine

(c)

The magnitude and the direction of the resultant velocity vector $2\overrightarrow{A}+\overrightarrow{B}$

Answer to Problem 39QAP

The resultant velocity vector $2\overrightarrow{A}+\overrightarrow{B}$ has a magnitude 92.7 m/s and is directed at an angle 62.8^oNorth of East.

Explanation of Solution

Given:

From part (a), the components of the vectors $\overrightarrow{A}\text{ and }\overrightarrow{B}$ along the x and y directions.

  $\begin{array}{c}{A}_x=21.2\text{ m/s}\\ A_y=21.2\text{ m/s}\\ B_x=0\\ B_y=40\text{ m/s}\end{array}$

Formula used:

If $\overrightarrow{E}=2\overrightarrow{A}+\overrightarrow{B}$, then the x and y components of the vector $\overrightarrow{E}$ are given by,

  $E_x=2A_x+B_x$

  $E_y=2A_y+B_y$

The magnitude of the vector $\overrightarrow{E}$ is given by,

  $E=\sqrt{E_x^2+E_y^2}$

The angle made by the vector with the x axis is given by,

  $\gamma ={\tan }^{-1}\left(\frac{E_y}{E_x}\right)$

Calculation:

Calculate the x and y components of the vector $\overrightarrow{E}$.

  $\begin{array}{c}{E}_x=2A_x+B_x\\ =2\times 21.2\text{ m/s}-0\\ =42.4\text{ m/s}\\ E_y=2A_y+B_y\\ =2\times 21.2\text{ m/s}+40\text{ m/s}\\ =82.4\text{ m/s}\end{array}$

Calculate the magnitude of the vector $\overrightarrow{E}$.

  $\begin{array}{c}E=\sqrt{E_x^2+E_y^2}\\ =\sqrt{{\left(42.4\text{ m/s}\right)}^2+{\left(82.4\text{ m/s}\right)}^2}\\ =92.7\text{ m/s}\end{array}$

Calculate the angle made by the vector $\overrightarrow{D}$ with East.

  $\begin{array}{c}\gamma ={\tan }^{-1}\left(\frac{E_y}{E_x}\right)\\ ={\tan }^{-1}\left(\frac{82.4\text{ m/s}}{42.4\text{ m/s}}\right)\\ =62.8°\end{array}$

Conclusion:

The resultant velocity vector $2\overrightarrow{A}+\overrightarrow{B}$ has a magnitude 92.7 m/s and is directed at an angle 62.8^oNorth of East.