Two beetles run across flat sand, starting at the same point, Beetle one runs 0.50mdue east, then 0.80mat $\mathbf{30}\mathbf{°}$north of due east. Beetle two also makes two runs; the first is 1.6mat $\mathbf{40}\mathbf{°}$east due to north. What must be (a) the magnitude and (b) direction of its second run if it is to end up at the new location of beetle 1?

The given vectors can be written as below:

$\overrightarrow{A}=0.50\mathrm{m}$due east

$\overrightarrow{B}=0.80\mathrm{m}$at $30°$north of due east

$\overrightarrow{c}=1.6\mathrm{m}$at$40°$ east of due north

The magnitude and direction of beetle 1 can be determined by vector algebra.

Formulae are as follows:

The unit vector form

$\overrightarrow{A}=x\stackrel{⏜}{i}+y\stackrel{⏜}{j}$ (i)

The formula for magnitude:

$\left|\overrightarrow{A}\right|=\sqrt{x^2+y^2}$ (ii)

The formula for angle:

$\theta ={\tan }^{-1}\frac{y}{x}$ (iii)

Where,

x,y,$\overrightarrow{A}$Are vectors and$\theta$is the angle between x and y .

Write the given vectors in the unit vector form.

For,$\overrightarrow{A}$

$\overrightarrow{A}=0.50\stackrel{⏜}{i}$

For $\overrightarrow{B}$,

$$\begin{gathered} \overrightarrow{B}=\left(0.80\times \cos 30°\right)\stackrel{⏜}{i}+\left(0.80\times \sin 30°\right)\stackrel{⏜}{j} \\ =0.69\stackrel{⏜}{i}+0.4\stackrel{⏜}{j} \end{gathered}$$

Adding both the vectors.

$$\begin{gathered} \overrightarrow{A}+\overrightarrow{B}=0.50\stackrel{⏜}{i}+\left(0.69\stackrel{⏜}{i}+0.4\stackrel{⏜}{j}\right) \\ =1.19\stackrel{⏜}{i}+0.40\stackrel{⏜}{j} \end{gathered}$$

Writing $\overrightarrow{C}$in unit vector form.

$$\begin{gathered} \overrightarrow{C}=\left(1.6\times \cos 50°\right)\stackrel{⏜}{i}+\left(1.6\times \sin 50°\right)\stackrel{⏜}{j} \\ =1.03\stackrel{⏜}{i}+1.23\stackrel{⏜}{j} \end{gathered}$$

Subtracting vector$\overrightarrow{C}$ from$\left(\overrightarrow{A}+\overrightarrow{B}\right)$ we get,

$\left(\overrightarrow{A}+\overrightarrow{B}\right)=\overrightarrow{C}+0.16\stackrel{⏜}{i}-0.83\stackrel{⏜}{j}$

Now, use equation (ii) to find the magnitude.Substituting the values,

$$\begin{gathered} \left|\left(\left(\overrightarrow{A}+\overrightarrow{B}\right)-\overrightarrow{c}\right)\right|=\sqrt{0.{16}^2+{\left(0.83\right)}^2} \\ =0.84m \end{gathered}$$

Therefore, the magnitude of second run of beetle two with respect of final position of beetle one is 0.84m.

Use equation (iii) to determine the angle.

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{-0.83}{0.16}\right) \\ ={\tan }^{-1}\left(-4\right) \\ =-79°\left(\mathrm{south}\mathrm{of}\mathrm{east}\right) \end{gathered}$$

Therefore, the magnitude and direction of second run of beetle two with respect of final position of beetle one is $-79°$(south of east).