91Ó°ÊÓ

The given vectors can be written as below:

$\stackrel{→}{A}=0.50\mathrm{m}$due east

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

$\stackrel{→}{c}=1.6\mathrm{m}$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

$\stackrel{→}{A}=x\stackrel{Áåœ}{i}+y\stackrel{Áåœ}{j}$ (i)

The formula for magnitude:

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

The formula for angle:

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

Where,

x,y,$\stackrel{→}{A}$Are vectors and$\mathrm{θ}$is the angle between x and y .

Write the given vectors in the unit vector form.

For,$\stackrel{→}{A}$

$\stackrel{→}{A}=0.50\stackrel{Áåœ}{i}$

For $\stackrel{→}{B}$,

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

Adding both the vectors.

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

Writing $\stackrel{→}{C}$in unit vector form.

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

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

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

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

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

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.

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

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