91Ó°ÊÓ

The length of each branch is x=2.0 cm

Here, to find magnitude and displacement, use the vector algebra method and Pythagoras theorem to find the direction of resultant displacement.

Formulae are as follows:

The formula for magnitude is,

$d=\sqrt{x^2+y^2}$ (i)

The formula for the angle is,

$\tan \mathrm{θ}=\frac{y}{x}$ (ii)

Where,

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

First entry at point A:

As, ants have to travel a minimum distance,so, they have to travel path h-i-v-w. So, while calculating resultant, consider only displacement along this path only.

As,w is inclined, so, horizontal component of w is as follow:

$$\begin{gathered} w_x=x×\cos 60 \\ =1\mathrm{cm} \end{gathered}$$

Vertical component of w is as follow:

$$\begin{gathered} w_y=x×\sin 60 \\ =1.732\mathrm{cm} \end{gathered}$$

Iis also inclined, horizontal component of I is as follow:

$$\begin{gathered} i_x=2.0\cos 120 \\ =-1\mathrm{cm} \end{gathered}$$

Vertical component of I is as follow:

$$\begin{gathered} i_x=2.0×\sin 120 \\ =1.732cm \end{gathered}$$

Now,x component of resultant is given below,

$$\begin{gathered} d_x=w_x+i_x \\ =1+1 \\ =0cm \end{gathered}$$

Now,y component of resultant is given below,

$$\begin{gathered} d_x=w_y+i_y+v+h \\ =1.732+1.732+2+2 \\ =7.463cm \end{gathered}$$

Use equation (i) to find the resultant displacement.

$$\begin{gathered} d=d_x\stackrel{Áåœ}{i}+d_y\stackrel{Áåœ}{j} \\ =7.464\stackrel{Áåœ}{j} \end{gathered}$$

So, magnitude is,

$$\begin{gathered} d=\sqrt{7.{464}^2} \\ =7.5cm \end{gathered}$$

Hence, the magnitude of displacement is 7.5 cm.

Use equation (ii) to find the angle.

Direction is,

$$\begin{gathered} \tan \mathrm{θ}=\frac{7.464}{0} \\ \mathrm{θ}=90° \end{gathered}$$

Hence, the angle of displacement is $90°$.

If entry is at B, then minimum path is o-p-j-v-w.

Here, p is inclined.So, horizontal and vertical vectors are as follow respectively:

$$\begin{gathered} p_x=2×\cos 30 \\ =1.732cm \\ {p}_y=2×\sin 30 \\ =1 \end{gathered}$$

Also,j is inclined.So, horizontal and vertical vectors are as follow respectively:

$$\begin{gathered} j_x=2×\cos 60 \\ =1cm \\ {j}_y=2×\sin 60 \\ =1.732cm \end{gathered}$$

Here,w is inclined.So, horizontal and vertical vectors are as follow respectively:

$$\begin{gathered} w_x=2×\cos 60 \\ =1cm \\ {w}_y=2×\sin 60 \\ =1.732cm \end{gathered}$$

Now, x components of resultant as follow:

$$\begin{gathered} d_x=p_x+j_x+w_x+0 \\ =1.732+1+1+2 \\ =5.732\mathrm{cm} \end{gathered}$$

Now,y components of resultant as follow:

$$\begin{gathered} d_x=p_x+j_x+w_x+0 \\ =1+1.732+1.732+2 \\ =6.464\mathrm{cm} \end{gathered}$$

So, resultant of displacement at entry B is as follow:

$$\begin{gathered} d=d_x+d_y \\ =5.732\stackrel{Áåœ}{i}+6.464\stackrel{Áåœ}{j} \end{gathered}$$

Use equation (i) to find the magnitude.

$$\begin{gathered} d=\sqrt{5.{732}^2+6.{464}^2} \\ =8.6cm \end{gathered}$$

Hence,the magnitude of displacement as it entre at Bis 8.6 cm.

Use the equation (ii) to find the direction.

$$\begin{gathered} \tan \mathrm{θ}=\frac{6.464}{5.732} \\ =48.43° \end{gathered}$$

Hence,the angle as it entre at B is $48.43°$.

Therefore, to find resultant displacement first, find x component of resultant by vector algebra then y component. Then find resultant vector. And to find magnitude of resultant, use Pythagoras theorem.