Question

Typical backyard ants often create a network of chemical trails for guidance. Extending outward from the nest, a trail branches (bifurcates) repeatedly, with$\mathbf{60}\mathbf{°}$between the branches. If a roaming ant chances upon a trail, it can tell the way to the nest at any branch point: If it is moving away from the nest, it has two choices of path requiring a small turn in its travel direction, either$\mathbf{30}\mathbf{°}$leftward$\mathbf{30}\mathbf{°}$or rightward. If it is moving toward the nest, it has only one such choice. Figure 3-29shows a typical ant trail, with lettered straight sections of 20 cmlength and symmetric bifurcation of$\mathbf{60}\mathbf{°}$. Path v is parallel to the y axis. What are the (a) magnitude and (b) angle (relative to the positive direction of the superimposed x axis) ofan ant’s displacement from the nest (find it in the figure) if the ant enters the trail at point A? What are the (c) magnitude and (d) angle if it enters at point B?

Short answer

(a) The magnitude of displacement is 7.5 cm.

(b) The angle of displacement is$90°$.

(c) The magnitude of displacement as it entre at Bis 8.6 cm.

(d) The angle as it entre at B is $48.43$.

Step-by-step solution

Step 1: Given

The length of each branch is x=2.0 cm

Determining the concept

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,

$\tan \theta =\frac{y}{x}$ (ii)

Where,

$x,y,d$Are vectors and $\theta$is the angle between x and y.

(a) Determining the magnitude of displacement

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:

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

Vertical component of w is as follow:

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

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:

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

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.

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

So, magnitude is,

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

Hence, the magnitude of displacement is 7.5 cm.

(b) Determining the angle of displacement

Use equation (ii) to find the angle.

Direction is,

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

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

(c) Determining the magnitude of displacement as it entre at B

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:

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

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

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

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

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

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:

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

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.

(d) Determining the angle as it entre at B

Use the equation (ii) to find the direction.

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

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.