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Question: A particle moves from A to P and from P to B, as shown in the figure. Find the path length and displ...

A particle moves from A to P and from P to B, as shown in the figure. Find the path length and displacement.

Explanation

Solution

Hint: We need to determine the length AP and PB to answer this question. From the figure, it is clear that AP=PB. Also, we should know the distinction between path length and displacement.

Complete step by step answer:

From the figure, we can see that the BPD=APC=60 (VOA)\angle BPD=\angle APC={{60}^{\circ }}\text{ (VOA)} because they are vertically opposite angles. We also know that the sum of all the angles will be 360{{360}^{\circ }}. So the other two angles which are equal (vertically opposite angles), can be found out by,

BPD+APC+APB+CPD=360 60+60+2APB=360 \begin{aligned} & \angle BPD+\angle APC+\angle APB+\angle CPD={{360}^{\circ }} \\\ & {{60}^{\circ }}+{{60}^{\circ }}+2\angle APB={{360}^{\circ }} \\\ \end{aligned}

APB=120\therefore \angle APB={{120}^{\circ }}

After figuring out APB=120\angle APB={{120}^{\circ }}, we will drop a perpendicular from P to E on AB, which will divide AB into two equal parts AE and EB, which are equal sections of length e2\dfrac{e}{2}. The line PE splits the angle APB\angle APB into two equal angles, i.e. APE=BPE=60\angle APE=\angle BPE={{60}^{\circ }}.

Considering the right triangle AEP, we can write

sin(60)=AEAP=(e2)AP\sin \left( {{60}^{\circ }} \right)=\dfrac{AE}{AP}=\dfrac{\left( \dfrac{e}{2} \right)}{AP}
AP=(e2)sin(60)AP=\dfrac{\left( \dfrac{e}{2} \right)}{\sin \left( {{60}^{\circ }} \right)}
AP=e3\therefore AP=\dfrac{e}{\sqrt{3}}

So, due to symmetry, PB will also be equal to e3\dfrac{e}{\sqrt{3}}.

So the path length is equal to the total distance travelled. So we can define path length as the sum of the lengths AP and PB,

P.L=AP+PBP.L=AP+PB

P.L=2e3\text{P}\text{.L}=\dfrac{2e}{\sqrt{3}}.

So the path length is the distance that the particle travelled from A to P and from P to B, which is equal to (P.L)=2e3\left( \text{P}\text{.L} \right)=\dfrac{2e}{\sqrt{3}}.

But, the displacement will always be the shortest distance travelled; in this case, it will be the length of AB = e.

Note: In mechanics, displacement is a vector whose length is the shortest distance from the initial to the final position of a point P undergoing motion.
In mechanics, the Path length is the total distance travelled by the body from an initial point to a final point.