Question

On an open ground , a motorist follows a track that turns to his left by an angle of 60° after every 500 m.Starting from a given turn,specify the displacement of the motorist at the third,sixth and eigth turn.Compare the magnitude of the displacement with the total path length covered by motorist in each case.​

Answer 1

Answer:

the answer of this question is 1500m

Answer 2

Given :-

The path followed by the motorist is a regular hexagon with side = 500 m

To Find :-

Compare the magnitude of the displacement with the total path length covered by motorist in each case.​

Solution :-

[Please refer to the attachment for better understanding.]

Let the motorist start from point P.

According to the question,

Magnitude of the displacement = PS = PV + VS

Given that,

Side = 500 m

Substituting them,

= 500 + 500

= 1000 m

Total path length = PQ + QR + RS

= 500 + 500 + 500

= 1500 m

Now,

The motorist takes the 6th turn at point P, which is the starting point.

Given that,

Magnitude of displacement = 0

Total path length = PQ + QR + RS + ST + TU + UP

By substituting,

= 500 + 500 + 500 + 500 + 500 + 500

= 3000 m

Next,

The motorist takes the eight turn at point R.

Magnitude of displacement = PR

As per the case,

$\sf \sqrt{PQ^2 + QR^2 + 2(PQ) \times 2(QR) \: cos \: 60^o}$

Substituting their values,

$\sf =\sqrt{ 500^2+500^2+2(500) \times 500 \: cos \: 60^o }$

$\sf =\sqrt{250000+250000 +\bigg(500000\times \dfrac{1}{2} \bigg)}$

$\sf =\sqrt{ 500000 +\bigg(500000\times \dfrac{1}{2} \bigg)}$

$\sf = 866.03 \ m$

$\sf \beta =tan^{-1} \bigg(\dfrac{500 \: sin \: 60^o}{500+500 \: cos \: 60^o} \bigg)$

$\sf = 30^o$

Therefore, the magnitude of displacement is 866.03 m at an angle of 30° with PR.

Now, we have

Total path length = Circumference of the hexagon + PQ + QR

Substituting them,

= 6 × 500 + 500 + 500

Total path length = 4000 m

Therefore, the total path length is 4000 m.