Question

On an open ground, a motorist follows a track that turns to his left by an angle of 600 after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.

Answer 1

Answer:

The path followed by the motorist is a regular hexagon with side 500 m, as shown in the given figure

Let the motorist start from point A.

The motorist takes the third turn at D.

Magnitude of displacement =AD=BC+EF=500+500=1000m

Total path length =AB+BC+CD=500+500+500=1500m

The motorist takes the sixth turn at point A, which is the starting point.

Magnitude of displacement =0

Total path length =6AB

=6×500=3000m

The motorist takes the eight turn at point C

The magnitude of displacement =AC

From triangle law, the magnitude of displacement is 866.03m at an angle of 30 with AC.

It means AC makes an angle 30 with the initial direction. Total path length =8×500=4000m.

Explanation:

please mark as branlist

Answer 2

Given :-

On an open ground, a motorist follows a track that turns to his left by an angle of 600 after every 500 m.

To Find :-

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

Solution :-

Given that,

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

Let us consider,

Let the motorist start from point P.

The motorist takes the third turn at S.

Magnitude of the displacement = PS = PV + VS

Substituting them,

$\sf = 500 + 500 = 1000 \ m$

Now,

Total path length = PQ + QR + RS

By substituting,

$\sf = 500 + 500 + 500 = 1500 \ m$

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

Then,

Magnitude of displacement = 0

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

Substituting their values,

$\sf = 500 + 500 + 500 + 500 + 500 + 500 = 3000 \ m$

The motorist takes the eight turn at point R.

Magnitude of displacement = PR

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

By substituting,

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

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

$\sf = 866.03 \ m$

$\sf \beta =tan^{-1} \bigg( \dfrac{500sin60^{o}}{500+500cos60^{o}} \bigg)$

$\sf =30^{o}$

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

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

Substituting them,

$\sf = 6 \times 500 + 500 + 500$

$\sf =4000 \ m$