Question

On an open ground, a biker follows a track that turns to his left by an angle of 60°after every 600 m. Starting from a given turn, specify the displacement of the biker at the third, sixth turn.
Compare the magnitude of the displacement with the total path length covered by the motorist in each case.​

Answer 1

$\LARGE\color{black}{\colorbox{#FF7968}{Q}\colorbox{#4FB3F6}{U}\colorbox{#FEDD8E}{E}\colorbox{#FBBE2E}{S}\colorbox{#60D399}{T}\colorbox{#6D83F3}{I}\colorbox{#FF7968}{O}\colorbox{#4FB3F6}{N}}$

On an open ground, a biker follows a track that turns to his left by an angle of 60°after every 600 m. Starting from a given turn, specify the displacement of the biker at the third, sixth turn.

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

$\huge\bf{Answer \: :- \: }$

Let us assume that the trip starts at point $\sf\fbox\orange{A}$

The third turn will be taken at $\sf\fbox\blue{D}$

So displacement will be = Distance AD

= 500 + 500 = $\sf\fbox\pink{1000 m}$

Total path covered = AB + BC + CD = 500 + 500 +

500 = $\sf\fbox\green{1500 m}$

The sixth turn is at A.

So the displacement will be Zero

and total path covered will be = 6 (500) = $\sf\fbox\blue{300 m}$

The eighth turn will be at C

So the displacement = AC

= √AB² + BC² + 2 (AB) (BC) Cos60°

= $\sf\fbox\orange{866.03 m}$

And the total distance covered = 3000 + 1000 = 4000 m=$\sf\fbox\red{4Km}$

Answer 2

the track is shown in the figure given below:-

$\huge\boxed{Attachment}$

Let us assume that the trip starts at point A.

The third turn will be taken at D.

 So displacement will be   =     Distance AD  =   500 + 500 =   1000 m

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

 

The sixth turn is at A.

 So the displacement will be Zero

and total path covered will be  =   6 (500)  =  3000 m

 

The eighth turn will be at C.

 So the displacement        =      AC

=$\red{\sqrt{AB²+BC²+2(AB)(BC)cos60°}}$

or

=$\red{\sqrt{(500)²+(500)²+2(500)(500)cos60°} }$

And the total dectance covered = 3000+1000=4000 m=4Km