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Q. 4.10 On an open ground, a motorist follows a track that turns to his left by an angle of 60^{\circ} 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.

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The track is shown in the figure given below:- 

                                     Motion in plane ,    20107

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

                                        =\ \sqrt{AB^2\ +\ BC ^2\ +\ 2(AB)(BC) \cos 60^{\circ}}

or                                     =\ \sqrt{(500)^2\ +\ (500) ^2\ +\ 2(500)(500) \cos 60^{\circ}}

                                        =\ 866.03\ m

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

Posted by

Devendra Khairwa

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