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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.

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Solution

Given: the motorist follows the track of an open ground and he turns to his lift by an angle of 60° after every 500m.

The path followed by the motorist can be shown as follows:



Let the motorist start his journey from point O. Then he takes a turn at point C.

The displacement for this turn is given by OC .

From the figure, displacement OC is,

OC= ( OB ) 2 + ( BC ) 2

However, OB=OF+FB, hence the above equation becomes,

OC= ( OB ) 2 + ( BC ) 2 = ( OF+FB ) 2 + ( BC ) 2 (1)

Substitute the values in equation (1),

OC= ( OF+FB ) 2 + ( BC ) 2 = ( 500cos30°+500cos30° ) 2 + ( 500 ) 2 =1000m

The total path length is equal to 500m+500m+500m=1500m.

Ratio of magnitude of displacement and total path length is,

Displacement( OC ) Path length = 1000 m 1500 m =0.67 .

Hence, the ratio of magnitude of displacement and the path length is 0.67.

Again, the motorist takes the sixth turn at O.

The displacement in this case will be equal to zero. So, the displacement vector is a null vector. Hence, the ratio of the magnitude of the displacement and the total path length is zero.

The motorist takes the 8thturn at point B.

The magnitude of displacement at 8th turn from the figure is,

2×BCcosθ=2×500cos30° =2×500× 3 2 m =0.866km

The total path length is equal to 8×500m=4km.

Ratio of magnitude of displacement and path length is given as,

0.866 4 =0.216

Hence, the ratio of magnitude of displacement and the path length is 0.216.


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