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, The path followed by the motorist is a regular hexagon with side `500 m`, as shown in the given figure specify the displacement of the motorist at the end of third turn

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Explanation

In the adjoining figure the given and required information shown are not to a scale. As motion diagram is a schematic representation only.
(a) Using the third equation of uniform acceleration motion, we have
`v_(t)^(2)=v_(o)^(2)+2a(x_(t)-x_(o))rarr a=(v_(t)^(2)-v_(o)^(2))/(2(x_(t)-x_(o)))=(10^(2)-20^(2))/(2(32-2))=-5 m//s^(2)`

(b) Using second equation of uniform acceleration motion, we have
`x_(t)=x_(o)+v_(o)t+1/2 at^(2) rarr x_(8)=2+20xx8+1/2 (-5)8^(2)=2 m`
Using the first equation of uniform acceleration motion, we have
`v_(t)=v_(o)+at rarr v_(8)=20+(-5)xx8=-20 m//s`
(c) Where the particle returns, its velocity must be zero. Using the third equation of uniform acceleration motion, we have
`v^(@)=v_(o)^(2)+2a(x-x_(o)) rarr x=x_(o)+(v^(2)-v_(o)^(2))/(2a)=2+(0-20^(2))/(2(-5))=42 m`
This location is shown in the adjoining modified motion diagram.
The distance `Deltas` is `Deltas=|x-x_(o)|+|x_(o)-x|=80 m`