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 sixth turn.

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`