In a ground, a cyclist follows a track that turn to his right by an angle of 60° after every 100 m. The displacement of the cyclist from the starting point to the fourth turn:

Question

In a ground, a cyclist follows a track that turn to his right by an angle of 60° after every 100 m. The displacement of the cyclist from the starting point to the fourth turn: (A) 200 √10 (B) 100 √2 (C) 2 √2 (D) 4 √6

Tardigrade Admin · Accepted Answer

Cyclist starts from A and reaches a point E <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/b7e7d9d71b64228580e6a22152addf8b-.png" /> So their displacement is AE AE =√ AF 2+ FE 2 (According to pythagoras theorem) A E=√(100)2+(100)2 AE =100 √2

Q. In a ground, a cyclist follows a track that turn to his right by an angle of $6 0^{\circ}$ after every $100 m$ . The displacement of the cyclist from the starting point to the fourth turn:

$20010$

$1002$

$22$

$46$

Cyclist starts from A and reaches a point $E$

So their displacement is $A E$ $A E = A F^{2} + F E^{2}$
(According to pythagoras theorem)
$A E = (100)^{2} + (100)^{2}$
$A E = 1002$

Q. In a ground, a cyclist follows a track that turn to his right by an angle of 60∘ after every 100m. The displacement of the cyclist from the starting point to the fourth turn:

20010​

1002​

22​

46​