Question

A girl moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the girl at the end of 2 minutes 20 seconds?

• A. $10\sqrt{2}$ m
• B. 10 m
• C. 3.14 m
• D. 20 m

Answer 1

The correct option is A

$10\sqrt{2}$ m

Distance covered in one round = 4 x 10 m = 40 m

Time taken = 40 s

Speed = $\frac{40}{40}$ = 1 m/s

So distance covered in 2 min 20 s = 1 m/s $\times (2\times 60+20)$ s = 140 m

No. of rounds completed = $\frac{140}{40}$ = 3.5

If it starts from point A, it will be at C after 3.5 rounds.

Hence displacement = AC = $\sqrt{({10}^2+{10}^2)}$ = $10\sqrt{2}$ m