Question

A player moves along the boundary of a square ground of side 50m in 200 sec.The magnitude of displacement of the farmer at the end of 11 min 40 sec from his initial position is
A) 50m
B) 150m
C) 200m
D) 50√2m

Answer 1

Explaination:time taken to travel 200m( perimeter of square is 4*side)=200s. i.e time taken to travel 1m is 1sec.i.e speed of player is 1m/s so The distance travelled is (1*(11*60)+40)=700m i.e 3.5 rounds of the field so displacement is AC(see in photo)=diagonal of square According to pythagorus theorem. (AD)^2+(DC)^2=(AC)^2. So AC=(2500+2500)^1/2. AC=50root2ANSWER:option D

Answer 2

50 √2  m is the displacement of the player at the end of 11  minutes 40 seconds

Explanation:

Square ground of 50 m  sides

=> perimeter = 4 * 50 = 200 m

one round is completed in 200 Sec

=> Hence speed = 1 m/s

Distance Covered in 11 min 40 secs = 700 sec    = 700 m

700 = 200 + 200 + 200 + 100

Hence he has moved 100 m from initial position

50 m on one side  and 50 m at right angle

Hence Distance of final position from initial position =  √50² + 50²

= 50√2

50 √2  m is the displacement of the player at the end of 11  minutes 40 seconds

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