A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from its initial position ?
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Case 1. Displacement = 10m
Time = 40 sec
Velocity = 10/40 = 0.25 m/sec
Case 2. Velocity = 0.25 m/sec
Time = 2 min 20 sec = 140 sec
Displacement = Velocity * Time
0.25 *140 = 35 m
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Time = 40 sec
Velocity = 10/40 = 0.25 m/sec
Case 2. Velocity = 0.25 m/sec
Time = 2 min 20 sec = 140 sec
Displacement = Velocity * Time
0.25 *140 = 35 m
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littlerose:
Isn't it helpful??????
well, answer is wrong
how????
i got the answer now I'm telling you
4*10m=40m(perimeter of square field )
then after 140 s he will be at opposite corner. or AC(if he starts from A)
by pythagorus theorem
(AC)^2=(AB)^2+(AB)^2
!! = (10M)^2+(10M)^2
!! = 100M^2 + 100M^2
(AC)^2= 200M^2
AC = SQUARE ROOT OF 200M^2
AC = (AFTER PRIME FACTORIZATION OF 200M^2) 10<ROOT>2
4*10m=40m(perimeter of square field )
then after 140 s he will be at opposite corner. or AC(if he starts from A)
by pythagorus theorem
(AC)^2=(AB)^2+(AB)^2
!! = (10M)^2+(10M)^2
!! = 100M^2 + 100M^2
(AC)^2= 200M^2
AC = SQUARE ROOT OF 200M^2
AC = (AFTER PRIME FACTORIZATION OF 200M^2) 10<ROOT>2
Answered by
1
SQ ABCD [figure][ diagonal AC ]
he starts from A then in 140 sec he will reach opp. corner
displacement = AC i.e. x
x^2= AB^2 +BC^2
x^2 = 100+100
x^2=200
x=√200
x=14.14m
he starts from A then in 140 sec he will reach opp. corner
displacement = AC i.e. x
x^2= AB^2 +BC^2
x^2 = 100+100
x^2=200
x=√200
x=14.14m
YA CORRECT BUT ... INSTEAD OF 14.14 IT IS 10<ROOT>2 BUT BOTH ARE SAME
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