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A person moves 30 m north and then 20 m towards east and then finally 30 √2 m in south - west direction. Find the dispalcement of the particle from origin.

10 years ago

The person goes 30 m in north, and then 20 m in east and then 30√2 m in south west. Now, if we draw this diagram on a paper, then it would form a right-angled triangle kind of thing. So, the final position would lie on the hypotenuse of the triangle. Now, studying the diagram a more, we get,

length of hypotenuse = root(30^{2} + 20^{2} ) = 10√13 m

net displacement vector = 10√13 - 30√2

==> 42-36.6

==>**5.4 m**

4 years ago

4 years ago

WHEN he starts from 30 m north then 20 m east then we can say that resultant displacement is sqrt(30^2+20^2)=10 sqrt(13) m

by addition of triagle vector.

he then finally moves 30 √2 m in south - west

so, displacement=final position-initial position

=30 √2 m in south - west - 10 √13 in south west

=42.43 m -36.05 m

=6.38 m

4 years ago

3 years ago

3 years ago

3 years ago

3 years ago

Apply phythagorean theorem;for 30m and 20m..you will get the dispalement 10*square root of 13.....then subract 30*square root of 2....and 10*square root of 13....the final answer is 6.4(approximately)...

3 years ago

3 years ago

3 years ago

3 years ago

3 years ago

First the man go in north direction 30m and than towards east direction 20m and than in southwest direction 30 root 2 than the right angle triangle is formed so, underoot 30x30 +20×20-30root2 appox answer is> 5.5 in west direction

one year ago

Please find below the solution to your problem.

30m North20m East

and the 30√2 SW can be rewritten as 30m in south and 30m in West because south and west being at 90 degree result

in √(30)^2 + (30)^2 = 30√2

so at last we have following data that in south XS=30m+north =30m+(−30m)=0,

negative becasue north is opposite to south.

and in West we have XW=30m+east=30m+(−20m)=10m

so the displacement is 10m due west

Thanks and Regards

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