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Sanchit Gupta Grade: 12

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.

6 years ago

Answers : (5)

Vikas TU
6741 Points


6 years ago
Ritvik Gautam
85 Points

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(302 + 202 ) = 10√13 m

net displacement vector = 10√13 - 30√2

==> 42-36.6

==>5.4 m

6 years ago
Erisha sayed
11 Points
										Let`s draw the figure.   Start from A to is 30 m and move east is 20 m now name the point as C. Therefore we get a triangle ABC but AC should be extended now we need to find AD which is CD -AC find AC  by pgt and subtract 30 root and 10root13 the answer will be 6.4
2 months ago
somi teez
100 Points
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
2 months ago
11 Points
										When he walks 30m north, vector obtained is 30j^Further When he walks 20m east,  vector obtained is 20i^ Further he walks 30√2 south west,  which cannot be written either along x-axis or y-axis so we take components of this movement along x-axis and y-axis *along x-axis = 30√2*-cos45°= -30i^*along Y-axis = 30√2*-sin45°=-30j^Now net movement, 30j^+20i^-30j^-30i^ = -10i^ This shows the displacement of 10m towards west from origin
23 days ago
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