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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 : (13)

Vikas TU
6872 Points
										

zero

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
										
8 months ago
somi teez
105 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
8 months ago
GuRI
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
										
6 months ago
Teju
17 Points
										10 m towards westX displacement​ = 30√2 cos 45°                             = 30√2×1÷√2                              =30Net displacement = 30-20=10mSo the displacement is 10 m towards west direction
										
5 months ago
Stonyarcade
13 Points
										Using vector30m North i.e +30j 20m east i.e +20i30√2 south west i.e (-30i)+(-30j)Therefore net displacement=30j+20i-30i-30j= -10i i.e 10m towards West
										
4 months ago
Uttam kumar
23 Points
										Here we have to find a vector value, so we have to solve the question as vector form..Now, here 1st man walks towards north so it is 30 j^ and then 20i^  and finally 30✓2 towards South-west which is = -30✓2cos thita + (-30✓2 sin thita)= -30i^ - 30j^.So we have-30j^ -30j^ + 20i^ - 30i^ = - 10i^.So it is 10m towards west..
										
4 months ago
Harshit Panwar
104 Points
										
√(302+202)=10√13
So, 30√2-10√13=6.37m 
Hope it helps you....................................................
4 months ago
moheen
13 Points
										
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)...
4 months ago
Siddharth thakkar
15 Points
										First of all draw the figure of this que .so that you can see simply right angle tringle .then use thereom of Pythagoras.like this.....(20)1+(√2)²=(ac)1There for your answer is ...20√2m
										
4 months ago
bhodard
13 Points
										
you all are fucking shit
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4 months ago
sagar kumar
100 Points
										
 
Here we have to find a vector value, so we have to solve the question as vector form..Now, here 1st man walks towards north so it is 30 j^ and then 20i^ and finally 30✓2 towards South-west which is = -30✓2cos thita + (-30✓2 sin thita)= -30i^ - 30j^.So we have-30j^ -30j^ + 20i^ - 30i^ = - 10i^.So it is 10m towards west..
4 months ago
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