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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.`
6 years ago

Vikas TU
6853 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
```
5 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) mby 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
```
5 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
```
3 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
```
2 months ago
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
```
one month 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..
```
one month ago
Harshit Panwar
99 Points
```										√(302+202)=10√13So, 30√2-10√13=6.37m Hope it helps you....................................................
```
one month 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)...
```
one month 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
```
one month ago
bhodard
13 Points
```										you all are fucking shithbugbfywgfvsygbcukrebvghyvbakvuibujhbhubhgbfyhubcyuzhceybquiL>vbhfrb eqriobhyuhrqeb fvyuv guhsudvbhyehkghuwgfyrgbeyrvurehnqiohivhjueilepuhiuvygbyeqrgbbsk kinc ghikahncvjksafbhcjno2iuyrcfbhreuwjksb xdcbewjghdfvbuwehjsih
```
one month 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..
```
one month ago
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