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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.

Vikas TU
14149 Points
10 years ago

zero

Ritvik Gautam
85 Points
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(302 + 202 ) = 10√13 m

net displacement vector = 10√13 - 30√2

==> 42-36.6

==>5.4 m

Erisha sayed
11 Points
4 years ago
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
somi teez
105 Points
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
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
GuRI
11 Points
4 years ago
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
Teju
17 Points
3 years ago
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
13 Points
3 years ago
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
Uttam kumar
23 Points
3 years ago
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..
Harshit Panwar
108 Points
3 years ago
√(302+202)=10√13
So, 30√2-10√13=6.37m
Hope it helps you
moheen
13 Points
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)...
Siddharth thakkar
15 Points
3 years ago
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
sagar kumar
99 Points
3 years ago

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..
Ankit guha
32 Points
3 years ago
AB= 30 MBC= 20MCD=30√2MSO NOW BREAK THE CD IN HORIZONTAL AND VERTICAL COMPONENTS SO vertical COMPONENT CANCELS AB AND NET HORIZONTAL DISPLACEMENT IS 30-20 i.e. 10 M towards west direction
Divyanshi
13 Points
3 years ago
Do it with the help of a rectangle diagram then find x by ptm. It will be 10mtowards west and it`s the correct answer.
Dnyandeo
15 Points
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

Rishi Sharma
one year ago
Dear Student,

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