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.

							
zero
						

							
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
						

							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
						

							
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
						

							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
						

							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
						

							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
						

							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..
						

							
√(302+202)=10√13
So, 30√2-10√13=6.37m 
Hope it helps you
						

							
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)...
						

							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
						

							
 
										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..										
						

							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
						

							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.
						

							
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