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.

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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 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..

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

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.

Dear Student,
Please find below the solution to your problem.

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