 ×     #### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
```
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.

```
9 years ago

```							zero
```
9 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
```
9 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
```
3 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) 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
```
3 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
```
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
```
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
```
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..
```
3 years ago
```							√(302+202)=10√13
So, 30√2-10√13=6.37m
Hope it helps you
```
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)...
```
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
```
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..
```
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
```
2 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.
```
2 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
```
2 years ago
```							Dear Student,Please find below the solution to your problem.30m North20m Eastand the 30√2​ SW can be rewritten as 30m in south and 30m in West because south and west being at 90 degree resultin √(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)=10mso the displacement is 10m due westThanks and Regards
```
3 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Mechanics

View all Questions »  ### Course Features

• 101 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions