Question

a body moves 30m North, 20m East and 30√2m South-west. Find displacement in vector form.

Answer 1

Displacement is the shortest distance between two points. In this case, the starting and finish point.

You need not make it very complex, you can solve it using simple Pythagorean Theorem,

Sum of Square of Sides = Square of Hypotenuse

Simply draw a diagram of the path taken by the person.



We nee AD,

At C, he turned South West, so it would be 45 Degrees to the current path of Travel, And It would normally intersect AB at 45 Degrees.

Now in Triangle BCE

We know Angle C is 45, and Angle E is 45 as Angle B is 90 (180–90–45)

As Angles E and C are 45, that implies, sides CB and BE are also identical in length. 20 kms. And say Side CE = a

==> a^2 = 20^2 + 20^2

==> a = 20√2

As we know, CD = 30√2

==>ED = CD-CE = 30√2 - 20√2 = 10√2

Now in Triangle ADE, ED = 10√2 and AE = (30–20) = 10

And we know Angle A = 90

S0,

DE^2 = AE^2 + AD^2

==> (10√2)^2 = 10^2 + AD^2

==> AD = Sqrt(200–100) = 10 Kms

Good Luck

Answer 2

Explanation:

30m North

20m East

and the 30

2

SW can be rewritten as 30m in south and 30m in West

because south and west being at 90

0

result

in

30

2

+30

2

=30

2

so at last we have following data that in south X

S

=30m+north=30m+(−30m)=0,

negative becasue north is opposite to south.

and in West we have X

W

=30m+east=30m+(−20m)=10m

so the displacement is 10m due west