Question

A man moves 10m in a direction of 37° East of North, then 5√2m in South East
direction and finally 20m in a direction of 53º South of West. Find the magnitude of
resultant displacement of man?​

Answer 1

Step-by-step explanation:

HELLO DEAR,

Given that a man moves 10 m in 37° East of North direction . Lets breat it in the form of component vector ,

$10\cos(53 ) = x$

10 × 3/5 = x

x= 6 m

And 10 cos 37° = y

10 × 4/5 = y

y= 8 m.

so, vector AB = ( 6i + 8j). .................(1)

Now , vector BC can be written in the form of vector components.

so, 5√2 cos 135° = x

5√2 ×(-1/√2) = x

5 ×(-1) = x

x= -5.

and 5√2 sin 135° = y

5√2 × 1/√2 = y

y= 5.

so,

victor BC = ( -5i , 5j).

Now, vector CD can be written in the form of vector component

CD = (16 i + 12j)

Therefore resultant vector => vector AB + BC + CD = AD

(6i + 8j) + ( -5i + 5j) + ( 16i+ 12j) = AD

(17i + 25j) =AD

And magnitude of resultantant vector AD

= √(17^2 + 25^2)

= √(289+625)

=√914. Answer.

I HOPE IT HELP YOU DEAR,

THANKS