Question

a particle has a displacement of 12 m towards the east and then 6 m vertically upwards . find the magnitude of the sum of these displacements <br />​

Answer 1

Answer:

please refer to the attachment

I hope it would help you

thank you

Answer 2

Answer:-

S = 6√5 m

Given :-

A body goes 12 m east and 6 m north.

To find :-

The magnitude of displacement covered by the particle.

Solution:-

Displacement:- The shortest distance between the final to initial position of a body is called displacement.

• It is a vector quantity.

• It is always less than or equal to distance.

Now ,

From given attachment

From given attachment We can see that it formed a right angle triangle at B.

Bu using Pythagoras theorem,

$\huge \boxed {H^2 = P^2 + B^2}$

→$AC^2 = AB^2 + BC^2$

→$AC^2 = (12)^2+(6)^2$

→$AC^2 = 144 + 36$

→$AC^2 = 180$

→$AC = \sqrt{180}$

→$AC = 6\sqrt{5}m$

hence,

The magnitude of displacement will be 6√5m.