Question

A particle moves along a path ABCD as shown in
the figure. Then the magnitude of net displacement
of the particle from position A to D is :
5m
OS
AK
3m
B​

Answer 1

Answer:

10 metres

Explanation:

Join AC.

In triangle ABC by Pythagoras' Theorem,

${ab}^{2} + { bc}^{2} = {ac}^{2}$

$= > {ac}^{2} = {4}^{2} + {3}^{2}$

$= > {ac}^{2} = 16 + 9$

$= > {ac}^{2} = 25$

$= > ac = \sqrt{25}$

$= > ac = 5 \: m$

Now, displacement

=AC+CD

=5+5

=10 m

Answer 2

As we know the displacement is the shortest route of the two points.So as it is givenCD=5m. Now we have to find AC and add AC with CD to find the displacement. By pythagorus theorum we can find AC. AC comes to be 5m . Therefore the total displacement is 10m

Answer:- 10m