Question

A particle moves through a distance of 8 m due east and then 6 m due north then magnitude of displacement is

Answer 1

Answer:

the displacement is equal to 10m

Explanation:

So here the shortest distance between point AC is the line joining it.

So we will find it out by using Pythagoras theorem

I.e. AC=(AB)2+(BC)2−−−−−−−−−−−−√ now put the data in the formula we get,

⇒AC=(8)2+(6)2−−−−−−−−−√=64+36−−−−−−√=100−−−√=10m

Hence the displacement is equal to 10m.

We will also solve such a question by using vector method,

Let the east direction to be i^ and north direction bej^.

Thus displacement of the man S→=8i^+6j^

And we know that the method to how to take the magnitude of the vector quantity we take mod of it for the magnitude.

So, magnitude of displacement ∣∣∣S→∣∣∣=82+62−−−−−−√=64+36−−−−−−√=10m

Answer 2

Answer:

abbe nhi aata toh kiu de raha hai nikal apni pehli fursat me hi