Question

A man moves towards east over a distance of 4m in 6s and then towards north over a
distance of 3m in 4s. What is the net velocity of the man in this journey?​

Answer 1

Answer:

Let the East direction be along  

i

^

 and the North direction be along  

j

^

​  

 and the person be initially at the origin .

Then  

k

^

 points in the vertically upward direction.

Therefore , final position vector of the particle=  

r

 = 3  

i

^

 + 4  

j

^

​  

 + 5  

k

^

 

Hence , the distance of the person from the origin = ∣  

r

∣

                                                                                    =  

3  

2

+4  

2

+5  

2

 

​  

 

                                                                                    = 5  

2

​  

 m

Explanation:

Hope it helps