Question

a particle moving with a velocity of 10m/s towards east. After 20 s it turns north with same velocity. its average acceleration is-

Answer 1

Answer:

acc is 0 because velocity is not change

acceleration is the rate change in velocity

or it can be also happen that

acceleration will be same after turning and before turning

that means v/t that is 0.5 m/s^2

Answer 2

$\huge\red{Answer}$

➡️➡️➡️➡️➡️➡️➡️➡️➡️➡️➡️➡️➡️➡️➡️➡️➡️

Average acceleration is defined as- (change in velocity/time elapsed)

Average acceleration is defined as- (change in velocity/time elapsed)Here,change in velocity =

Average acceleration is defined as- (change in velocity/time elapsed)Here,change in velocity = √102 + 102 or 10√2ms

2 or 10√2ms So,average acceleration is

10√210ms2

i.e √2ms2

i.e √2ms2 As,both the velocity vectors are equal in magnitude,their resultant will be at angle of 45 degrees w.r.t each other. And that will be the direction of the average acceleration as well,i.e along north west

i.e √2ms2 As,both the velocity vectors are equal in magnitude,their resultant will be at angle of 45 degrees w.r.t each other. And that will be the direction of the average acceleration as well,i.e along north westNote: **Change in acceleration means,final velocity vector( v ) - initial velocity vector (v' )

= v+(−v)

= v+(−v) As v' vector is located towards east,so −v'

vector is located towards east,so −v' will be located towards west,so a resultant of two vectors of equal magnitude one directed along north and one along west will be along north west**

jai siya ram☺ __/\__

➡️➡️➡️➡️➡️➡️➡️➡️➡️➡️➡️➡️➡️➡️➡️➡️➡️

¯\_(ツ)_/¯