Question

A particle is moving at speed 5m/s towards east. After 10sec, it's Velocity changes and becomes 5m/s towards north. What is the average acceleration during this interval​

Answer 1

Answer:

1/^2 towards n- w...

Explanation:

Initial velocity =v1=5m/s seast wards.

Initial velocity =v1=5m/s seast wards.then after time t=10 sec

Initial velocity =v1=5m/s seast wards.then after time t=10 sectowards east =5m/s

Initial velocity =v1=5m/s seast wards.then after time t=10 sectowards east =5m/sv2=5m/s north wards.

Initial velocity =v1=5m/s seast wards.then after time t=10 sectowards east =5m/sv2=5m/s north wards.towards east =5m/s

Initial velocity =v1=5m/s seast wards.then after time t=10 sectowards east =5m/sv2=5m/s north wards.towards east =5m/sV resultant =v12+v22

Initial velocity =v1=5m/s seast wards.then after time t=10 sectowards east =5m/sv2=5m/s north wards.towards east =5m/sV resultant =v12+v22=25+25=52

Initial velocity =v1=5m/s seast wards.then after time t=10 sectowards east =5m/sv2=5m/s north wards.towards east =5m/sV resultant =v12+v22=25+25=52acceleration a=v=velocity/time

Initial velocity =v1=5m/s seast wards.then after time t=10 sectowards east =5m/sv2=5m/s north wards.towards east =5m/sV resultant =v12+v22=25+25=52acceleration a=v=velocity/time=1052=22×22=21towardsn−w

pls. mark as the brainlist pls..