## Direction of acceleration when direction of moving particle changes by 90 degrees?

1. The problem statement, all variables and given/known data

A particle is moving eastwards with a velocity of 5m/s. In 10 seconds, the velocity changes to 5 m/s northwards. What is the average acceleration time? What is the direction of accleration?

2. Relevant equations

a=v-u/t

3. The attempt at a solution

initial velocity northwards = u = 0 m/s
final velocity northwards = v = 5 m/s
t = 10 s
a = v-u/t= 1/2 m/s^2

Now initial direction = eastwards = 0 degree
final direction = northwards = resultant = 90 degree
therefore, using vector rules, direction of acceleration = north-west = 120 degree

Am I right?

Mr V

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 Imagine, for the sake of simplicity, that the particle was undergoing uniform circular motion, then everything will be easy.

Recognitions:
Homework Help
 Quote by bel Imagine, for the sake of simplicity, that the particle was undergoing uniform circular motion, then everything will be easy.
Don't do that. Just subtract the initial velocity vector from the final and divide by delta t (though the final division won't affect the direction). NW is ok but why 120 degrees?

## Direction of acceleration when direction of moving particle changes by 90 degrees?

 Don't do that. Just subtract the initial velocity vector from the final and divide by delta t (though the final division won't affect the direction). NW is ok but why 120 degrees?
Oh, I just thought that if E is 0 degrees, then north west will be 120 degrees.

Thanks a lot.

Mr V

 Recognitions: Gold Member Science Advisor Staff Emeritus No, it would be 90+ 45= 125 degrees. (Normally North is at 0 degrees and Northwest at 360- 45= 315 degrees.)
 Recognitions: Gold Member Science Advisor Staff Emeritus No, it would be 90+ 45= 125 degrees. (Normally North is at 0 degrees and Northwest at 360- 45= 315 degrees.)
 Oh, yeah. My mistake. Mr V