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

Mr Virtual

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

bel

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

Dick

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?

Mr Virtual

Oh, I just thought that if E is 0 degrees, then north west will be 120 degrees.

Thanks a lot.

Mr V

HallsofIvy

No, it would be 90+ 45= 125 degrees. (Normally North is at 0 degrees and Northwest at 360- 45= 315 degrees.)

HallsofIvy

No, it would be 90+ 45= 125 degrees. (Normally North is at 0 degrees and Northwest at 360- 45= 315 degrees.)

Mr Virtual

Oh, yeah. My mistake.

Mr V