Find average acceleration from two given velocities each at a given ti

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To find the average acceleration of an object transitioning from a velocity of 40 m/s North to 30 m/s East over 5 seconds, vector components must be analyzed. The change in velocity in the North-South direction is -40 m/s, and in the East-West direction, it is +30 m/s. The average acceleration can be calculated by determining the change in velocity for each direction, dividing by the time interval of 5 seconds. The resultant average acceleration magnitude can be found using the Pythagorean theorem with these components. The direction of the average acceleration will be determined by the angle formed between the resultant vector and the North direction.
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The velocity of an object at t=0 seconds is 40 m/s directed due North. The velocity of the same object at t=5 seconds is 30 m/s directed due east.

Find the magnitude of the average acceleration of the object over the time interval from t=0 to t=5 seconds.

Also, show the direction of the average acceleration. So... I am not quite sure how to even approach this problem. I originally tried it and did it with Pythagorean Theorem, but my answer was incorrect.

The answer I got using Pythagorean Theorem was 50 m/s at 36.9 East of North.

Please help!
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Judging from your diagrams, you did a vector addition of the two velocities. You want the change involved in going from one velocity to the other. What operation should you use?
 
First consider the component of acceleration in the north-south direction. What is the change in the velocity component in this direction? From this, what is the acceleration (don't forget to divide by the time)?

Now do the same thing for the east-west direction.
 

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