If a car changes its velocity for 5 secs. from 35.0m/s to the north to 20.0m/s to the east, what is its averag

If a car changes its velocity for 5 secs. from 35.0m/s to the north to 20.0m/s to the east, what is its average acceleration? in what direction is this average acceleration interms of the N,E,S and W directions.

Can anyone show me the formula please.

Acceleration = (Final velocity – Initial velocity) ÷ time
Since velocity and acceleration are vectors, draw the vectors. North is positive y direction, and East is the positive x direction.
Final velocity = 20.0 m/s east
The final velocity is represented by a 2 cm long arrow from the origin to the point (2, 0).

To subtract a vector, you add the opposite. To subtract 35.0m/s to the north, add 35.0m/s to the south. Draw a 3.5 cm long arrow in the negative y direction. The arrow will start at the end point of the first vector, (2,0), and end 3.5 cm below, at the point, (2, -3.5). This arrow represents the opposite of the initial velocity.

Finally draw an arrow from the origin to the point (2, -3.5). This arrow represents the change of the velocity. The 3 vectors form a right triangle. The change of the velocity vector is the hypotenuse of the right triangle.

The magnitude of change of the velocity = (20^2 + 35^2)^0.5
The magnitude of the acceleration = magnitude of change of the velocity ÷ time
The magnitude of the acceleration = (20^2 + 35^2)^0.5 ÷ 5 = 8.06 m/s^2
The tangent of the angle south of east = 35/20
The angle ≈ 60.3˚ South of East

The average is 3..