Calculate change in velocity resulting from its turn

Veronica drives her car north at 36km/h and does a left-hand turn without changing speed.

Determine the magnitude of the car's change in velocity resulting from its making this turn?


For the answer they have written: DeltaV=Vf-Vi
=Vf+(-Vi)

Then they have written: DeltaV= SQRT(Vf^2 + Vi^2) = 36 SQRT(2)

The answer is 50.9 km/h or 14.14m/s

Can someone please explain to me what the second formula is and the reasoning behind it?

Think of it like a triangle and remember that with velocity, the direction does matter!
A change in direction is a change in velocity.
With that in mind, Veronica is moving north (straight up on a piece of paper) at 36 km/h.
She then turns left (left on a piece of paper) and stays at 36 km/h.
With vectors we measure changes from the start point of the first vector to the end point of the second one.

If you draw a line from the beginning of the north vector to the end of the one going left, You get a right triangle.

Knowing that both the north and left legs are 36 mph, you can use pythagoras theorem to get the change in velocity

DeltaV^2 = Vf^2+Vi^2