To find the change in velocity of the car, we need to calculate the vector difference between the initial and final velocities. We can do this by subtracting the initial velocity vector from the final velocity vector.
Let's assume that the positive x-axis points east and the positive y-axis points north. Then the initial velocity of the car can be represented as:
v1 = 15 m/s * north
And the final velocity of the car can be represented as:
v2 = 18 m/s * east
To calculate the change in velocity, we need to subtract the initial velocity vector from the final velocity vector:
Δv = v2 - v1
Δv = (18 m/s * east) - (15 m/s * north)
To add or subtract vectors, we need to resolve them into their x and y components. The x component represents the vector's magnitude in the east-west direction, while the y component represents the vector's magnitude in the north-south direction.
The magnitude of the x component of Δv is:
Δvx = 18 m/s * cos(90°) - 15 m/s * cos(0°)
Δvx = 0 - 15 m/s
Δvx = -15 m/s
The magnitude of the y component of Δv is:
Δvy = 18 m/s * sin(90°) - 15 m/s * sin(0°)
Δvy = 18 m/s - 0
Δvy = 18 m/s
Therefore, the change in velocity of the car is:
Δv = -15 m/s * north + 18 m/s * east
This means that the car's velocity changed by 15 m/s in the northward direction and 18 m/s in the eastward direction. The magnitude of Δv can be calculated using the Pythagorean theorem:
|Δv| = sqrt((-15 m/s)^2 + (18 m/s)^2)
|Δv| = 22.8 m/s
So the magnitude of the change in velocity is 22.8 m/s, and the direction of the change is 53.1 degrees east of north.