To find the magnitude and direction of the velocity of car A relative to car B, we can use the formula v = √((vx_A - vx_B)^2 + (vy_A - vy_B)^2), where v is the magnitude of the velocity, vx_A is the x-component of the velocity of car A, vy_A is the y-component of the velocity of car A, vx_B is the x-component of the velocity of car B, and vy_B is the y-component of the velocity of car B.
Since car A is moving due north and car B is moving due east, the x-component of the velocity of car A is 0 and the y-component is 6 m/s, while the x-component of the velocity of car B is 8 m/s and the y-component is 0. Plugging these values into the formula, we get:
v = √((0 - 8)^2 + (6 - 0)^2)
= √(64 + 36)
= √100
= 10 m/s
To find the direction of the velocity, we can use the inverse tangent function (arctan) to calculate the angle between the velocity vector and the positive x-axis. The angle is given by the formula θ = arctan((vy_A - vy_B)/(vx_A - vx_B)). In this case, we get:
θ = arctan((6 - 0)/(0 - 8))
= arctan(-6/8)
= -53.13 degrees
So the magnitude of the velocity of car A relative to car B is 10 m/s, and the direction is 53.13 degrees counterclockwise from the positive x-axis.