Relative velocity problems are handled in a similar way in two dimensions - it's just more challenging to add or subtract vectors in two dimensions.

Let's change the 1-D example to 2-D. The truck still moves at 40 km/h west, but the car turns on to a road going 40 degrees south of east, and travels at 30 km/h. What is the velocity of the car relative to the truck now?

The relative velocity equation for this situation looks like this:

v_CT = v_CG + v_GT

Method 1 - Sine Law and Cosine Law

In the vector diagram shown at the bottom right we know the lengths of two of the sides and we know one of the angles. We can then apply the cosine law to get the length of the third side, which is the magnitude of the velocity of the car with respect to the truck.

c² = a² + b² - 2 ab cos(θ_c)

(v_CT)² = 30² + 40² - 2*30*40 * cos(140°)

v_CT = 65.9 km/h

Use the sine law to get the angle between the two vectors v_CT and v_GT

sin(θa) a

=

sin(θc) c

sin(θa)

=

30*sin(140°) 65.9

=

0.293

This gives an angle of 17.0°

Thus, the velocity of the car with respect to the truck is 65.9 km/h at an angle of 17° south of east.

Method 2 - The Component Method

To calculate v_CT write out two equations, one for the x direction and one for the y direction. Take +x to be east and +y south.

Glue the components together to find the magnitude and direction:

v_CT² = v_CTx² + v_CTy²

v_CT = 66 km/hr.

The angle is given by the inverse tangent of 19.3 / 63.0, which is 17 degrees.

So, the velocity of the car relative to the truck is 66 km/hr, 17 degrees south of east.