Relative velocity problems are handled in a similar way in two dimensions - it's just harder to add and subtract the vectors because you have to use components.
Let's change the 1-D example to 2-D. The truck still moves at 40 km/hr west, but the car turns on to a road going 40 degrees south of east, and travels at 30 km/hr. What is the velocity of the car relative to the truck now?
| Vector | x component | y component |
|---|---|---|
| vCG | vCGx = +30 cos(40) vCGx = +23 km/hr | vCGy = +30 sin(40) vCGy = +19.3 km/hr |
| vGT | vGTx = 40 km/hr | vGTy = 0 |
| vCT | vCTx = vCGx + vGTx
vCTx = +63 km/hr | vCTy = vCGy + vGTy vCTy = +19.3 km/hr |
The relative velocity equation for this situation looks like this:
vCT = vCG + vGT
To calculate vCT write out two equations, one for the x direction and one for the y direction. Take +x to be east and +y south.
| Vector | x component | y component |
|---|---|---|
| vCG | vCGx = +30 cos(40) vCGx = +23 km/hr | vCGy = +30 sin(40) vCGy = +19.3 km/hr |
| vGT | vGTx = 40 km/hr | vGTy = 0 |
| vCT | vCTx = vCGx + vGTx
vCTx = +63 km/hr | vCTy = vCGy + vGTy vCTy = +19.3 km/hr |
Glue the components together to find the magnitude and direction:
vCT2 = vCTx2 + vCTy2
vCT = 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.