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.
You have a boat and you're trying to cross a river that has parallel banks. If there is no current in the river, the quickest way to cross is to aim your boat directly across the river
If the water is flowing down the river, how should you aim your boat to reach the far shore in the shortest time?
If the current in the river is 3 m/s and your boat can travel at 5 m/s relative to the water, how should you aim your boat so you land at the point directly across the river? If the river is 100 m wide, how long does it take you to cross?