Relative velocity in 2-D

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?

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.

Vectorx componenty 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.

Relative velocity questions

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?

  1. Angle the boat upstream, and go against the current, landing somewhere upstream.
  2. Angle the boat upstream in such a way that you land at the point directly across from where you started.
  3. Point your boat directly across the river and get carried some way downstream.
  4. Angle your boat downstream and get carried even further downstream.

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?