Posted by **Patricia** on Tuesday, September 8, 2009 at 3:54am.

You drive a car 1400 to the east, then 2400 to the north. The trip took 3.0 minutes.

What was the direction of your average velocity?

What was the magnitude of your average velocity?

For some reason I seem to be lacking ability to answer these questions. I tried displacement/time but I got it wrong what do you guys think?

- Physics -
**MathMate**, Tuesday, September 8, 2009 at 7:40am
Displacement take into account of the direction, and does not take into account of the intermediate steps. Distance is the linear measurement.

For example, if you go east 10 km, and then north 10 km, and then west 10 km. You finish at at point at 10 km north of the initial position.

The displacement is 10 km north.

The distance is 30 km.

The same applies to your problem.

The separate displacements are

1400 units east, 2400 units north.

If we take the cartesian plan with east=x-axis, and north=y-axis, the displacement is (1400,2400).

The magnitude of the displacement is

√(1400²2400²)

=2778.5 units

The direction is given by

tan^{-1}(y/x)

=tan^{-1}(2400/1400)

=59.74°

- Physics - erratum -
**MathMate**, Tuesday, September 8, 2009 at 7:41am
The magnitude of the displacement is

√(1400²**+**2400²)

- Physics -
**MathMate**, Tuesday, September 8, 2009 at 7:43am
The average velocity is the displacement divided by time, in this case, in units/min, or any other that you see fit.

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