# A car travels east 2 km for 5 minutes

A car travels east 2 km for 5 minutes, then north 3 km for 8 minutes, and then west 4 km for 10 minutes. Determine the total distance traveled and the magnitude of displacement of the car. Also, what is the magnitude of the average velocity and the average speed?

#### Solution:

Let us draw out the travel path of the car to better visualize the question like so:

Note that our origin is at A

The total distance traveled by the car is:

Total distance = AB + BC + CD = 2 + 3 + 4 = 9 km = 9000 m

To find the displacement, we can write a position vector from A to D and calculate it’s magnitude.

$r=\left\{(2-4)i+3j\right\}$ km

$r_{magnitude}=\sqrt{(-2)^2+(3)^2}=3.6$ km = 3600 m

To figure out the average velocity and speed, we need to first determine the length of time the car was in motion for.

Total time = 5 min + 8 min + 10 min = 23 min = 1380 s

The average velocity is displacement divided by time:

$v_{avg}=\dfrac{displacement}{t}=\dfrac{3600}{1380}=2.61$ m/s

The average speed is total distance divided by the total time of travel:

Average speed = $\dfrac{distance}{time}=\dfrac{9000}{1380}=6.52$ m/s

Total distance = 9 km = 9000 m

Displacement = 3.6 km = 3600 m

Average velocity = 2.61 m/s

Average speed = 6.52 m/s