Answer
The distance from the starting point is $~~2.7~km$
Work Step by Step
We can assume that the initial direction is north.
We can find the distance and direction of each section of the trip:
$x_1 = (50~km/h)(\frac{2.0~min}{60~min}) = \frac{5}{3}~km~(north)$
$x_2 = (20~km/h)(\frac{4.0~min}{60~min}) = \frac{4}{3}~km~(east)$
$x_3 = (20~km/h)(\frac{1.0~min}{60~min}) = \frac{1}{3}~km~(south)$
$x_4 = (50~km/h)(\frac{1.0~min}{60~min}) = \frac{5}{6}~km~(east)$
$x_5 = (20~km/h)(\frac{2.0~min}{60~min}) = \frac{2}{3}~km~(south)$
$x_6 = (50~km/h)(\frac{0.5~min}{60~min}) = \frac{5}{12}~km~(east)$
We can find the east component
$x_2+x_4+x_6 = \frac{4}{3}~km+\frac{5}{6}~km+\frac{5}{12}~km = \frac{31}{12}~km$
We can find the north component
$x_1+x_3+x_5 = \frac{5}{3}~km-\frac{1}{3}~km-\frac{2}{3}~km = \frac{2}{3}~km$
We can find the distance from the starting point:
$d = \sqrt{(\frac{31}{12}~km)^2+(\frac{2}{3}~km)^2} = 2.7~km$
The distance from the starting point is $~~2.7~km$
